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程序代写案例-MAS 381

时间：2021-01-09

MAS 381:

Mathematics III

University of Sheffield, 2020

1 Complex functions

1.1 Introduction

The term complex analysis refers to the calculus of complex-valued functions,

f(z), depending on a single complex variable z. To an un-initiated, it may

seem that this subject should merely be a simple reworking of standard real

variable theory that you learned in first and second year calculus. However, this

naive first impression could not be further from the truth! Complex analysis is

the culmination of a deep and far-ranging study of the fundamental notions

of complex differentiation and integration, and has an elegance and beauty

not found in the real domain. For instance, complex functions are necessarily

analytic, meaning that they can be represented by convergent power series, and

hence are infinitely differentiable. Thus, difficulties with degree of smoothness,

strange discontinuities, subtle convergence phenomena, and other pathological

properties of real functions never arise in the complex world.

Complex functions have a very wide applicability in electrical engineering.

The calculation of harmonically varying currents in circuits consisting of re-

sistors, capacitors, induction, coils, etc. require functions of complex variable.

Secondly, functions of complex variables are used for solving Laplace and Pois-

son equations describing electrostatic potential.

The driving force behind many of the applications of complex analysis is the

remarkable connection between complex functions and harmonic functions of

two variables, a.k.a. solutions of the planar Laplace equation. To wit, the real

and imaginary parts of any complex analytic function are automatically har-

monic. In this manner, complex functions provide a rich spectrum of additional

solutions to the two-dimensional Laplace equation, which can be exploited in a

wide range of physical, mathematical, and engineering applications.

One of the most useful consequences stems from the elementary observation

that the composition of two complex functions is also a complex function. We

re-interpret this operation as a complex change of variables, producing a confor-

mal mapping that preserves (signed) angles in the Euclidean plane. Conformal

mappings can be effectively used for constructing solutions to the Laplace equa-

tion on complicated planar domains that appear in a wide range of physical

problems, including fluid mechanics, aerodynamics, thermomechanics, electro-

statics, and elasticity. In particular, conformal mapping is a widespread math-

ematical technique to approach the electromagnetic study of electric machines,

e.g. energy conservation in magnetostatic and time-harmonic problems; current

source assignment in the transformed domain; transformation of the flux density

components through conformal mapping, etc.

1.2 Properties of complex functions

Our principal objects of study are complex-valued functions, f(z), depending

on the complex variable z = x + iy and they are defined over of the complex

plain, C.

2

Any complex function f(z) can be uniquely written as a complex combina-

tion

f(z) = f(x+ iy) = u(x, y) + iv(x, y), (1)

of two real functions both depending on the two real variables, x and y with

Re(f) = u(x, y) and Im(f) = u(x, y). For example, the function f(z) = z2 can

be expanded and written as

z2 = (x+ iy)2 = x2 − y2 + 2ixy

so

Re(f) = x2 − y2, Im(f) = 2xy

Many of the well-known functions appearing in real-variable calculus (polyno-

mials, rational functions, exponentials, trigonometric functions, logarithms, and

many more) have natural complex extensions. For example, complex polynomi-

als

p(z) = anz

n + an−1zn−1 + · · ·+ a1z + a0 (2)

are complex linear combinations (meaning that the coefficients ak are allowed

to be complex numbers) of the basic monomial functions zk = (x+ iy)k.

As we will see the notion of harmonic function is very important for our

module. By definition a function is called harmonic if it is at least twice

differentiable and it satisfies the well-known Laplace equation. In 2D this partial

differential equation takes the form

∂2f

∂x2

+

∂2f

∂y2

= 0.

Before going deeper into the many remarkable properties of complex func-

tions, let us look at some of the most basic examples. In each case, you can

directly check that the harmonic functions provided by the real and imagi-

nary parts of the complex function are indeed solutions to the two-dimensional

Laplace equation.

Examples of complex functions

(a) Harmonic Polynomials: As noted above, any complex polynomial is a linear

combination, as in (2), of the basic complex monomials

zn = (x+ iy)n = un(x, y) + ivn(x, y). (3)

Their real and imaginary parts, un(x, y), vn(x, y), are harmonic polynomials.

(b) Rational functions: Ratios of the form

f(z) =

p(z)

q(z)

, (4)

3

where p(z) and q(z) are complex polynomials provide a large variety of harmonic

functions. The simplest case is

1

z

=

1

x+ iy

=

x− iy

(x+ iy)(x− iy) =

x− iy

x2 + y2

=

x

x2 + y2

− i y

x2 + y2

. (5)

Apart from an interesting behaviour at the origin (x = 0, y = 0), these functions

are harmonic everywhere. It is easy to see that they are twice differentiable and

they satisfy the 2D Laplace equation. For the real part

∂

∂x

(

x

x2 + y2

)

= − x

2 − y2

(x2 + y2)2

,

∂2

∂x2

(

x

x2 + y2

)

=

2x(x2 − 3y2)

(x2 + y2)3

∂

∂y

(

x

x2 + y2

)

= − 2xy

(x2 + y2)2

,

∂2

∂y2

x

x2 + y2

=

−2x(x2 − 3y2)

(x2 + y2)3

.

It is obvious that the Laplace equation is satisfied. The same can be shown to

happen to the imaginary part.

A slightly more complicated example is the function

f(z) =

z − 1

z + 1

. (6)

To write (6) in the standard form (1) we multiply and divide by the complex

conjugate of the denominator, so

f(z) =

z − 1

z + 1

=

(z − 1)(z + 1)

(z + 1)(z + 1)

=

=

|z|2 + z − z − 1

|z + 1|2 =

x2 + y2 − 1

(x+ 1)2 + y2

+ i

2y

(x+ 1)2 + y2

, (7)

where z denotes the complex conjugate of z. Again, it is easy to show that these

functions (the real and imaginary parts) are harmonic functions.

(c) Complex exponential functions: Euler’s formula

ez = ex+iy = ex(cos y + i sin y), (8)

for the complex exponential yields two important harmonic functions: ex cos y

and ex sin y. For example

dex cos y

dx

=

d2ex cos y

dx2

= ex cos y

dex cos y

dy

= −ex sin y; d

2ex cos y

dy2

= −ex cos y

and it is obvious why the Laplace equation is satisfied. Similar simple calculation

can be used for the imaginary part. More generally, writing out ecz for a complex

constant c = a+ ib, produces the complex exponential function

ecz = e(a+ib)(x+iy) = e(ax−by)+i(ay+bx) = eax−by[cos(ay+bx)+i sin(ay+bx)] (9)

4

whose real and imaginary parts are harmonic functions for arbitrary a, b ∈ R.

(d) Complex trigonometric functions: These are defined in terms of the complex

exponential according to

cos z =

eiz + e−iz

2

, sin z =

eiz − e−iz

2i

. (10)

Similarly, hyperbolic functions can be also expressed with the help of the com-

plex exponential function, so

cosh z =

ez + e−z

2

, sinh z =

ez − e−z

2

. (11)

Example 1.: Show that

sin z = sinx cosh y + i cosx sinh y

is a harmonic function

(e) The complex logarithmic function: In a similar fashion, the complex loga-

rithm is a complex extension of the usual real natural (i.e., base e) logarithm.

In terms of polar coordinates z = reiθ, the complex logarithm has the form

ln z = ln reiθ = ln r + ln eiθ = ln r + iθ. (12)

Thus, the logarithm of a complex number has a real part

Re{ln z} = ln r = ln |z| = ln

√

x2 + y2 =

1

2

ln(x2 + y2), (13)

which is a well-defined harmonic function. The imaginary part

Im{ln z} = θ = arg(z),

of the complex logarithm is the polar angle, also known as the argument of the

complex number (note that the complex logarithm is not defined for x = y = 0).

5

1.3 Complex differentiation

The notion of complex derivatives stays at the very foundation of complex func-

tion theory. Complex differentiation is defined in the same manner as the usual

calculus limit definition of the derivative of a real function. Yet, despite a

superficial similarity, complex differentiation is a profoundly different theory,

displaying an elegance and depth not shared by its real counterpart.

Definition: A complex function, f(z), is differentiable at a point z ∈ C if and

only if the limiting ratio quotient

f ′(z) = lim

w→z

f(w)− f(z)

w − z (14)

exists.

The key feature of this definition is that the limiting value f ′(z) of the

difference quotient must be independent of how w converges to z. On the real

line, there are only two directions to approach a limiting point: either from

the left or from the right. These two ways to approach a value lead to the

concepts of left- and right-handed derivatives and their equality is required for

the existence of the usual derivative of a real function.

In the complex plane, there are an infinite variety of directions to approach

the point z, and the definition requires that all of these directional derivatives

must agree. This requirement imposes severe restrictions on complex deriva-

tives, and is the source of their remarkable properties. To understand the con-

sequences of this definition, let us first see what happens when we approach z

along the two simplest directions: horizontal and vertical (N.B. in a complex

plane, the real axis is the horizontal axis, while imaginary numbers are put on

the vertical axis). If we set

w = z + h = (x+ h) + iy, h ∈ R,

then w → z along a horizontal line as h→ 0. If we write f(z) = u(x, y)+iv(x, y),

then we must have

f ′(z) = lim

h→0

f(z + h)− f(z)

h

= lim

h→0

f(x+ h+ iy)− f(x+ iy)

h

= lim

h→0

[

u(x+ h, y)− u(x, y)

h

+ i

v(x+ h, y)− v(x, y)

h

]

=

∂u

∂x

+ i

∂v

∂x

=

∂f

∂x

which follows from the usual definition of the (real) partial derivative. On the

other hand, if we set

w = z + ik = x+ i(y + k), k ∈ R

then w → z along the vertical line as k → 0. Therefore we must have

f ′(z) = lim

k→0

f(z + ik)− f(z)

ik

= lim

k→0

[

−if(x+ i(y + k))− f(x+ iy)

k

]

=

6

lim

k→0

[

v(x, y + k)− v(x, y)

k

− iu(x, y + k)− u(x, y)

k

]

=

∂v

∂y

− i∂u

∂y

= −i∂f

∂y

When we equate the real and imaginary parts of these two distinct formulae for

the complex derivative f ′(z), and take into account that the two formulations

denote the same operation, we discover that the real and imaginary components

of f(z) must satisfy a linear system of partial differential equations, called the

Cauchy-Riemann equations.

Theorem: A complex function f(z) = u(x, y)+iv(x, y) depending on z = x+iy

has a complex derivative f ′(z) if and only if its real and imaginary parts are

continuously differentiable and satisfy the Cauchy-Riemann equations

∂u

∂x

=

∂v

∂y

,

∂u

∂y

= −∂v

∂x

(15)

In this case, the complex derivative of f(z) is equal to any of the following

expressions:

f ′(z) =

∂f

∂x

=

∂u

∂x

+ i

∂v

∂x

= −i∂f

∂y

= −i∂u

∂y

+

∂v

∂y

(16)

A pair of functions that satisfy the system (15) is called conjugate.

Example 2.: Consider the elementary function

z3 = (x3 − 3xy2) + i(3x2y − y3)

Show that Its real part u = x3− 3xy2 and imaginary part v = 3x2y− y3 satisfy

the Cauchy-Riemann equations.

Solution

Fortunately, the complex derivative obeys all of the usual rules that you learned

in real-variable calculus. For example

d

dz

zn = nzn−1,

d

dz

ecz = cecz,

d

dz

ln z =

1

z

, (17)

7

and so on. Here, the power n can be non-integral or even, in view of the identity

zn = en ln z, complex, while c is any complex constant. The exponential formulae

(10) for the complex trigonometric functions imply that they also satisfy the

standard rules

d

dz

cos z = − sin z, d

dz

sin z = cos z. (18)

The formulae for differentiating sums, products, ratios, inverses, and compo-

sitions of complex functions are all identical to their real counterparts, with

similar proofs. Thus, thankfully, you do not need to learn any new rules for

performing complex differentiation.

Remark: There are many examples of seemingly reasonable functions which do

not have a complex derivative. The simplest is the complex conjugate function

f(z) = z = x− iy.

Its real and imaginary parts do not satisfy the Cauchy-Riemann equations, and

hence z does not have a complex derivative. More generally, any function f(z, z)

that explicitly depends on the complex conjugate variable z is not complex-

differentiable.

Example 3.: You are given that

Re(f(z)) =

x

x2 + y2

= u(x, y)

and f(1) = 1. Find f(z).

Solution: Let f(z) = u(x, y) + iv(x, y). Using the Cauchy-Riemann equations

∂v

∂y

=

∂u

∂x

=

(x2 + y2)− 2x2

(x2 + y2)2

=

y2 − x2

(x2 + y2)2

, (1)

∂v

∂x

= −∂u

∂y

= − −2xy

(x2 + y2)2

=

2xy

(x2 + y2)2

(2)

From (2) we have

v =

∫

2xy

(x2 + y2)2

dx

Let us change the variable and introduce w = x2 + y2, so dw = 2xdx, therefore

v =

∫

y

w2

dw = − y

w

+ C(y) = − y

x2 + y2

+ C(y)

Let us differentiate this, so

∂v

∂y

= − (x

2 + y2)− 2y2

(x2 + y2)2

+

dC

dy

=

y2 − x2

(x2 + y2)2

+

dC

dy

.

8

Using (1) we have

y2 − x2

(x2 + y2)2

+

dC

dy

=

y2 − x2

(x2 + y2)2

−→ dC

dy

= 0 −→ C = const.

Therefore

f(z) = u(x, y) + iv(x, y) =

x

x2 + y2

− iy

x2 + y2

+ iC =

x− iy

(x+ iy)(x− iy) + iC =

1

x+ iy

+ iC =

1

z

+ iC.

Using the given condition

f(1) = 1 = 1 + iC =⇒ C = 0 =⇒ f(z) = 1

z

.

***********************

There is a connection between Cauchy-Riemann equations and the Laplace equa-

tion we discussed earlier. Let us differentiate the first equation of (15) with

respect to x and the second equation with respect to y, so we have

∂2u

∂x2

=

∂2v

∂x∂y

,

∂2u

∂y2

= − ∂

2v

∂x∂y

.

Adding these two equations we obtain

∂2u

∂x2

+

∂2u

∂y2

= 0,

which is exactly the Laplace equation applied to the real part of f(z). Now

taking again Eq. (15) and differentiating the first equation with respect to y

and the second equation with respect to x we obtain

∂2u

∂x∂y

=

∂2v

∂y2

,

∂2u

∂x∂y

= −∂

2v

∂x2

.

Subtracting these two equations we arrive to

∂2v

∂x2

+

∂2v

∂y2

= 0,

which is the Laplace equation written for the imaginary part of f(z). These

two relations show that the real and imaginary part of a complex number are

harmonic functions.

9

1.4 Analytical functions

If f(z) = u(x, y) + iv(x, y) is a complex function and u(x, y) and v(x, y) are

conjugate in the same domain, then the function f(z) is called analytical.

A complex function is said to be analytical over a certain domain if it is com-

plex differentiable at every point of the domain. The necessary and sufficient

conditions that a complex function f = u+ iv is analytical are

• The four derivatives of its real and imaginary parts

∂u

∂x

,

∂v

∂y

,

∂u

∂y

,

∂v

∂x

satisfy the Cauchy-Riemann equations (15) and, in addition, they are

continuous

A remarkable feature of complex differentiation is that the existence of one com-

plex derivative automatically implies the existence of infinitely many!

A complex function f(z) is called analytic at a point z0 ∈ C if it has a power

series expansion

f(z) = a0 + a1(z − z0) + a2(z − z0)2 + · · · =

∞∑

n=0

an(z − z0)n (19)

that converges to all z sufficiently close to z0. We note that if f(z) and g(z)

are analytic at a point z0, so is their sum f(z) + g(z), product f(z)g(z) and,

provided g(z0) 6= 0, ratio f(z)/g(z).

The simplest test to check whether a complex series is convergent or not is

the ratio test. Let us suppose that the n-th term of a series is cn. Then, by

definition, the test makes use of the limit

L = lim

n→∞

∣∣∣∣cn+1cn

∣∣∣∣ . (20)

According to the value of L we can distinguish three cases:

• if L < 1, the series is convergent

• if L > 1, the series is divergent

• if L = 1, or the limit fails to exist, then the test is inconclusive, because

there exist both convergent and divergent series that satisfy this case.

All of the real power series found in elementary calculus carry over to the com-

plex versions of the functions. For example,

ez = 1 + z +

z2

2

+

z3

3!

+ · · · =

∞∑

n=0

zn

n!

,

10

is the power series for the exponential function based at z0 = 0. A straight-

forward application of the ratio test proves that the series converges for all z,

so

L = lim

n→∞

∣∣∣∣zn+1/(n+ 1)!zn/n!

∣∣∣∣ = limn→∞ zn+ 1 → 0.

On the other hand, the power series

1

z2 + 1

= 1− z2 + z4 − z6 + · · · =

∞∑

n=0

(−1)nz2n,

has a restricted convergence domain because

L = lim

n→∞

∣∣∣∣ (−1)n+1z2n+2(−1)nz2n

∣∣∣∣ = |z|2 = (x2 + y2).

In order to be convergent we must have L < 1, i.e. x2 + y2 < 1, i.e. the series is

convergent inside a circle of radius 1, and diverges outside the circle for |z| > 1.

In general, there are three possible options for the domain of convergence of

a complex power series (19)

• The series converges for all z

• The series converges inside a disk |z − z0| < ρ of radius ρ > 0 centred at

z0 diverges for all |z− z0| > ρ outside the disk. The series might converge

for at some (but not all) of the points on the boundary of the disk where

|z − z0| = ρ

• The series only converges, trivially, at z = z0

The number ρ is known as the radius of convergence of the series. In the first

case, we say ρ =∞, while in last case, ρ = 0, and the series does not represent

an analytic function.

Let us suppose that the sequence {cn}, n = 1, 2, . . . converges with a limit

L. If L = 0, then the radius of convergence is infinity, i.e. ρ =∞, i.e. the power

series converges for all z. If L 6= 0 (hence L > 0), then

ρ =

1

L

= lim

n→∞

∣∣∣∣ anan+1

∣∣∣∣ , (21)

here an is the coefficient of the nth term in the series.

Remarkably, the radius of convergence for the power series of a known an-

alytic function f(z) can be determined by inspection, without recourse to any

fancy convergence tests! Namely, ρ is equal to the distance from z0 to the near-

est singularity of f(z), meaning a point where the function fails to be analytic.

In particular, the radius of convergence ρ =∞ if and only if f(z) is analytic for

11

all z ∈ C and has no singularities; examples include polynomials, ez, cos z, and

sin z.

On the other hand, the rational function

f(z) =

1

z2 + 1

=

1

(z + i)(z − i)

has singularities at z = ±i, so its power series shown earlier has a radius of

convergence ρ = 1, which is the distance from z0 = 0 to the singularities. If we

expand (z2 + 1)−1 in a power series at some other point, say z0 = 1 + 2i, then

we need to determine which singularity is closest.

We compute |i−z0| = |−1−i| =

√

2, while |−i−z0| = |−1−3i| =

√

10, and

so the radius of convergence ρ =

√

2 is the smaller. This allows us to determine

the radius of convergence in the absence of any explicit formula for its (rather

complicated) Taylor expansion at z0 = 1 + 2i.

There are, in fact, only four possible types of singularities of a complex

function f(z):

• A singular point z = z0 is called pole of order n if and only if

f(z) =

h(z)

(z − z0)n , (22)

where h(z) is an analytical function at z = z0 and h(z0) 6= 0. The function

f(z) = 1/(z−2)3 has a pole of order 3 at z = 2, and f(z) = (3z−2)/[(z−

1)2(z+1)(z−4)] has a pole of order 2 at z = 1 and simple poles at z = −1

and z = 4. In the case of poles, it is true that we can find a positive

integer n (identical to the order of the pole) so that

lim

z→z0

(z − z0)nf(z) 6= 0

is analytical

• Branch points: refer to multi-valued functions. We have already encoun-

tered the two basic types: algebraic branch points, such as the function n

√

z

at z0 = 0 and logarithmic branch point such as lnz at z0 = 0. For example

f(z) = (z − 3)1/2 has a branch point at z = 3 and f(z) = ln(z2 + z − 2)

has branch points where z2 + z − 2 = 0, i.e. at z = 1 and z = −2

• Essential singularity. By definition, a singularity is essential if it is not

a pole or a branch point. The quintessential example is the essential

singularity of the function e1/z at z0. The behaviour of a complex function

near an essential singularity is quite complicated

• Singularities at infinity. This type of singularity of f(z) at z = ∞ is the

same as that of f(1/w) at w = 0. For example, the function f(z) = z3

has a pole of order 3 at z =∞ since f(1/w) = 1/w3 has a pole of order 3

at w = 0. Equally, a function has a singularity at infinity if

lim

z→∞ f(z) =∞

12

The nature and number of singularities will be very important later on when

we will discuss the integration of complex functions.

The complex function

f(z) =

ez

z3 − z2 − 5z − 3 =

ez

(z − 3)(z + 1)2

is analytic everywhere except for singularities at the points z = 3 and z = −1,

where its denominator vanishes. Therefore z = 3 is a simple (order 1) pole and

z = −1 is a double (order 2) pole.

A complicated complex function can have a variety of singularities. For

example, the function

f(z) =

e−1/(z−1)

2

(z2 + 1)(z + 2)2/3

has simple poles at z = ±i, a branch point of degree 3 at z = −2, and an

essential singularity at z = 1.

If a function is single-valued and has a singularity, then this singularity is

either a pole or an essential singularity. For this reason a pole is sometimes

called non-essential singularity.

Example 4.: Locate and name all the singularities of

f(z) =

z8 + z4 + 2

(z − 1)3(3z + 2)2

Determine where f(z) is analytic.

Solution:

13

Taylor’s Theorem: Let f(z) be analytic inside and on a simple closed curve

C. Let a and a+ h two points inside C. Then

f(a+ h) = f(a) + hf ′(a) +

h2

2!

f ′′(a) + · · ·+ h

n

n!

f (n)(a) + . . . (23)

or writing z = a+ h and h = z − a

f(z) = f(a) + (z − a)f ′(a) + (z − a)

2

2!

f ′′(a) + · · ·+ (z − a)

n

n!

f (n)(a) + . . . (24)

The series (23) or (24) are called the Taylor series or expansion for f(a+ h) or

f(z). The region of convergence of the series (24) is given by |z− a| < ρ, where

the radius of convergence ρ is the distance from a to the nearest singularity of

the function f(z). On |z − a| = ρ, the series may or may not converge. For

|z − a| > ρ, the series diverges. Obviously, if the nearest singularity of f(z) is

at infinity, the radius of convergence is infinite, i.e. the series converges for all

z. For a = 0 the above series are also called MacLaurin series.

Example 5.: Let f(z) = ln(1 + z) (with some special considerations, that are

not so important here).

(a) Expand f(z) in Taylor series about z = 0

(b) Determine the region of convergence for the series determined in (a)

(c) Expand ln

(

1+z

1−z

)

in Taylor series about z = 0

Solution:

14

Example 6.: Find the radius of convergence of the series:

1− (z − i) + 2(z − i)2 − 3(z − i)3 + · · ·+ (−1)nn(z − i)n + . . .

Solution: In this case the radius of the convergence is

ρ = lim

n→∞

∣∣∣∣ (−1)nn(−1)n+1(n+ 1)

∣∣∣∣ = limn→∞ nn+ 1 = 1

Hence the series is convergent for |z − i| < 1 and divergent for |z − i| > 1. Let

us write |z − i| as |x + i(y − 1)| which can be transformed to √x2 + (y − 1)2.

Therefore the region where this series is convergent satisfies the condition

x2 + (y − 1)2 < 1

i.e. is the circle in the complex plane centred at (0,1) and having a radius 1.

The power series discussed above and the convergence criteria are also true for

negative powers of (z − z0). We can also consider series of the form

∞∑

n=−∞

an(z − z0)n = · · ·+ a−n

(z − z0)n + · · ·+

a−2

(z − z0)2 +

a−1

z − z0 +

+a0 + a1(z − z0) + a2(z − z0)2 + · · ·+ an(z − z0)n + . . . (25)

The part of of this series containing positive powers of z − z0 is convergent for

|z − z0| < ρ, where ρ is given in the standard form

ρ = lim

n→∞

∣∣∣∣ anan+1

∣∣∣∣ .

15

Let us find where the part of series containing negative powers of z − z0 is

convergent. First of all, it is convergent when a−n = 0 for n > m, so that is

finite. Consider now the case when the series having negative powers is infinite.

Once again, we use the ratio test and obtain that it is convergent provided

lim

n→∞

∣∣∣∣a−n−1(z − z0)−n−1a−n(z − z0)−n

∣∣∣∣ < 1

We can rewrite this as

1

|z − z0| limn→∞

∣∣∣∣a−n−1a−n

∣∣∣∣ < 1.

Introducing

R = lim

n→∞

∣∣∣∣a−n−1a−n

∣∣∣∣ , (26)

(we assume that this limit exists) we eventually obtain that the condition for

convergence is that |z − z0| > R. In summary, taking now all powers together

we obtain that the complete series (25) is convergent when z is in the annulus

R < |z − z0| < ρ (27)

Obviously this condition makes sense is R < ρ.

Example 7.: Find the region of convergence of the following series

(i) + · · ·+ 2

n

zn

+ · · ·+ 2

2

z2

+

2

z

+ 1 +

z

3

+

z2

32

+ · · ·+ z

n

3n

(ii) + · · ·+ 1

n!zn

+ · · ·+ 1

2z2

+

1

z

+ 1 + z + z2 + · · ·+ zn + . . .

Solution:

16

Now we can formulate a new theorem that will be useful when discussing inte-

gration

Laurent’s theorem: Let the function f(z) be analytical in the annulus R <

|z − z0| < ρ. Then, the function f(z) can be expanded in convergent series

f(z) =

∞∑

n=−∞

an(z − z0)n = · · ·+ a−n

(z − z0)n + · · ·+

a−2

(z − z0)2 +

a−1

z − z0 +

+a0 + a1(z − z0) + a2(z − z0)2 + · · ·+ an(z − z0)n + . . . (28)

and this is called the Laurent series of the function f(z) about the point z0.

To calculate the Laurent series we use the standard and modified geometric

series which are

1

1− z =

{ ∑∞

n=0 z

n, |z| < 1

−∑∞n=1 1zn , |z| > 1 (29)

Here the function is analytic everywhere except in the singularity at z = 1.

The expressions (29) are the expansions for f in the regions inside and outside

the circle of radius 1, centred on z = 0, where z| < 1 is the region inside

the circle and |z| > 1 is the region outside the circle. In reality to find the

Laurent series for even very simple functions is not trivial, therefore we combine

simple known expansions (Taylor or Laurent) to obtain Laurent expansions of

more complicated functions. In applications, often only the first few terms in a

Laurent expansion are needed.

Example 8.: Determine the Laurent series of f(z) = 1/(z + 5) that are valid

in the regions (i) |z| < 5 and (ii) |z| > 5.

Solution:

17

Example 9.: Determine the Laurent series for

f(z) =

1

z(z + 5)

valid in the region |z| < 5.

Solution:

We know from the previous example that for |z| < 5 the series expansion of

1/(z + 5) is

1

z + 5

=

∞∑

n=0

(−1)nzn

5n+1

,

It follows from this that we can calculate the series expansion of f(z) since

f(z) =

1

z

· 1

z + 5

=

1

z

∞∑

n=0

(−1)nzn

5n+1

=

∞∑

n=0

(−1)nzn−1

5n+1

Example 10.: Find the Laurent series expansion of

f(z) =

1

z(z + 2)

valid in the region 1 < |z − 1| < 3

Solution:

The region where the function has to be expanded is an open annulus be-

tween circles of radius 1 and 3, centred on z = 1. We want a series expansion

about z = 1; to do this we make the substitution w = z − 1, and look for the

expansion in w, where 1 < |w| < 3. In terms of w the function becomes

f(z) =

1

(w + 1)(w + 3)

In the new variable the domain of interest is still an open annulus of radii 1 and

3, bit they are centred on 0. To make the series expansion easier to calculate

we can manipulate our function f(z) into a similar form to the series expansion

shown in Eq. (29). To do this we will split the function using partial fractions,

and then manipulate each of the fractions into a form based on Eq. (29), so we

obtain

f(z) =

1

2

(

1

w + 1

− 1

w + 3

)

=

1

2

(

1

1− (−w) −

1

3

(

1− (−w3 ))

)

Using the standard and modified geometric series given by Eq. (29), we calculate

that

1

1− (−w) =

∞∑

n=0

(−w)n =

∞∑

n=0

(−1)nwn, |w| < 1

−

∞∑

n=1

1

(−w)n = −

∞∑

n=1

(−1)n

wn |w| > 1

18

and

1

3

(

1− (−w3 )) =

1

3

∞∑

n=0

(−w

3

)n

= 13

∞∑

n=0

(−1)nwn

3n =

∞∑

n=0

(−1)nwn

3n+1 , |w| < 3

− 13

∞∑

n=1

1

(−w3 )

n = − 13

∞∑

n=1

(−3)n

wn = −

∞∑

n=1

(−1)n3n−1

wn , |w| > 3

We require the expansion in w where 1 < |w| < 3, so we use the expansions

for |w| > 1 and |w| < 3, which we can substitute back into our f(z) in partial

fractions to obtain

f(z) =

1

2

(

−

∞∑

n=1

(−1)n

wn

−

∞∑

n=0

(−1)nwn

3n+1

)

= −1

2

( ∞∑

n=1

(−1)n

wn

+

∞∑

n=0

(−1)nwn

3n+1

)

Substituting back in w = z − 1 we obtain the Laurent series, valid within the

region 1 < |z − 1| < 3,

f(z) = −1

2

( ∞∑

n=1

(−1)n

(z − 1)n +

∞∑

n=0

(−1)n(z − 1)n

3n+1

)

Example 11.: Obtain the series expansion of

f(z) =

1

z2 + 4

valid in the region |z − 2i| > 4

19

Key Points:

1. First check to see if you need to make a substitution for the region you

are working with, a substitution is useful if the region is not centred on

z = 0

2. Then you will need to manipulate the function into a form where you can

use the series expansions shown in Eq. (29): this may involve splitting by

partial fractions first.

3. Find the series expansions for each of the fractions you have in your func-

tion within the specified region, then substitute these back into your func-

tion.

4. If part of the function cannot be expanded into Laurent series (not in

the standard form), then use Taylor or binomial series to expand those

functions

5. Finally, simplify the function and, if you made a substitution, change it

back into the original variable

Example 12.: ( to be solved at home) Expand

f(z) =

1

(z + 1)(z + 3)

in a Laurent series valid for (a) 1 < |z| < 3 and (b) 0 < |z + 1| < 2.

*******

Before embarking into the study of more challenging topics of complex analysis,

let us first discuss some preliminaries, more precisely the possible equations of

a straight line and circle It is well known that the most simplistic equation of a

straight line is

ax+ by = c (30)

where x, y are real variables and a, b, c are real numbers. In the case of complex

numbers, this equation changes to

|z − a| = |z − b| (31)

where now a and b are complex numbers. Another way of writing the equation

of a straight line is in the parametric form

z = at+ b, (32)

where the parameter t is real and a and b are complex quantities.

20

Example 14.: A straight line passes through the points A(1, 2) and B(−1, 0).

Write down the three different form of equations of this line.

Solution:

21

xz

R

y

z

0

0

y0

x

ϕ

Figure 1: The connectivity between the two ways to describe the equation of a

circle (cartesian and polar)

A circle in the real plane will have the equation

(x− x0)2 + (y − y0)2 = R2, (33)

which represents a circle of radius R centred on (x0, y0). In the complex plane,

the equation of a circle is written as

|z − z0| = R, z0 = x0 + iy0 (34)

which represents a circle of radius R centred on the complex z0. Another con-

venient form to write the equation of a circle is the polar form (used a lot

throughout complex analysis). The transition between the two ways to describe

the equation of a circle is shown in Fig (1.4). In the new coordinate system the

variables are R and φ. In the Cartesian system a point in the complex plane

is z = x + iy while the centre of the circle is at z0 = x0 + iy0. The connection

between the Cartesian coordinates (x, y) and polar coordinates is made via

x = x0 +R cosφ and y = y0 +R sinφ.

In the complex plane we can write

z = x0 + iy0 +R(cosφ+ i sinφ) = z0 +Re

iφ.

Hence the equation of a circle in the complex plane is

z = z0 +Re

iφ, 0 ≤ φ < 2pi (35)

where the centre is situated at z0 and the circle has a radius R.

Example 15.: A circle passes through the points A(0, 2), B(1, 0) and C(2,−1).

Write down the three different forms of equations of this circle.

22

Solution:

23

2 Complex integration and Cauchy’s Theorem

We can ingrate complex functions along smooth curves in complex plane. It is

similar to integration of real function. Let us consider a complex function f(z)

and a curve C in the complex plane given by z = z(t) with a ≤ t ≤ b.

Newton-Leibnitz formula: Let the equation of C be z = z(t), a ≤ t ≤ b,

then if

f(z) =

dF (z)

dz

, then

∫

C

f(z) dz = F (b)− F (a).

Example 18.: Calculate the integral of f(z) = z along an interval of straight

line z = (2 + i)t, 0 ≤ t ≤ 1.

Solution:

Obviously

z =

dF (z)

dz

=⇒ F (z) = z

2

2

and a = z(0) = 0, b = z(1) = 2 + i, so∫

C

z dz =

z2

2

∣∣∣∣2+i

0

=

1

2

(2 + i)2 =

3 + 4i

2

2.1 Relation between complex and linear integration

If f(z) = u(x, y) + iv(x, y) = u+ iv, and z = x+ iy, the complex line integral∫

C

f(z) dz

can be expressed in terms of real line integrals as∫

C

f(z) dz =

∫

C

(u+ iv)(dx+ idy) =

∫

C

[udx− vdy] + i

∫

C

[vdx+ udy] (36)

A slightly different approach is needed when the curve is given in parametric

form. Let us calculate the same integral as in the previous example but now

using the above relation. In what follows we denote the derivatives with respect

to t by dot, i.e. dx/dt = x˙. In that example we had x = 2t, y = t (after the

identification of real and imaginary parts), so that dx = x˙dt = 2dt, dy = y˙dt =

dt, u = x, v = y and∫

C

zdz =

∫ 1

0

(2xdt− ydt) + i

∫ 1

0

(2ydt+ xdt) = 3

∫ 1

0

t dt+ 4i

∫ 1

0

t dt =

(3 + 4i)

∫ 1

0

t dt = (3 + 4i)

t2

2

∣∣∣∣1

0

=

3 + 4i

2

In general, if the curve C is given in parametric form as x = x(t), y = y(t) with

a ≤ t ≤ b, then Eq. (36) can be written as∫

C

f(z)dz =

∫ b

a

(ux˙− vy˙)dt+ i

∫ b

a

(vx˙+ uy˙)dt (37)

24

Example 17.: Evaluate∫ (2,4)

(0,3)

(2y + x2)dx+ (3x− y)dy

along (a) the parabola x = 2t, y = t2 + 3; (b) straight lines from (0, 3) to (2, 3)

and then from (2, 3) to (2, 4); (c) the straight line from (0, 3) to (2, 4).

Solution:

25

2.2 Cauchy’s theorem

Let C be simple (i.e. without self-intersection) piece-wise smooth closed contour

and f(z) an analytic function defined in the domain D enclosed by the curve C

and continuous at all points in D and on C. Then∮

C

f(z)dz = 0 (38)

Example 18.: Calculate

∮

C

zdz, where C is the circle of radius R centred on

the coordinate origin.

Solution:

The parametric equations of this circle are x = R cosφ and y = R sinφ. Because

the circle is a closed contour, the variable φ covers the range [0, 2pi]. Then

dx = −R sinφ dφ, dy = R cosφ dφ and∮

C

zdz =

∮

C

(x+ iy)(dx+ idy) =

∮

C

(xdx− ydy) + i

∮

C

(ydx+ xdy) =

−

∫ 2pi

0

(R cosφ×R sinφdφ+R sinφ×R cosφ dφ)+i

∫ 2pi

0

(−R sinφ×R sinφ dφ+R cosφ×R cosφ dφ)

= −2R2

∫ 2pi

0

(cosφ sinφ)dφ+iR2

∫ 2pi

0

(− sin2 φ+cos2 φ)dφ = −R2

∫ 2pi

0

sin 2φ dφ+iR2

∫ 2pi

0

cos 2φdφ

=

1

2

R2 cos 2φ

∣∣2pi

0

+

1

2

iR2 sin 2φ|2pi0 =

1

2

R2(cos 4pi−cos 0)+1

2

iR2(sin 4pi−sin 0) = 0

There is a very important consequence of the Cauchy’s theorem: if f(z) is

an analytic function in a domain D and if a and z are any two points in D then∫ z

a

f(z)dz (39)

is independent of the path in D joining a and z.

Example 19.: If the curve C is joining the points (1, 1) and (2, 3), find the

value of ∫

C

(12z2 − 4iz)dz

Solution:

26

2.3 Cauchy integral formula and Taylor series

Let f(z) be analytic inside a domain D and on a simple closed curve C around

this domain. If a is any point inside C, then∮

C

f(z)

z − a dz =

{

2piif(a), a ∈ D

0, a /∈ D

(40)

where C is traversed in the positive (counterclockwise) sense. Similarly, the nth

derivative of f(z) at z = a is given by

f (n)(a) =

n!

2pii

∮

C

f(z)

(z − a)n+1 dz n = 1, 2, 3 . . . (41)

The above two results are called Cauchy’s integral formulae and are quite re-

markable because they show that if a function f(z) is known on the simple

closed curve C, then the value of the function and all its higher derivatives can

be found at all points inside C.

Example 20.: Evaluate

(a)

∮

C

sinpiz2 + cospiz2

(z − 1)(z − 2) dz and (b)

∮

C

e2z

(z + 1)4

dz

where C is the circle |z| = 3

Solution: Since

1

(z − 1)(z − 2) =

1

z − 2 −

1

z − 1

we have∮

C

sinpiz2 + cospiz2

(z − 1)(z − 2) dz =

∮

C

sinpiz2 + cospiz2

z − 2 dz −

∮

C

sinpiz2 + cospiz2

z − 1 dz

By Cauchy’s integral formula with a = 2 and a = 1, respectively we have∮

C

sinpiz2 + cospiz2

z − 2 dz = 2pii[sinpi2

2 + cospi22] = 2pii

∮

C

sinpiz2 + cospiz2

z − 1 dz = 2pii[sinpi1

2 + cospi12] = −2pii

Since z = 1 and z = 2 are inside C and sinpiz2 + cospiz2 is analytic inside C.

Then the required integral has the value 2pii− (−2pii) = 4pii.

b.

27

One of the consequences of Cauchy’s formulae is that if a function of complex

variable has a first derivative, i.e. analytic in a certain domain, all its higher

derivatives exist in that domain. This means that any analytical function can be

expanded in an infinite series in the vicinity of any point z0 inside that domain,

i.e. it can be expanded into Taylor series with

a0 =

1

2pii

∮

C

f(ζ) dζ

ζ − z0 , a1 =

1

2pii

∮

C

f(ζ) dζ

(ζ − z0)2 ,

a2 =

1

2pii

∮

C

f(ζ) dζ

(ζ − z0)3 , . . . , an =

1

2pii

∮

C

f(ζ) dζ

(ζ − z0)n+1 , . . .

So, we eventually obtain that

f(z) = a0 + a1(z − z0) + a2(z − z0)2 + · · ·+ an(z − z0)n + . . . (42)

This expansion is called Taylor series expansion for function f(z). Let us find

expressions for coefficients an. Substituting z = z0 in (42) we obtain that

a0 = f(z0)

Now we differentiate (42) to obtain

f ′(z) = a1 + 2a2(z − z0) + 3a3(z − z0)2 + · · ·+ nan(z − z0)n−1 + . . .

and once again substitute z = z0 to arrive at

a1 = f

′(z0)

Once again differentiate

f ′′(z) = 2a2 + 3 · 2a3(z − z0) + · · ·+ n(n− 1)an(z − z0)n−2 + . . .

and substitute z = z0:

a2 =

1

2

f ′′(z0)

Once again differentiate

f ′′′(z) = 3 · 2a3 + · · ·+ n(n− 1)(n− 2)an(z − z0)n−3 + . . .

and again substitute z = z0:

a3 =

1

3 · 2 f

′′′(z0)

which can be continued as n→∞. After differentiating n times we obtain

f (n)(z) = n(n− 1)(n− 2) . . . 3 · 2an + . . .

28

Substituting z = z0 yields

an =

1

n!

f (n)(z0) (43)

This is the same series as discussed earlier in (24), so the entire discussion

presented there on convergence of series (Taylor and Laurent) should be ap-

plied here, too. The above discussion shows that there is a connection between

Cauchy’s integral formulae and Taylor series. Now the notion of singularity of

analytical function will have a slightly different connotation and will be essential

in future discussion.

If the Laurent series of a function contains at least one term with negative

power of (z − z0), then the point z0 is called isolated singularity of function

f(z). Isolated singularities are divided in two types. If the Laurent series of

f(z) contains infinitely many terms with negative powers of (z − z0), then z0

is called essential singularity. If the series contains only finite number of terms

with negative powers of (z − z0), then z0 is called a pole. When the Laurent

series expansion has the form

f(z) =

a−m

(z − z0)m + . . .

a−2

(z − z0)2 +

a−1

z − z0 +

+ a0 + a1(z − z0) + a2(z − z0)2 + · · ·+ an(z − z0)n + . . .

i.e. it starts from a term proportional to (z − z0)−m, the pole z0 is called the

mth order pole. In the particular case when m = 1 it is called a simple pole.

Example 21.: Find the Laurent series expansions of the following functions

about the singularity z0 = 0 and determine the type of this singularity.

(i) f(z) =

z2 + 2

z(z2 + 1)

; (ii) f(z) = e1/z

Solution:

(i) We decompose f(z) in partial fractions:

z2 + 2

z(z2 + 1)

=

A

z

+

Bz + C

z2 + 1

=⇒ z2 + 2 = A(z2 + 1) + (Bz + C)z

Take z = 0 =⇒ A = 2. Then we have

z2 = (2 +B)z2 + Cz =⇒ B = −1, C = 0

so finally

z2 + 2

z(z2 + 1)

=

2

z

− z

z2 + 1

=

2

z

− z

1− (−z2) =

2

z

− z[1 + (−z2) + (−z2)2 + · · ·+ (−z2)n + . . . ]

=

2

z

− z + z3 − z5 + · · ·+ (−1)n+1z2n+1 + . . .

29

We see that z = 0 is simple pole.

(ii)

2.4 Residue Theorem

Let z = z0 be a singularity of the function f(z) and (28) its expansion in Laurent

series. Then the coefficient a−1 is called the residue of the function f(z) at the

singularity z0 and we denote

a−1 = res

z=z0

f(z)

Let the simple closed contour C be the boundary of domain D, and f(z) analytic

inside D except isolated singularities z1, z2, . . . , zn, and continuous up to the

domain boundary C. Then∮

C

f(ζ) dζ = 2pii

(

res

z=z1

f(z) + res

z=z2

f(z) + · · ·+ res

z=zn

f(z)

)

(44)

To obtain the residue of a function f(z) at z = z0, it may appear that first the

Laurent expansion of f(z) must be obtained. However, in the case where z = z0

is a pole of order k, there is a simple formula for a−1 given by

a−1 = lim

z→z0

1

(k − 1)!

dk−1

dzk−1

[

(z − z0)kf(z)

]

If k = 1 (z0 is a simple pole) the result is especially simple and is given by

a−1 = lim

z→z0

(z − z0)f(z)

Example 22.: If

f(z) =

z

(z − 1)(z + 1)2

then z = 1 and z = −1 are poles of order one and two, respectively. Therefore

using the standard formula we have:

30

For the residue at z = 1

lim

z→1

(z − 1)

[

z

(z − 1)(z + 1)2

]

=

1

4

For the residue at z = −1

lim

z→−1

1

1!

d

dz

[

(z + 1)2

(

z

(z − 1)(z + 1)2

)]

= −1

4

Example 23.: Evaluate the integral

∮

C

f(z) dz, where C is the circle of radius

1/2 centred at the origin, and f(z) is given by

(i) f(z) =

z2 + 2

z(z2 + 1)

; (ii) f(z) = e1/z

Solution:

(i) This function has three singularities: 0 and ±i, but only 0 is inside the

circle. In previous example we found the Laurent expansion of this function

about 0. In particular, we found that a−1 = 2. Hence∮

z2 + 2

z(z2 + 1)

dz = 2pii res

z=0

f(z) = 4pii

(ii) This function has only one singularity at z = 0. In the previous example we

found the Laurent expansion of this function about 0. In particular, we found

that a−1 = 1. Hence ∮

e1/z dz = 2pii res

z=0

f(z) = 2pii

Example 24.: Evaluate ∮

C

ezt

z2(z2 + 2z + 2)

dz

around the circle C with equation |z| = 3.

31

2.5 Using the residue theorem for calculating real inte-

grals

The residue theorem can be used to calculate real integrals. We consider this

topic using examples.

Example 25.: Calculate ∫ ∞

−∞

dx

x2 − 6x+ 10 .

Solution

3+j

−b b

C

y

x

Figure 2: Representation for Example 33

Consider contour Γ that consists of half-circle C of radius b centred at the

origin and the interval on the real axis [−b, b], i.e. Γ = C ∪ [−b, b]. The roots of

x2 − 6x+ 10 are

x1,2 = 3±

√

9− 10 = 3±√−1 = 3± i

32

Let us consider the complex function 1/(z2 − 6z + 10). This function has two

simple poles at z = 3± i. However, only 3 + i is inside contour Γ. Hence,∮

Γ

dz

z2 − 6z + 10 = 2pii resz=3+i

1

z2 − 6z + 10 = 2pii limz→3+i

z − 3− i

(z − 3− i)(z − 3 + i)

= 2pii lim

z→3+i

1

z − 3 + i =

2pii

3 + i− 3 + i =

2pii

2i

= pi

On the other hand,∮

Γ

dz

z2 − 6z + 10 =

∫ b

−b

dx

x2 − 6x+ 10 +

∫

C

dz

z2 − 6z + 10

So that ∫ b

−b

dx

x2 − 6x+ 10 +

∫

C

dz

z2 − 6z + 10 = pi

Let us take the limit b→∞. In the second integral we take z = beiθ, so∫

C

dz

z2 − 6z + 10 =

∫ pi

0

ibeiθ dθ

b2e2iθ − 6beiθ + 10 → 0 as b→∞∫ b

−b

dx

x2 − 6x+ 10 →

∫ ∞

−∞

dx

x2 − 6x+ 10 as b→∞

As a result we obtain ∫ ∞

−∞

dx

x2 − 6x+ 10 = pi

Of course, this integral can be easily calculated by the traditional method:∫ ∞

−∞

dx

x2 − 6x+ 10 =

∫ ∞

−∞

dx

1 + (x− 3)2 =

∫ ∞

−∞

du

1 + u2

= tan−1 u

∣∣∣∞

−∞

=

pi

2

−

(

−pi

2

)

= pi

Example 26.: Calculate ∫ ∞

−∞

cosx

x2 + 1

dx.

Solution

33

3 Advanced vector calculus: double operators,

polar coordinates

In many applications, we do not consider individual vectors or scalars, but func-

tions that give a vector or scalar at every point. Such functions are called vector

fields or scalar fields. For example:

(a) Suppose we want to model the flow of air around an aeroplane. The velocity

of the air flow at any given point is a vector. These vectors will be different at

different points, so they are functions of position (and also of time). Thus, the

air velocity is a vector field. Similarly, the pressure and temperature are scalar

quantities that depend on position, or in other words, they are scalar fields.

(b) The magnetic field inside an electrical machine is a vector that depends on

position, or in other words a vector field. The electric potential is a scalar field.

34

Although we will mainly be concerned with scalar and vector fields in three-

dimensional space, we will sometimes use two-dimensional examples because

they are easier to visualise.

3.1 The gradient of a scalar field

If f is a scalar field, then we define ∇(f) = (fx, fy, fz) =

(

∂f

∂x ,

∂f

∂y ,

∂f

∂z

)

. The

result of this operator is a vector field. Often the gradient of f is written as

grad(f) rather than ∇(f).

Example 27.:

(a) for the function f = x3 + y4 + z5, we have ∇(f) = (3x2, 4y3, 5z4).

(b) for the function f = sinx sin y sin z we have

∇(f) = (cosx sin y sin z, sinx cos y sin z, sinx sin y cos z)

(c) for the function r =

√

x2 + y2 + z2 we have

rx =

1

2

2x√

x2 + y2 + z2

=

x√

x2 + y2 + z2

=

x

r

and similarly ry = y/r and rz = z/r. This means that

∇(r) =

(x

r

,

y

r

,

z

r

)

(a) We write E for the electric field (which is a vector field) and φ for the

electric potential (which is a scalar field). These are related by the equation

E = ∇(φ). (All this is valid only when there are no significant time-varying

magnetic fields.)

(b) Similarly, there is a gravitational potential function, Ψ, and the gravita-

tional force field is proportional to ∇(Ψ).

(c) The net force on a particle of air involves ∇(p), where p is the pressure.

If we have a single charge at the origin, then the resulting electric potential

function is φ = Ar−1 for some constant A, where r = (x2 +y2 +z2)1/2, as usual.

Note that

(r−1)x = −1

2

2x

(x2 + y2 + z2)3/2

= − x

r3

and similarly (r−1)y = −y/r3 and (r−1)z = −z/r3. This gives the electric field

as

E = ∇(φ) = −A (x, y, z)

r3

= −A r

r3

3.2 The div and curl operators

Now suppose we have a vector field u = (f, g, h), so that f , g, and h are all

functions of the coordinates. We can think of ∇ as itself being a strange kind

35

of vector, in which the components are differential operators

∇ =

(

∂

∂x

,

∂

∂y

,

∂

∂z

)

.

This means that in the case of the div and curl operators we can make use of

the dot and cross operators as follows:

∇ · u =

(

∂

∂x

,

∂

∂y

,

∂

∂z

)

· (f, g, h) = ∂f

∂x

+

∂g

∂y

+

∂h

∂z

= fx + gy + hz

∇× u =

∣∣∣∣∣∣

i j k

∂

∂x

∂

∂y

∂

∂z

f g h

∣∣∣∣∣∣ = (hy − gz, fz − hx, gx − fy),

Let u and v be vector fields, let f be a scalar field, and let p be a function of

one variable. Then:

∇(f + g) = ∇(f) +∇(g), ∇(fg) = f∇(g) + g∇(f)

∇ · (u + v) = ∇ · u +∇ · v, ∇ · (fu) = f∇ · (u) +∇(f) · u

∇× (u + v) = ∇× u +∇× v, ∇× (fu) = f∇× u +∇(f)× u

∇(p(f)) = p′(f)∇(f), ∇ · (u× v) = v · (∇× u)− u · (∇× v)

3.3 Second-order operators

There are several ways to combine the three operators

scalar field ∗ grad→ vector field ∗ div → scalar field

scalar field ∗ grad→ vector field ∗ curl→ vector field

vector field ∗ div → scalar field ∗ grad→ vector field

vector field ∗ curl→ vector field ∗ div → scalar field

vector field ∗ curl→ vector field ∗ curl→ vector field

No other combinations make sense. For example we cannot define curl(div(u)),

because div(u) is a scalar field, and we can only take the curl of a vector field.

It is important that two of the above combinations are automatically zero, e.g.

• For any scalar field f we have curl(grad(f)) = ∇× (∇(f)) = 0

• For any vector field u we have div(curl(u)) = ∇ · (∇× u) = 0

36

These can be checked directly. For a scalar field f , we have ∇(f) = (fx, fy, fz).

Taking into account that fxy = fyx and so on we find that

∇× (∇(f)) =

∣∣∣∣∣∣

i j k

∂

∂x

∂

∂y

∂

∂z

fx fy fz

∣∣∣∣∣∣ = (fzy − fyz, fxz − fzx, fyx − fxy) = (0, 0, 0)

Now consider instead a vector field u = (p, q, r) We have

∇× u =

∣∣∣∣∣∣

i j k

∂

∂x

∂

∂y

∂

∂z

p q r

∣∣∣∣∣∣ = (ry − qz, pz − rx, qx − py)

so

∇·(∇×u) = (ry−qz)x+(pz−rx)y+(qx−py)z = ryx−qzx+pzy−rxy+qxz−pyz

= pzy − pyz + qxz − qzx + ryx − rxy = 0

There are three more possible combinations

• For a scalar field f we have div(grad(f)) = ∇ · (∇(f)) = fxx + fyy + fzz.

This is usually written as ∇2(f), and called the Laplacian of f . Note

that the Laplacian of a scalar field is a scalar field. We can also define the

Laplacian of a vector field by the rule

∇2(p, q, r) = (∇2(p),∇2(q),∇2(r)) = (pxx+pyy+pzz, qxx+qyy+qzz, rxx+ryy+rzz)

with the Laplacian of a vector field being again a vector field

• For a vector field u = (p, q, r) we have

grad(div(u)) = ∇(∇ · u) = ∇(px + qy + rz) =

= (pxx + qyx + rzx, pxy + qyy + rzy, pxz + qyz + rzz)

• The last remaining combination can be expressed in terms of the above

two by the equation

curl(curl(u)) = ∇× (∇× u) = ∇(∇ · u)−∇2(u)

In two dimensions the situation is similar but simpler.

• For any scalar field f we have

div(grad(f)) = ∇ · (∇(f)) = fxx + fyy

which is again called the Laplacian and denoted by ∇2(f)

• We also have

curl(grad(f)) = curl(fx, fy) = fyx − fxy = 0

37

A vector field u is it incompressible (or solenoidal) if div(u) = 0, and that it is

irrotational (or conservative) if curl(u) = 0.

- For any scalar field f (in two or three dimensions) we have a vector field

∇(f) = grad(f). The rule curl(grad(f)) = 0 tells us that grad(f) is irrota-

tional

- For any vector field v in three dimensions we have another vector field curl(v).

The rule div(curl(v)) = ∇ · (∇× v) = 0 tells us that curl(v) is incompressible.

Example 28: For the two-dimensional vector field u = (x2 − y2 + 2xy, x2 −

y2 − 2xy) calculate div(u) and curl(u).

Solution:

38

3.4 Potential functions

If u is an irrotational vector field, a potential function for u is a scalar field , p

such that∇(p) = u. (Because curl(grad(p)) = 0, only irrotational field can have

a potential.) Potential functions always exist (but they may be multi-valued),

and it is often useful to find them.

Consider the vector field u = (y + z, z + x, x+ y). This has

∇× u =

∣∣∣∣∣∣

i j k

∂

∂x

∂

∂y

∂

∂z

y + z z + x x+ y

∣∣∣∣∣∣ = (1− 1, 1− 1, 1− 1) = 0

so it is irrotational. It therefore makes sense to look for a potential function, or

in other words a function p(x, y, z) with (px, py, pz) = (y + z, z + x, x + y). As

we want px = y + z, we must have

p =

∫

(y + z)dx = xy + xz + C1 = xy + xz + q(x, y)

where the constant C1 is independent of x, i.e. q(y, z). We thus have qx = 0,

and the equation py = z + x becomes x+ qy = z + x, or in other words qy = z.

Integrating this we get

q =

∫

z dy = yz + C2 = yz + r(z)

where now the constant C2 is independent of x and y, say C2 = r(z). We now

have p = xy + xz + q = xy + xz + yz + r, so the equation pz = x+ y becomes

x + y + rz = x + y, so rz = 0. As r can only depend on z and we have rz = 0

we see that r is a genuine constant. We can choose it to be zero, and we find

that the function p = xy + xz + yz is a potential function for u.

Now consider the vector field u = (0, 0, x2). This has

∇× u =

∣∣∣∣∣∣

i j k

∂

∂x

∂

∂y

∂

∂z

0 0 x2

∣∣∣∣∣∣ = (0,−2x, 0) 6= 0

so it is not irrotational, so it cannot have a potential function. We will nonethe-

less try to find one, and see what goes wrong. A potential function p would

have to have (px, py, pz) = (0, 0, x

2). As px = py = 0, we see that p can only

depend on z. That means that the derivative pz also depends only on z, so we

cannot have pz = x

2. Thus, there is no potential function.

39

4 Line integrals, surface integrals, Stokes’ theo-

rem, volume integrals, divergence theorem

Engineering applications often have to deal with curves in 3-dimensional space.

For example

• A wire in an electrical machine is a curve. To calculate the magnetic field

created by a current in the wire, or the force exerted on the wire by an

externally applied magnetic field, we need equations for the curve.

• The path of a moving particle over time defines a curve. If the particle

is charged then it will feel a force from any electric or magnetic fields;

to understand the effect of this, we need various equations relating the

position, velocity, force and acceleration to the fields.

We can describe a curve by giving the x, y and z coordinates (or equivalently,

the position vector r = (x, y, z)) in terms of another parameter t. (In the case

of a moving particle we often take t to be time, but that is not compulsory.)

For example the equation

r = (x, y, z) = (at, b cos(t), b sin(t))

describes a helix winding around the x-axis.

This is the path followed by an electron moving in a uniform magnetic field, but

it could also describe a wire wound round a cylinder.

4.1 Integration along curves

To integrate along a curve C, we divide C into many small pieces, each running

from some position r to a nearby position r + δr. Each such piece will give a

contribution to the integral, and we add up the contributions to get an approx-

imation to the required value. For the exact value, we pass to the limit where

the length of the small pieces tends to zero.

• The length of the curve is approximately the sum of the lengths |δr| over

all the small pieces. The exact length is denoted by

∫

C

|dr|.

• If a particle moves along a curve C through a force field F, then the work

done against the force is − ∫

C

F · dr.

40

For the last integral it makes a difference which direction we follow when travers-

ing the curve: the answer we get when traversing the curve backwards will be

the negative of the answer we get when traversing the curve in the forward

direction.

In practice, we calculate these integrals as follows. We parametrise the curve

as r = (x(t), y(t), z(t)) for some range of values of t (say a ≤ t ≤ b), and we

write x˙ = dx/dt and so on. We then have

dr =

dr

dt

dt = r˙dt = (x˙ dt, y˙ dt, z˙ dt)

|dr| =

√

x˙2 + y˙2 + z˙2 dt,

so

length(C) =

∫

C

|dr| =

∫ b

t=a

√

x˙2 + y˙2 + z˙2 dt

work =

∫

C

F · dr =

∫ b

t=a

F · r˙ dt

and so on.

Example 30.: Let C be the curve given by

r = (x, y, z) = (6t, 3

√

2t2, 2t3)

for 0 ≤ t ≤ 1. Calculate the length of this curve.

Solution: According to the definition we have

dr = (6, 6

√

2t, 6t2) dt =⇒ |dr| =

√

36 + 72t2 + 36t4 dt = 6

√

1 + 2t2 + t4 dt = 6(1+t2) dt,

so

length =

∫

C

|dr| =

∫ 1

t=0

6(1 + t2) dt =

[

6t+ 2t3

]1

t=0

= 8.

Example 31.: Consider a particle moving along a path r = (x, y, z) =

(t, 0, t/2) (for 0 ≤ t ≤ 1) against a force field F = (y2 + z2 − 1, 0, 0) (This could

reasonably model the wind force in a wind tunnel of radius one centred on the

x-axis). Find the work done against the force.

Solution:

41

If f is a function of one variable, then

∫ b

x=a

f ′(x) dx = f(b) − f(a) in the

virtue of the Fundamental Theorem of Calculus. For any curve C from a to b,

and any scalar field p, we have∫

C

∇(p) · dr = p(b)− p(a).

Let us suppose we have a curve C from a to b, and we want to calculate the

integral I =

∫

C

F · dr for some vector field F. Suppose that F is conservative

(i.e. curl(F) = 0). We can then find a potential function p with ∇(p) = F and

it will follow that

∫

C

F · dr = p(b)− p(a).

Note that in this method, we do not need to know anything about C except

where it starts and ends. This often makes calculations much easier.

If we have trouble finding a potential function, it may be better to use the

following approach: Suppose we have a curve C from a to b, and we want to

calculate the integral I =

∫

C

F · dr for some vector field F. Suppose that F is

conservative. We can then find a different curve C ′ from a to b for which the

calculation is easier, and then I will be equal to

∫

C′ F · dr. The reason why this

method works is that both

∫

C

F · dr and ∫

C′ F · dr are equal to p(b) − p(a),

where p is the potential function. For this to be valid, we need to know that p

exists (so we must check that F is conservative) but we do not actually need to

find p.

Example 32.: Let C be the helical path given by r = (t, cos(10pit), sin(10pit))

for 0 ≤ t ≤ 1, which runs from a = (0, 1, 0) to b = (1, 1, 0). Let F be the vector

field (yz, xz, xy). Calculate

∫

C

F · dr.

Solution:

42

Now let us discuss a case where the vector space is not conservative. Now

for the same helical path as before (see above) consider the vector field G =

(0,−z, y).

43

Let F be an conservative vector field. We can then define a potential function

p for F by the rule

p(a, b, c) = the integral

∫

C

F · dr, for any curve C from (0, 0, 0) to (a, b, c) .

The answer will not depend on the choice of curve, so we can choose whichever

curve makes the integral easiest. A straight line is often good, but sometimes

a broken line (from (0, 0, 0) to (a, 0, 0) to (a, b, 0) to (a, b, c), for example) is

better. Note again that this is only valid for conservative fields. Fields that are

not conservative do not have a potential function.

Example 33. The vector field F = (yz, xz, xy) was earlier shown to be conser-

vative, i.e. curl(F) = 0. Find a potential function p that satisfies grad(p) = F.

Solution: To find p(a, b, c), we evaluate

∫

L

F · dr, where L is the straight line

from (0, 0, 0) to (a, b, c). This line can be parametrised by r = (x, y, z) =

(ta, tb, tc) for 0 ≤ t ≤ 1, which gives

dr = (a, b, c)dt

F = ((tb)(tc), (ta)(tc), (ta)(tb)) = (t2bc, t2ac, t2ab)

F · dr = 3t2abc dt =⇒ p(a, b, c) =

∫

L

F · dr =

∫ 1

t=0

3t2abc dt =

[

t3abc

]1

t=0

= abc.

It is convenient to write this calculation in terms of a, b and c, to avoid confusion

between the end of the path (where (x, y, z) = (a, b, c)) and the points along the

path (where (x, y, z) = (ta, tb, tc)). However, we can restate the final answer as

p(x, y, z) = xyz, which is more convenient for later use.

Let us consider a vector field F and a curve C, with the vector dr pointing

along the curve. We can construct another vector dn of the same length per-

pendicular to dr. The integral

∫

C

F · dr measures the extent to which F points

along the curve. For some purposes, however, we want to measure the flow of

F across the curve, in which case we want to evaluate

∫

C

F · dn rather than∫

C

F · dr.

Note that dr = (dx, dy) = (x˙, y˙)dt, and dn is obtained by rotating this a

quarter turn clockwise, so dn = (dy,−dx) = (y˙,−x˙)dt (after all the two vectors

are perpendicular).

Example 34.: Let L be the straight line from (1, 0) to (0, 1), so r = (x, y) =

(1 − t, t) for 0 ≤ t ≤ 1. If F be the vector field (x2 − y2, 2xy), find the flux of

the vector F through the line L.

Solution: Since dr = (−1, 1)dt, it means that dn = (1, 1)dt. On L we have

F = ((1− t)2 − t2, 2t(1− t)) = (1− 2t+ t2 − t2, 2t− 2t2) = (1− 2t, 2t− 2t2),

44

so

F · dn = ((1− 2t) + (2t− 2t2))dt = (1− 2t2)dt

so ∫

C

F · dn =

∫ 1

t=0

(1− 2t2) dt =

[

t− 23 t3

]1

t=0

= 1− 2

3

=

1

3

Example 35.: Let us calculate the flow of the field

F = (x+ 2y, 3x+ 4y)

out of the unit circle C.

Solution:

4.2 Surface Integrals

As well as considering curved paths, we also need to consider curved sur-

faces in three-dimensional space. Such a surface can be parametrised as r =

(x(s, t), y(s, t), z(s, t)) for some pair of parameters s and t.

As typical surface we can talk about is a hemisphere, i.e. the surface that

covers half of a sphere. For instance the upper half of a spherical shell of radius

2 can be described in terms of parameters φ and θ by

(x, y, z) = (2 sin(φ) cos(θ), 2 sin(φ) sin(θ), 2 cos(φ))

45

(for 0 ≤ θ ≤ 2pi and 0 ≤ φ ≤ pi/2).

Another example is the off-centre cylinder. Let S be a cylindrical surface of ra-

dius 1, centred on the line joining (1, 1,−1) to (1, 1, 1). Then S can be described

in terms of parameters s and t by

(x, y, z) = (1 + cos(s), 1 + sin(s), t)

(for 0 ≤ s ≤ 2pi and −1 ≤ t ≤ 1).

46

For any function f(x, y), the equation z = f(x, y) defines a surface.

We can use the variables x and y themselves as parameters, and then the full

parametrisation is

(x, y, z) = (x, y, f(x, y)).

To integrate over S, we need a formula for the area of a small piece of S

in terms of a parametrisation r = (x(s, t), y(s, t), z(s, t)). If s and t vary by δs

and δt, then the corresponding part of the surface will be a small parallelogram

spanned by the vectors rs δs = (xsδs, ys δs) and rt δt = (xtδt, yt δt).

We write δA for the area of this parallelogram. We also write δA for the vector

(rs × rt)δs δt. This is perpendicular to rs and rt which means that it is normal

to the surface), and |δA| = δA. In the limit we get dA = (rs × rt)ds dt and

dA = |dA| = |rs × rt|ds dt. Also dA = n dA, where n is the unit normal to S.

Consider again a hemispherical shell of radius a. We have

r = (a sin(φ) cos(θ), a sin(φ) sin(θ), a cos(φ))

47

rφ = (a cos(φ) cos(θ), a cos(φ) sin(θ), −a sin(φ))

rθ = (−a sin(φ) sin(θ), a sin(φ) cos(θ), 0)

rφ × rθ =

∣∣∣∣∣∣

i j k

a cos(φ) cos(θ) a cos(φ) sin(θ) −a sin(φ)

−a sin(φ) sin(θ) a sin(φ) cos(θ) 0

∣∣∣∣∣∣ =

= (a2 sin2(φ) cos(θ), a2 sin2(φ) sin(θ), a2 sin(φ) cos(φ)) = a2 sin(φ)er

dA = a2 sin(φ)er dφ dθ, dA = |dA| = a2 sin(φ)dθ dφ.

It follows that the area of the surface is

A =

∫∫

S

1 dA =

∫ 2pi

θ=0

∫ pi

2

φ=0

a2 sin(φ)dθ dφ =

= 2a2pi

∫ pi

2

φ=0

sin(φ) dφ = 2a2pi

[

− cos(φ)

]pi

2

φ=0

= 2a2pi.

Now let us consider the case of a cylinder. In this case we have

r = (1 + cos(s), 1 + sin(s), t) (0 ≤ s ≤ 2pi, −1 ≤ t ≤ 1)

rs = (− sin(s), cos(s), 0), rt = (0, 0, 1)

rs × rt =

∣∣∣∣∣∣

i j k

− sin(s) cos(s) 0

0 0 1

∣∣∣∣∣∣ = (cos(s), sin(s), 0)

dA = (cos(s), sin(s), 0) ds dt

|rs × rt| = |(cos(s), sin(s), 0)| =

√

cos2(s) + sin2(s) = 1

dA = |rs × rt| ds dt = ds dt.

It follows that the area of the surface is∫∫

S

1 dA =

∫ 2pi

s=0

∫ 1

t=−1

1 ds dt = 2pi(1− (−1)) = 4pi.

For a general surface given by z = f(x, y) we have

r = (x, y, f(x, y)), rx = (1, 0, fx), ry = (0, 1, fy)

rx × ry =

∣∣∣∣∣∣

i j k

1 0 fx

0 1 fy

∣∣∣∣∣∣ = (−fx,−fy, 1)

dA = (rx × ry) dx dy = (−fx,−fy, 1) dx dy

|rx × ry| =

√

f2x + f

2

y + 1

48

and

dA =

√

f2x + f

2

y + 1 dx dy

Now let us apply these relations to a particular case when we consider that

z = f(x, y) = cosh(x+ y)/

√

2 for 0 ≤ x, y ≤ 1. The task is to calculate the area

of this surface for the values of x and y given earlier. It is easy to see that

fx = sinh(x+ y)/

√

2 fy = sinh(x+ y)/

√

2

So now we can apply the formulae just derived to obtain√

1 + f2x + f

2

y =

√

1 +

1

2

sinh2(x+ y) +

1

2

sinh2(x+ y) =

√

1 + sinh2(x+ y) =

=

√

cosh2(x+ y) = cosh(x+ y)

With this relation we can calculate the area element as

dA =

√

1 + f2x + f

2

y dx dy = cosh(x+ y) dx dy.

It follows that the area of the surface is

A =

∫∫

S

1 dA =

∫ 1

x=0

∫ 1

y=0

cosh(x+ y)dy dx =

∫ 1

x=0

[

sinh(x+ y)

]1

y=0

dx =

=

∫ 1

x=0

sinh(x+ 1)− sinh(x) dx =

[

cosh(x+ 1)− cosh(x)

]1

x=0

= (cosh(2)− cosh(1))− (cosh(1)− cosh(0)) = cosh(2)− 2 cosh(1) + 1 ' 1.676

Let us now look into the problem of the flow across a surface. Consider a

surface S in a region where there is a vector field F.

49

We want to calculate the flow of F across S. At each point on S there is a

normal vector n. The flow across S only involves the component of F in the

normal direction, or in other words F · n. We need to integrate this with re-

spect to area, giving

∫∫

S

F · ndA. However, ndA is the same as the vector dA

considered earlier, which can be calculated from a parametrisation by the rule

dA = (rs × rt)ds dt.

Example 36.: Let S be the surface given by z = f(x, y) = xy for 0 ≤ x, y ≤ 1,

and let F be the vector field (x+ y+ z, x+ y+ z, x+ y+ z). Calculate the flow

of the vector field through the surface S.

Solution

Now let us concentrate on the surface integration. This topic has a special

role in electromagnetism since it is closely connected to the Maxwell’s equations.

• For any three-dimensional region, the total electric field crossing the bound-

ary of the region is −10 times the total charge in the region

50

• On the other hand, the magnetic field crossing the boundary always can-

cels out to give a total of zero

• Now suppose we have a surface S in three-dimensional space. Suppose that

has a boundary that is a closed curve C (so the surface could be a disk or

a hemispherical bowl, but not a complete sphere). Then the circulation of

E around C is minus the rate of change of the total magnetic field passing

through S.

• Similarly, the circulation of B around C is µ0 times the rate of change

of the current passing through S (including the ’”displacement current”

0E˙).

Maxwell’s equations told us about the values of scalar and vector fields and their

derivatives at every point in space.The above statements are about various kinds

of integrals of such scalar and vector fields over curves, surfaces and three-

dimensional regions. The main point of this final section of the course is to

understand why these integral statements are the same as the earlier differential

statements.

Let D be a region in the plane. The edge of the region will be a curve,

which we call C. For any vector field u, we can consider the integral

∫

C

u · dn

measuring the flux of u across C.

4.3 The two-dimensional divergence theorem

Let D be a region in the plane whose boundary is a closed curve C. The two-

dimensional divergence theorem says that for any vector field u, we have∫∫

D

div(u) dA =

∫

C

u · dn.

Functions like 1/(x2 + y2) (which blows up to infinity at the origin) are allowed

if the origin lies outside D, but disallowed if the origin is inside D.

Let us consider the above divergence theorem, i.e.∫∫

D

div(u) dA =

∫

C

u · dn.

where C is the boundary of D and the contour is traversed in the anticlockwise

direction. Then let u be (p, q). We then have

div(u) = px + qy

so

u · dn = (p, q) · (dy,−dx) = p dy − q dx.

Therefore ∫∫

D

(px + qy) dA =

∫

C

(p dy − q dx).

51

Example 37. Let D be the disc where x2 + y2 ≤ m2, so C is a circle of radius

m. Take u = (ax + by, cx + dy) for some constants a, b, c and d. Prove the

divergence theorem.

Solution:

Given the expression of u we have

div(u) = (ax+ by)x + (cx+ dy)y = a+ d,

so ∫∫

D

div(u)dA = (a+ d)area(D) = pim2(a+ d)

On the other hand, we can parametrise C by r = (x, y) = (m cos(t),m sin(t)),

so dn = (y˙,−x˙)dt = (m cos(t),m sin(t)) dt. On C we also have

u = (ax+ by, cx+ dy) = (am cos(t) + bm sin(t), cm cos(t) + dm sin(t))

so

u·dn = (am cos(t)+bm sin(t))(m cos(t))dt+(cm cos(t)+dm sin(t))(m sin(t))dt =

= m2(a cos2(t)+(b+c) sin(t) cos(t)+d sin2(t))dt =

m2

2

(a+a cos(2t)+(b+c) sin(2t)+d−d cos(2t)) =

=

m2

2

((a+ d) + (a− d) cos(2t) + (b+ c) sin(2t))

As a result we have∫

C

u · dn = m

2

2

[

(a+ d)t+

1

2

(a− d) sin(2t)− 1

2

(b+ c) cos(2t)

]2pi

t=0

=

=

m2

2

2pi(a+ d) = pim2(a+ d).

Example 38. Let D be the rectangle as shown below. The boundary of the

domain consists of C1, . . . , C4. If u is the horizontal vector field u = (e

−x−y, 0),

prove the divergence theorem.

Solution:

52

Example 39.: The figure shows the deltoid curve C The curve C can be written

Figure 3: The deltoid curve used for Example 39

in parametric form as

x = 2 cos(t) + cos(2t) y = 2 sin(t)− sin(2t).

Calculate the area of this object.

Solution: Obviously it is hard to find the area of D directly. However, we can

evaluate it by a trick using the divergence theorem. Consider the vector field

53

F = (x, 0), so

div(F) =

∂x

∂x

+

∂0

∂y

= 1

As a consequence we have that

∫∫

D

div(F)dA = area(D).

The Divergence Theorem tells us that this is the same as

∫

C

F · dn, where

dn = (y˙,−x˙) dt = (2 cos(t)− 2 cos(2t), 2 sin(t) + 2 sin(2t)) dt

Furthermore

F = (x, 0) = (2 cos(t) + cos(2t), 0)

so

F·dn = (2 cos(t)−2 cos(2t))(2 cos(t)+cos(2t)) = 4 cos2(t)−2 cos(t) cos(2t)−2 cos2(2t) =

= (2+2 cos(2t))−(cos(3t)+cos(t))−(1+cos(4t)) = 1−cos(t)+2 cos(2t)−cos(3t)−cos(4t)

Therefore

area =

∫ 2pi

t=0

F · dn = 2pi

4.4 Green’s theorem

Let D be a region in the plane whose boundary is a closed curve C. Green’s

theorem says that for any vector field u that is well-behaved (i.e. is not discon-

tinuous) everywhere in D, we have∫∫

D

curl(u) · dA =

∫

C

u · dr.

Since D is plane dA = ndA, where n is unit normal vector to D. Accordingly,

if u = (P,Q), then∫∫

D

curl(u) · dA =

∫∫

D

(

∂Q

∂x

− ∂P

∂y

)

dA

Example 40.: Let D be the unit disc, so the boundary curve C is the unit

circle. Let u be the vector field (x3, x3). Verify the Green’s Theorem.

Solution:

54

4.5 The (three-dimensional) Divergence Theorem

The two dimensional knowledge can be expanded to study three-dimensional

cases. Let E be the three-dimensional solid region enclosed by a surface S. Let

u be a vector field that is well-behaved everywhere in E. Then∫∫∫

E

div(u) dV =

∫∫

S

u · dA

This can be proved by an argument similar to that used for the two-dimensional

version. The physical interpretation is also similar: in a steady state, the rate

of flow of particles escaping through S must balance the rate of creation of par-

ticles in E.

Example 41.: Let S be the unit sphere, and let E be the solid ball enclosed

by S. Consider the vector field u = (x, 0, 0). Verify the divergence theorem in

3D.

Solution: The given vector has div(u) = ∂x/∂x+ ∂0/∂y + ∂0/∂z = 1, so∫∫∫

E

div(u)dV =

∫∫∫

E

dV = volume of E = 4pi/3.

On S we have

r = (x, y, z) = (sin(φ) cos(θ), sin(φ) sin(θ), cos(φ))

55

and

u = (x, 0, 0) = (sin(φ) cos(θ), 0, 0).

The unit normal vector is n = er = r, so u · n = x2 = sin2(φ) cos2(θ). We have

also seen before that dA = sin(φ) dφ dθ, so∫∫

S

u · dA =

∫∫

S

u · n dA =

∫ 2pi

θ=0

∫ pi

φ=0

sin3(φ) cos2(θ) dφ dθ =

=

(∫ 2pi

θ=0

cos2(θ) dθ

)(∫ pi

φ=0

sin3(φ) dφ

)

= pi

∫ pi

φ=0

1

4

(3 sin(φ)− sin(3φ))dφ =

=

pi

4

[

− 3 cos(φ) + 1

3

cos(3φ)

]pi

φ=0

=

pi

4

((3− 1

3

)− (−3 + 1

3

)) =

4pi

3

As expected, this is the same as

∫∫∫

E

div(u)dV .

Example 42.: Let E be the solid vertical cylinder of radius a and height 2b

centred at the origin, and let S be the surface of E. Consider the vector field

u = (−y, x, z3). Verify the divergence theorem in 3D.

Solution:

56

4.6 Stokes’ Theorem

Stokes’ Theorem is analogous to Green’s Theorem, but it applies to curved

surfaces as well as to flat regions in the plane. Suppose we have a surface S

whose boundary is a closed curve C, and a well-behaved vector field u. Then∫∫

S

curl(u) · dA = ±

∫

C

u · dr.

We need a little more discussion to eliminate the ambiguity in the sign. To

make sense of the right hand side, we need to specify the direction in which

we move around C. The integral in one direction will be the negative of the

integral in the opposite direction. Similarly, on the left hand side we have the

integral of curl(u) ·n dA, where n is a unit vector normal to the surface. There

are two possible directions for n (each opposite to the other) and there is no

natural rule to choose between them. However, the choice of n can be linked to

the choice of direction around the curve as follows: if you walk in the specified

direction with your feet on C and your head pointing in the direction of n, then

the surface S should be on your left. Provided that we follow this convention,

we will have ∫∫

S

curl(u) · dA =

∫∫

S

curl(u) · n dA = +

∫

C

u · dr.

Example 43.: Consider the surface S given by z = x2 − y2 with x2 + y2 ≤ 1.

Consider the vector field given as u = (−y, x, 0). Check Stokes’ Theorem.

Solution:

First we parametrise S as

r = (x, y, z) = (r cos(s), r sin(s), r2 cos2(s)− r2 sin2(s))

with 0 ≤ r ≤ 1 and 0 ≤ s ≤ 2pi. Using cos2(s)− sin2(s) = cos(2s):

r = (x, y, z) = (r cos(s), r sin(s), r2 cos(2s))

which gives

rr = (cos(s), sin(s), 2r cos(2s)) rs = (−r sin(s), r cos(s), −2r2 sin(2s))

Now, we can calculate

rr × rs =

∣∣∣∣∣∣

i j k

cos(s) sin(s) 2r cos(2s)

−r sin(s) r cos(r) −2r2 sin(2s)

∣∣∣∣∣∣

= (−2r2 sin(s) sin(2s)−2r2 cos(s) cos(2s), 2r2 cos(s) sin(2s)−2r2 sin(s) cos(2s), r cos2(s)+r sin2(s))

Using the identities

sin(a) sin(b) + cos(a) cos(b) = cos(a− b) = cos(b− a)

57

and

sin(a) cos(b)− cos(a) sin(b) = sin(a− b) = − sin(b− a)

the result becomes rr × rs = (−2r2 cos(s), 2r2 sin(s), r), so

dA = (rr × rs) dr ds = (−2r2 cos(s), 2r2 sin(s), r) dr ds.

Next, we have

curl(u) =

∣∣∣∣∣∣

i j k‘

∂

∂x

∂

∂y

∂

∂z

−y x 0

∣∣∣∣∣∣

so ∫∫

S

curl(u) · dA =

∫ 2pi

s=0

∫ 1

r=0

2r dr ds =

∫ 2pi

s=0

1 ds = 2pi.

On the other hand, we can parametrise the boundary curve C (where r = 1) as

r = (x, y, z) = (cos(s), sin(s), cos(2s)).

On this curve we have

u = (−y, x, 0) = (− sin(s), cos(s), 0), dr = (− sin(s), cos(s),−2 sin(2s)) ds

therefore

u · dr = (sin2(s) + cos2(s)) ds = ds

and ∫

C

u · dr =

∫ 2pi

s=0

ds = 2pi.

As expected, this is the same as

∫∫

S

curl(u) · dA.

Example 44. Let us consider the cylinder with r = a and −b ≤ z ≤ b with

0 ≤ θ ≤ 2pi.

Check Stokes’s Theorem for the vector field u = (−zy, zx, z2).

58

Solution:

Let us denote by S the lateral surface of the cylinder. We can parametrise S as

r = (x, y, z) = (a cos(θ), a sin(θ), z)

so

rθ = (−a sin(θ), a cos(θ), 0), rz = (0, 0, 1)

Now we can calculate

rθ × rz =

∣∣∣∣∣∣

i j k

−a sin(θ) a cos(θ) 0

0 0 1

∣∣∣∣∣∣ = (a cos(θ), a sin(θ), 0)

so

dA = (rθ × rz)dθdz = a(cos(θ), sin(θ), 0) dθ dz.

Note that dA points outwards, away from the z-axis. Also

curl(u) =

∣∣∣∣∣∣

i j k

∂

∂x

∂

∂y

∂

∂z

−zy zx z2

∣∣∣∣∣∣ = (0− x, −y − 0, z − (−z)) = (−x,−y, 2z).

On the surface S this becomes

curl(u) = (−a cos(θ), −a sin(θ), 2z)

so

curl(u) · dA = (−a2 cos2(θ)− a2 sin2(θ)) dθ dz = −a2 dθ dz

meaning that∫∫

S

curl(u) · dA = −a2

∫ 2pi

θ=0

∫ b

z=−b

dθ dz = −a2 × 2pi × 2b = −4pia2b.

The boundaries of S are C1 and C2. Directions as shown keep S on the left when

walking with head in the direction of dA, away from the z-axis. Compatible

parametrisations of the two boundaries are

C1 (x, y, z) = (a cos(t),−a sin(t), b)

and

C2 (x, y, z) = (a cos(t), a sin(t),−b).

On C1 we have

dr = (−a sin(t),−a cos(t), 0) dt u = (−zy, zx, z2) = (ab sin(t), ab cos(t), b2)

so

u · dr = −a2b sin2(t)− a2b cos2(t) = −a2b

59

and ∫

C1

u · dr =

∫ 2pi

t=0

−a2b dt = −2pia2b

The discussion of C2 is similar:

dr = (−a sin(t), a cos(t), 0) dt u = (ab sin(t),−ab cos(t), b2),

therefore

u · dr = −a2b sin2(t)− a2b cos2(t) = −a2b

and ∫

C2

u · dr = −2pia2b.

Putting these together, we get

∫

C

u · dr = −4pia2b, which is the same as∫∫

S

curl(u).dA, as expected.

60

Mathematics III

University of Sheffield, 2020

1 Complex functions

1.1 Introduction

The term complex analysis refers to the calculus of complex-valued functions,

f(z), depending on a single complex variable z. To an un-initiated, it may

seem that this subject should merely be a simple reworking of standard real

variable theory that you learned in first and second year calculus. However, this

naive first impression could not be further from the truth! Complex analysis is

the culmination of a deep and far-ranging study of the fundamental notions

of complex differentiation and integration, and has an elegance and beauty

not found in the real domain. For instance, complex functions are necessarily

analytic, meaning that they can be represented by convergent power series, and

hence are infinitely differentiable. Thus, difficulties with degree of smoothness,

strange discontinuities, subtle convergence phenomena, and other pathological

properties of real functions never arise in the complex world.

Complex functions have a very wide applicability in electrical engineering.

The calculation of harmonically varying currents in circuits consisting of re-

sistors, capacitors, induction, coils, etc. require functions of complex variable.

Secondly, functions of complex variables are used for solving Laplace and Pois-

son equations describing electrostatic potential.

The driving force behind many of the applications of complex analysis is the

remarkable connection between complex functions and harmonic functions of

two variables, a.k.a. solutions of the planar Laplace equation. To wit, the real

and imaginary parts of any complex analytic function are automatically har-

monic. In this manner, complex functions provide a rich spectrum of additional

solutions to the two-dimensional Laplace equation, which can be exploited in a

wide range of physical, mathematical, and engineering applications.

One of the most useful consequences stems from the elementary observation

that the composition of two complex functions is also a complex function. We

re-interpret this operation as a complex change of variables, producing a confor-

mal mapping that preserves (signed) angles in the Euclidean plane. Conformal

mappings can be effectively used for constructing solutions to the Laplace equa-

tion on complicated planar domains that appear in a wide range of physical

problems, including fluid mechanics, aerodynamics, thermomechanics, electro-

statics, and elasticity. In particular, conformal mapping is a widespread math-

ematical technique to approach the electromagnetic study of electric machines,

e.g. energy conservation in magnetostatic and time-harmonic problems; current

source assignment in the transformed domain; transformation of the flux density

components through conformal mapping, etc.

1.2 Properties of complex functions

Our principal objects of study are complex-valued functions, f(z), depending

on the complex variable z = x + iy and they are defined over of the complex

plain, C.

2

Any complex function f(z) can be uniquely written as a complex combina-

tion

f(z) = f(x+ iy) = u(x, y) + iv(x, y), (1)

of two real functions both depending on the two real variables, x and y with

Re(f) = u(x, y) and Im(f) = u(x, y). For example, the function f(z) = z2 can

be expanded and written as

z2 = (x+ iy)2 = x2 − y2 + 2ixy

so

Re(f) = x2 − y2, Im(f) = 2xy

Many of the well-known functions appearing in real-variable calculus (polyno-

mials, rational functions, exponentials, trigonometric functions, logarithms, and

many more) have natural complex extensions. For example, complex polynomi-

als

p(z) = anz

n + an−1zn−1 + · · ·+ a1z + a0 (2)

are complex linear combinations (meaning that the coefficients ak are allowed

to be complex numbers) of the basic monomial functions zk = (x+ iy)k.

As we will see the notion of harmonic function is very important for our

module. By definition a function is called harmonic if it is at least twice

differentiable and it satisfies the well-known Laplace equation. In 2D this partial

differential equation takes the form

∂2f

∂x2

+

∂2f

∂y2

= 0.

Before going deeper into the many remarkable properties of complex func-

tions, let us look at some of the most basic examples. In each case, you can

directly check that the harmonic functions provided by the real and imagi-

nary parts of the complex function are indeed solutions to the two-dimensional

Laplace equation.

Examples of complex functions

(a) Harmonic Polynomials: As noted above, any complex polynomial is a linear

combination, as in (2), of the basic complex monomials

zn = (x+ iy)n = un(x, y) + ivn(x, y). (3)

Their real and imaginary parts, un(x, y), vn(x, y), are harmonic polynomials.

(b) Rational functions: Ratios of the form

f(z) =

p(z)

q(z)

, (4)

3

where p(z) and q(z) are complex polynomials provide a large variety of harmonic

functions. The simplest case is

1

z

=

1

x+ iy

=

x− iy

(x+ iy)(x− iy) =

x− iy

x2 + y2

=

x

x2 + y2

− i y

x2 + y2

. (5)

Apart from an interesting behaviour at the origin (x = 0, y = 0), these functions

are harmonic everywhere. It is easy to see that they are twice differentiable and

they satisfy the 2D Laplace equation. For the real part

∂

∂x

(

x

x2 + y2

)

= − x

2 − y2

(x2 + y2)2

,

∂2

∂x2

(

x

x2 + y2

)

=

2x(x2 − 3y2)

(x2 + y2)3

∂

∂y

(

x

x2 + y2

)

= − 2xy

(x2 + y2)2

,

∂2

∂y2

x

x2 + y2

=

−2x(x2 − 3y2)

(x2 + y2)3

.

It is obvious that the Laplace equation is satisfied. The same can be shown to

happen to the imaginary part.

A slightly more complicated example is the function

f(z) =

z − 1

z + 1

. (6)

To write (6) in the standard form (1) we multiply and divide by the complex

conjugate of the denominator, so

f(z) =

z − 1

z + 1

=

(z − 1)(z + 1)

(z + 1)(z + 1)

=

=

|z|2 + z − z − 1

|z + 1|2 =

x2 + y2 − 1

(x+ 1)2 + y2

+ i

2y

(x+ 1)2 + y2

, (7)

where z denotes the complex conjugate of z. Again, it is easy to show that these

functions (the real and imaginary parts) are harmonic functions.

(c) Complex exponential functions: Euler’s formula

ez = ex+iy = ex(cos y + i sin y), (8)

for the complex exponential yields two important harmonic functions: ex cos y

and ex sin y. For example

dex cos y

dx

=

d2ex cos y

dx2

= ex cos y

dex cos y

dy

= −ex sin y; d

2ex cos y

dy2

= −ex cos y

and it is obvious why the Laplace equation is satisfied. Similar simple calculation

can be used for the imaginary part. More generally, writing out ecz for a complex

constant c = a+ ib, produces the complex exponential function

ecz = e(a+ib)(x+iy) = e(ax−by)+i(ay+bx) = eax−by[cos(ay+bx)+i sin(ay+bx)] (9)

4

whose real and imaginary parts are harmonic functions for arbitrary a, b ∈ R.

(d) Complex trigonometric functions: These are defined in terms of the complex

exponential according to

cos z =

eiz + e−iz

2

, sin z =

eiz − e−iz

2i

. (10)

Similarly, hyperbolic functions can be also expressed with the help of the com-

plex exponential function, so

cosh z =

ez + e−z

2

, sinh z =

ez − e−z

2

. (11)

Example 1.: Show that

sin z = sinx cosh y + i cosx sinh y

is a harmonic function

(e) The complex logarithmic function: In a similar fashion, the complex loga-

rithm is a complex extension of the usual real natural (i.e., base e) logarithm.

In terms of polar coordinates z = reiθ, the complex logarithm has the form

ln z = ln reiθ = ln r + ln eiθ = ln r + iθ. (12)

Thus, the logarithm of a complex number has a real part

Re{ln z} = ln r = ln |z| = ln

√

x2 + y2 =

1

2

ln(x2 + y2), (13)

which is a well-defined harmonic function. The imaginary part

Im{ln z} = θ = arg(z),

of the complex logarithm is the polar angle, also known as the argument of the

complex number (note that the complex logarithm is not defined for x = y = 0).

5

1.3 Complex differentiation

The notion of complex derivatives stays at the very foundation of complex func-

tion theory. Complex differentiation is defined in the same manner as the usual

calculus limit definition of the derivative of a real function. Yet, despite a

superficial similarity, complex differentiation is a profoundly different theory,

displaying an elegance and depth not shared by its real counterpart.

Definition: A complex function, f(z), is differentiable at a point z ∈ C if and

only if the limiting ratio quotient

f ′(z) = lim

w→z

f(w)− f(z)

w − z (14)

exists.

The key feature of this definition is that the limiting value f ′(z) of the

difference quotient must be independent of how w converges to z. On the real

line, there are only two directions to approach a limiting point: either from

the left or from the right. These two ways to approach a value lead to the

concepts of left- and right-handed derivatives and their equality is required for

the existence of the usual derivative of a real function.

In the complex plane, there are an infinite variety of directions to approach

the point z, and the definition requires that all of these directional derivatives

must agree. This requirement imposes severe restrictions on complex deriva-

tives, and is the source of their remarkable properties. To understand the con-

sequences of this definition, let us first see what happens when we approach z

along the two simplest directions: horizontal and vertical (N.B. in a complex

plane, the real axis is the horizontal axis, while imaginary numbers are put on

the vertical axis). If we set

w = z + h = (x+ h) + iy, h ∈ R,

then w → z along a horizontal line as h→ 0. If we write f(z) = u(x, y)+iv(x, y),

then we must have

f ′(z) = lim

h→0

f(z + h)− f(z)

h

= lim

h→0

f(x+ h+ iy)− f(x+ iy)

h

= lim

h→0

[

u(x+ h, y)− u(x, y)

h

+ i

v(x+ h, y)− v(x, y)

h

]

=

∂u

∂x

+ i

∂v

∂x

=

∂f

∂x

which follows from the usual definition of the (real) partial derivative. On the

other hand, if we set

w = z + ik = x+ i(y + k), k ∈ R

then w → z along the vertical line as k → 0. Therefore we must have

f ′(z) = lim

k→0

f(z + ik)− f(z)

ik

= lim

k→0

[

−if(x+ i(y + k))− f(x+ iy)

k

]

=

6

lim

k→0

[

v(x, y + k)− v(x, y)

k

− iu(x, y + k)− u(x, y)

k

]

=

∂v

∂y

− i∂u

∂y

= −i∂f

∂y

When we equate the real and imaginary parts of these two distinct formulae for

the complex derivative f ′(z), and take into account that the two formulations

denote the same operation, we discover that the real and imaginary components

of f(z) must satisfy a linear system of partial differential equations, called the

Cauchy-Riemann equations.

Theorem: A complex function f(z) = u(x, y)+iv(x, y) depending on z = x+iy

has a complex derivative f ′(z) if and only if its real and imaginary parts are

continuously differentiable and satisfy the Cauchy-Riemann equations

∂u

∂x

=

∂v

∂y

,

∂u

∂y

= −∂v

∂x

(15)

In this case, the complex derivative of f(z) is equal to any of the following

expressions:

f ′(z) =

∂f

∂x

=

∂u

∂x

+ i

∂v

∂x

= −i∂f

∂y

= −i∂u

∂y

+

∂v

∂y

(16)

A pair of functions that satisfy the system (15) is called conjugate.

Example 2.: Consider the elementary function

z3 = (x3 − 3xy2) + i(3x2y − y3)

Show that Its real part u = x3− 3xy2 and imaginary part v = 3x2y− y3 satisfy

the Cauchy-Riemann equations.

Solution

Fortunately, the complex derivative obeys all of the usual rules that you learned

in real-variable calculus. For example

d

dz

zn = nzn−1,

d

dz

ecz = cecz,

d

dz

ln z =

1

z

, (17)

7

and so on. Here, the power n can be non-integral or even, in view of the identity

zn = en ln z, complex, while c is any complex constant. The exponential formulae

(10) for the complex trigonometric functions imply that they also satisfy the

standard rules

d

dz

cos z = − sin z, d

dz

sin z = cos z. (18)

The formulae for differentiating sums, products, ratios, inverses, and compo-

sitions of complex functions are all identical to their real counterparts, with

similar proofs. Thus, thankfully, you do not need to learn any new rules for

performing complex differentiation.

Remark: There are many examples of seemingly reasonable functions which do

not have a complex derivative. The simplest is the complex conjugate function

f(z) = z = x− iy.

Its real and imaginary parts do not satisfy the Cauchy-Riemann equations, and

hence z does not have a complex derivative. More generally, any function f(z, z)

that explicitly depends on the complex conjugate variable z is not complex-

differentiable.

Example 3.: You are given that

Re(f(z)) =

x

x2 + y2

= u(x, y)

and f(1) = 1. Find f(z).

Solution: Let f(z) = u(x, y) + iv(x, y). Using the Cauchy-Riemann equations

∂v

∂y

=

∂u

∂x

=

(x2 + y2)− 2x2

(x2 + y2)2

=

y2 − x2

(x2 + y2)2

, (1)

∂v

∂x

= −∂u

∂y

= − −2xy

(x2 + y2)2

=

2xy

(x2 + y2)2

(2)

From (2) we have

v =

∫

2xy

(x2 + y2)2

dx

Let us change the variable and introduce w = x2 + y2, so dw = 2xdx, therefore

v =

∫

y

w2

dw = − y

w

+ C(y) = − y

x2 + y2

+ C(y)

Let us differentiate this, so

∂v

∂y

= − (x

2 + y2)− 2y2

(x2 + y2)2

+

dC

dy

=

y2 − x2

(x2 + y2)2

+

dC

dy

.

8

Using (1) we have

y2 − x2

(x2 + y2)2

+

dC

dy

=

y2 − x2

(x2 + y2)2

−→ dC

dy

= 0 −→ C = const.

Therefore

f(z) = u(x, y) + iv(x, y) =

x

x2 + y2

− iy

x2 + y2

+ iC =

x− iy

(x+ iy)(x− iy) + iC =

1

x+ iy

+ iC =

1

z

+ iC.

Using the given condition

f(1) = 1 = 1 + iC =⇒ C = 0 =⇒ f(z) = 1

z

.

***********************

There is a connection between Cauchy-Riemann equations and the Laplace equa-

tion we discussed earlier. Let us differentiate the first equation of (15) with

respect to x and the second equation with respect to y, so we have

∂2u

∂x2

=

∂2v

∂x∂y

,

∂2u

∂y2

= − ∂

2v

∂x∂y

.

Adding these two equations we obtain

∂2u

∂x2

+

∂2u

∂y2

= 0,

which is exactly the Laplace equation applied to the real part of f(z). Now

taking again Eq. (15) and differentiating the first equation with respect to y

and the second equation with respect to x we obtain

∂2u

∂x∂y

=

∂2v

∂y2

,

∂2u

∂x∂y

= −∂

2v

∂x2

.

Subtracting these two equations we arrive to

∂2v

∂x2

+

∂2v

∂y2

= 0,

which is the Laplace equation written for the imaginary part of f(z). These

two relations show that the real and imaginary part of a complex number are

harmonic functions.

9

1.4 Analytical functions

If f(z) = u(x, y) + iv(x, y) is a complex function and u(x, y) and v(x, y) are

conjugate in the same domain, then the function f(z) is called analytical.

A complex function is said to be analytical over a certain domain if it is com-

plex differentiable at every point of the domain. The necessary and sufficient

conditions that a complex function f = u+ iv is analytical are

• The four derivatives of its real and imaginary parts

∂u

∂x

,

∂v

∂y

,

∂u

∂y

,

∂v

∂x

satisfy the Cauchy-Riemann equations (15) and, in addition, they are

continuous

A remarkable feature of complex differentiation is that the existence of one com-

plex derivative automatically implies the existence of infinitely many!

A complex function f(z) is called analytic at a point z0 ∈ C if it has a power

series expansion

f(z) = a0 + a1(z − z0) + a2(z − z0)2 + · · · =

∞∑

n=0

an(z − z0)n (19)

that converges to all z sufficiently close to z0. We note that if f(z) and g(z)

are analytic at a point z0, so is their sum f(z) + g(z), product f(z)g(z) and,

provided g(z0) 6= 0, ratio f(z)/g(z).

The simplest test to check whether a complex series is convergent or not is

the ratio test. Let us suppose that the n-th term of a series is cn. Then, by

definition, the test makes use of the limit

L = lim

n→∞

∣∣∣∣cn+1cn

∣∣∣∣ . (20)

According to the value of L we can distinguish three cases:

• if L < 1, the series is convergent

• if L > 1, the series is divergent

• if L = 1, or the limit fails to exist, then the test is inconclusive, because

there exist both convergent and divergent series that satisfy this case.

All of the real power series found in elementary calculus carry over to the com-

plex versions of the functions. For example,

ez = 1 + z +

z2

2

+

z3

3!

+ · · · =

∞∑

n=0

zn

n!

,

10

is the power series for the exponential function based at z0 = 0. A straight-

forward application of the ratio test proves that the series converges for all z,

so

L = lim

n→∞

∣∣∣∣zn+1/(n+ 1)!zn/n!

∣∣∣∣ = limn→∞ zn+ 1 → 0.

On the other hand, the power series

1

z2 + 1

= 1− z2 + z4 − z6 + · · · =

∞∑

n=0

(−1)nz2n,

has a restricted convergence domain because

L = lim

n→∞

∣∣∣∣ (−1)n+1z2n+2(−1)nz2n

∣∣∣∣ = |z|2 = (x2 + y2).

In order to be convergent we must have L < 1, i.e. x2 + y2 < 1, i.e. the series is

convergent inside a circle of radius 1, and diverges outside the circle for |z| > 1.

In general, there are three possible options for the domain of convergence of

a complex power series (19)

• The series converges for all z

• The series converges inside a disk |z − z0| < ρ of radius ρ > 0 centred at

z0 diverges for all |z− z0| > ρ outside the disk. The series might converge

for at some (but not all) of the points on the boundary of the disk where

|z − z0| = ρ

• The series only converges, trivially, at z = z0

The number ρ is known as the radius of convergence of the series. In the first

case, we say ρ =∞, while in last case, ρ = 0, and the series does not represent

an analytic function.

Let us suppose that the sequence {cn}, n = 1, 2, . . . converges with a limit

L. If L = 0, then the radius of convergence is infinity, i.e. ρ =∞, i.e. the power

series converges for all z. If L 6= 0 (hence L > 0), then

ρ =

1

L

= lim

n→∞

∣∣∣∣ anan+1

∣∣∣∣ , (21)

here an is the coefficient of the nth term in the series.

Remarkably, the radius of convergence for the power series of a known an-

alytic function f(z) can be determined by inspection, without recourse to any

fancy convergence tests! Namely, ρ is equal to the distance from z0 to the near-

est singularity of f(z), meaning a point where the function fails to be analytic.

In particular, the radius of convergence ρ =∞ if and only if f(z) is analytic for

11

all z ∈ C and has no singularities; examples include polynomials, ez, cos z, and

sin z.

On the other hand, the rational function

f(z) =

1

z2 + 1

=

1

(z + i)(z − i)

has singularities at z = ±i, so its power series shown earlier has a radius of

convergence ρ = 1, which is the distance from z0 = 0 to the singularities. If we

expand (z2 + 1)−1 in a power series at some other point, say z0 = 1 + 2i, then

we need to determine which singularity is closest.

We compute |i−z0| = |−1−i| =

√

2, while |−i−z0| = |−1−3i| =

√

10, and

so the radius of convergence ρ =

√

2 is the smaller. This allows us to determine

the radius of convergence in the absence of any explicit formula for its (rather

complicated) Taylor expansion at z0 = 1 + 2i.

There are, in fact, only four possible types of singularities of a complex

function f(z):

• A singular point z = z0 is called pole of order n if and only if

f(z) =

h(z)

(z − z0)n , (22)

where h(z) is an analytical function at z = z0 and h(z0) 6= 0. The function

f(z) = 1/(z−2)3 has a pole of order 3 at z = 2, and f(z) = (3z−2)/[(z−

1)2(z+1)(z−4)] has a pole of order 2 at z = 1 and simple poles at z = −1

and z = 4. In the case of poles, it is true that we can find a positive

integer n (identical to the order of the pole) so that

lim

z→z0

(z − z0)nf(z) 6= 0

is analytical

• Branch points: refer to multi-valued functions. We have already encoun-

tered the two basic types: algebraic branch points, such as the function n

√

z

at z0 = 0 and logarithmic branch point such as lnz at z0 = 0. For example

f(z) = (z − 3)1/2 has a branch point at z = 3 and f(z) = ln(z2 + z − 2)

has branch points where z2 + z − 2 = 0, i.e. at z = 1 and z = −2

• Essential singularity. By definition, a singularity is essential if it is not

a pole or a branch point. The quintessential example is the essential

singularity of the function e1/z at z0. The behaviour of a complex function

near an essential singularity is quite complicated

• Singularities at infinity. This type of singularity of f(z) at z = ∞ is the

same as that of f(1/w) at w = 0. For example, the function f(z) = z3

has a pole of order 3 at z =∞ since f(1/w) = 1/w3 has a pole of order 3

at w = 0. Equally, a function has a singularity at infinity if

lim

z→∞ f(z) =∞

12

The nature and number of singularities will be very important later on when

we will discuss the integration of complex functions.

The complex function

f(z) =

ez

z3 − z2 − 5z − 3 =

ez

(z − 3)(z + 1)2

is analytic everywhere except for singularities at the points z = 3 and z = −1,

where its denominator vanishes. Therefore z = 3 is a simple (order 1) pole and

z = −1 is a double (order 2) pole.

A complicated complex function can have a variety of singularities. For

example, the function

f(z) =

e−1/(z−1)

2

(z2 + 1)(z + 2)2/3

has simple poles at z = ±i, a branch point of degree 3 at z = −2, and an

essential singularity at z = 1.

If a function is single-valued and has a singularity, then this singularity is

either a pole or an essential singularity. For this reason a pole is sometimes

called non-essential singularity.

Example 4.: Locate and name all the singularities of

f(z) =

z8 + z4 + 2

(z − 1)3(3z + 2)2

Determine where f(z) is analytic.

Solution:

13

Taylor’s Theorem: Let f(z) be analytic inside and on a simple closed curve

C. Let a and a+ h two points inside C. Then

f(a+ h) = f(a) + hf ′(a) +

h2

2!

f ′′(a) + · · ·+ h

n

n!

f (n)(a) + . . . (23)

or writing z = a+ h and h = z − a

f(z) = f(a) + (z − a)f ′(a) + (z − a)

2

2!

f ′′(a) + · · ·+ (z − a)

n

n!

f (n)(a) + . . . (24)

The series (23) or (24) are called the Taylor series or expansion for f(a+ h) or

f(z). The region of convergence of the series (24) is given by |z− a| < ρ, where

the radius of convergence ρ is the distance from a to the nearest singularity of

the function f(z). On |z − a| = ρ, the series may or may not converge. For

|z − a| > ρ, the series diverges. Obviously, if the nearest singularity of f(z) is

at infinity, the radius of convergence is infinite, i.e. the series converges for all

z. For a = 0 the above series are also called MacLaurin series.

Example 5.: Let f(z) = ln(1 + z) (with some special considerations, that are

not so important here).

(a) Expand f(z) in Taylor series about z = 0

(b) Determine the region of convergence for the series determined in (a)

(c) Expand ln

(

1+z

1−z

)

in Taylor series about z = 0

Solution:

14

Example 6.: Find the radius of convergence of the series:

1− (z − i) + 2(z − i)2 − 3(z − i)3 + · · ·+ (−1)nn(z − i)n + . . .

Solution: In this case the radius of the convergence is

ρ = lim

n→∞

∣∣∣∣ (−1)nn(−1)n+1(n+ 1)

∣∣∣∣ = limn→∞ nn+ 1 = 1

Hence the series is convergent for |z − i| < 1 and divergent for |z − i| > 1. Let

us write |z − i| as |x + i(y − 1)| which can be transformed to √x2 + (y − 1)2.

Therefore the region where this series is convergent satisfies the condition

x2 + (y − 1)2 < 1

i.e. is the circle in the complex plane centred at (0,1) and having a radius 1.

The power series discussed above and the convergence criteria are also true for

negative powers of (z − z0). We can also consider series of the form

∞∑

n=−∞

an(z − z0)n = · · ·+ a−n

(z − z0)n + · · ·+

a−2

(z − z0)2 +

a−1

z − z0 +

+a0 + a1(z − z0) + a2(z − z0)2 + · · ·+ an(z − z0)n + . . . (25)

The part of of this series containing positive powers of z − z0 is convergent for

|z − z0| < ρ, where ρ is given in the standard form

ρ = lim

n→∞

∣∣∣∣ anan+1

∣∣∣∣ .

15

Let us find where the part of series containing negative powers of z − z0 is

convergent. First of all, it is convergent when a−n = 0 for n > m, so that is

finite. Consider now the case when the series having negative powers is infinite.

Once again, we use the ratio test and obtain that it is convergent provided

lim

n→∞

∣∣∣∣a−n−1(z − z0)−n−1a−n(z − z0)−n

∣∣∣∣ < 1

We can rewrite this as

1

|z − z0| limn→∞

∣∣∣∣a−n−1a−n

∣∣∣∣ < 1.

Introducing

R = lim

n→∞

∣∣∣∣a−n−1a−n

∣∣∣∣ , (26)

(we assume that this limit exists) we eventually obtain that the condition for

convergence is that |z − z0| > R. In summary, taking now all powers together

we obtain that the complete series (25) is convergent when z is in the annulus

R < |z − z0| < ρ (27)

Obviously this condition makes sense is R < ρ.

Example 7.: Find the region of convergence of the following series

(i) + · · ·+ 2

n

zn

+ · · ·+ 2

2

z2

+

2

z

+ 1 +

z

3

+

z2

32

+ · · ·+ z

n

3n

(ii) + · · ·+ 1

n!zn

+ · · ·+ 1

2z2

+

1

z

+ 1 + z + z2 + · · ·+ zn + . . .

Solution:

16

Now we can formulate a new theorem that will be useful when discussing inte-

gration

Laurent’s theorem: Let the function f(z) be analytical in the annulus R <

|z − z0| < ρ. Then, the function f(z) can be expanded in convergent series

f(z) =

∞∑

n=−∞

an(z − z0)n = · · ·+ a−n

(z − z0)n + · · ·+

a−2

(z − z0)2 +

a−1

z − z0 +

+a0 + a1(z − z0) + a2(z − z0)2 + · · ·+ an(z − z0)n + . . . (28)

and this is called the Laurent series of the function f(z) about the point z0.

To calculate the Laurent series we use the standard and modified geometric

series which are

1

1− z =

{ ∑∞

n=0 z

n, |z| < 1

−∑∞n=1 1zn , |z| > 1 (29)

Here the function is analytic everywhere except in the singularity at z = 1.

The expressions (29) are the expansions for f in the regions inside and outside

the circle of radius 1, centred on z = 0, where z| < 1 is the region inside

the circle and |z| > 1 is the region outside the circle. In reality to find the

Laurent series for even very simple functions is not trivial, therefore we combine

simple known expansions (Taylor or Laurent) to obtain Laurent expansions of

more complicated functions. In applications, often only the first few terms in a

Laurent expansion are needed.

Example 8.: Determine the Laurent series of f(z) = 1/(z + 5) that are valid

in the regions (i) |z| < 5 and (ii) |z| > 5.

Solution:

17

Example 9.: Determine the Laurent series for

f(z) =

1

z(z + 5)

valid in the region |z| < 5.

Solution:

We know from the previous example that for |z| < 5 the series expansion of

1/(z + 5) is

1

z + 5

=

∞∑

n=0

(−1)nzn

5n+1

,

It follows from this that we can calculate the series expansion of f(z) since

f(z) =

1

z

· 1

z + 5

=

1

z

∞∑

n=0

(−1)nzn

5n+1

=

∞∑

n=0

(−1)nzn−1

5n+1

Example 10.: Find the Laurent series expansion of

f(z) =

1

z(z + 2)

valid in the region 1 < |z − 1| < 3

Solution:

The region where the function has to be expanded is an open annulus be-

tween circles of radius 1 and 3, centred on z = 1. We want a series expansion

about z = 1; to do this we make the substitution w = z − 1, and look for the

expansion in w, where 1 < |w| < 3. In terms of w the function becomes

f(z) =

1

(w + 1)(w + 3)

In the new variable the domain of interest is still an open annulus of radii 1 and

3, bit they are centred on 0. To make the series expansion easier to calculate

we can manipulate our function f(z) into a similar form to the series expansion

shown in Eq. (29). To do this we will split the function using partial fractions,

and then manipulate each of the fractions into a form based on Eq. (29), so we

obtain

f(z) =

1

2

(

1

w + 1

− 1

w + 3

)

=

1

2

(

1

1− (−w) −

1

3

(

1− (−w3 ))

)

Using the standard and modified geometric series given by Eq. (29), we calculate

that

1

1− (−w) =

∞∑

n=0

(−w)n =

∞∑

n=0

(−1)nwn, |w| < 1

−

∞∑

n=1

1

(−w)n = −

∞∑

n=1

(−1)n

wn |w| > 1

18

and

1

3

(

1− (−w3 )) =

1

3

∞∑

n=0

(−w

3

)n

= 13

∞∑

n=0

(−1)nwn

3n =

∞∑

n=0

(−1)nwn

3n+1 , |w| < 3

− 13

∞∑

n=1

1

(−w3 )

n = − 13

∞∑

n=1

(−3)n

wn = −

∞∑

n=1

(−1)n3n−1

wn , |w| > 3

We require the expansion in w where 1 < |w| < 3, so we use the expansions

for |w| > 1 and |w| < 3, which we can substitute back into our f(z) in partial

fractions to obtain

f(z) =

1

2

(

−

∞∑

n=1

(−1)n

wn

−

∞∑

n=0

(−1)nwn

3n+1

)

= −1

2

( ∞∑

n=1

(−1)n

wn

+

∞∑

n=0

(−1)nwn

3n+1

)

Substituting back in w = z − 1 we obtain the Laurent series, valid within the

region 1 < |z − 1| < 3,

f(z) = −1

2

( ∞∑

n=1

(−1)n

(z − 1)n +

∞∑

n=0

(−1)n(z − 1)n

3n+1

)

Example 11.: Obtain the series expansion of

f(z) =

1

z2 + 4

valid in the region |z − 2i| > 4

19

Key Points:

1. First check to see if you need to make a substitution for the region you

are working with, a substitution is useful if the region is not centred on

z = 0

2. Then you will need to manipulate the function into a form where you can

use the series expansions shown in Eq. (29): this may involve splitting by

partial fractions first.

3. Find the series expansions for each of the fractions you have in your func-

tion within the specified region, then substitute these back into your func-

tion.

4. If part of the function cannot be expanded into Laurent series (not in

the standard form), then use Taylor or binomial series to expand those

functions

5. Finally, simplify the function and, if you made a substitution, change it

back into the original variable

Example 12.: ( to be solved at home) Expand

f(z) =

1

(z + 1)(z + 3)

in a Laurent series valid for (a) 1 < |z| < 3 and (b) 0 < |z + 1| < 2.

*******

Before embarking into the study of more challenging topics of complex analysis,

let us first discuss some preliminaries, more precisely the possible equations of

a straight line and circle It is well known that the most simplistic equation of a

straight line is

ax+ by = c (30)

where x, y are real variables and a, b, c are real numbers. In the case of complex

numbers, this equation changes to

|z − a| = |z − b| (31)

where now a and b are complex numbers. Another way of writing the equation

of a straight line is in the parametric form

z = at+ b, (32)

where the parameter t is real and a and b are complex quantities.

20

Example 14.: A straight line passes through the points A(1, 2) and B(−1, 0).

Write down the three different form of equations of this line.

Solution:

21

xz

R

y

z

0

0

y0

x

ϕ

Figure 1: The connectivity between the two ways to describe the equation of a

circle (cartesian and polar)

A circle in the real plane will have the equation

(x− x0)2 + (y − y0)2 = R2, (33)

which represents a circle of radius R centred on (x0, y0). In the complex plane,

the equation of a circle is written as

|z − z0| = R, z0 = x0 + iy0 (34)

which represents a circle of radius R centred on the complex z0. Another con-

venient form to write the equation of a circle is the polar form (used a lot

throughout complex analysis). The transition between the two ways to describe

the equation of a circle is shown in Fig (1.4). In the new coordinate system the

variables are R and φ. In the Cartesian system a point in the complex plane

is z = x + iy while the centre of the circle is at z0 = x0 + iy0. The connection

between the Cartesian coordinates (x, y) and polar coordinates is made via

x = x0 +R cosφ and y = y0 +R sinφ.

In the complex plane we can write

z = x0 + iy0 +R(cosφ+ i sinφ) = z0 +Re

iφ.

Hence the equation of a circle in the complex plane is

z = z0 +Re

iφ, 0 ≤ φ < 2pi (35)

where the centre is situated at z0 and the circle has a radius R.

Example 15.: A circle passes through the points A(0, 2), B(1, 0) and C(2,−1).

Write down the three different forms of equations of this circle.

22

Solution:

23

2 Complex integration and Cauchy’s Theorem

We can ingrate complex functions along smooth curves in complex plane. It is

similar to integration of real function. Let us consider a complex function f(z)

and a curve C in the complex plane given by z = z(t) with a ≤ t ≤ b.

Newton-Leibnitz formula: Let the equation of C be z = z(t), a ≤ t ≤ b,

then if

f(z) =

dF (z)

dz

, then

∫

C

f(z) dz = F (b)− F (a).

Example 18.: Calculate the integral of f(z) = z along an interval of straight

line z = (2 + i)t, 0 ≤ t ≤ 1.

Solution:

Obviously

z =

dF (z)

dz

=⇒ F (z) = z

2

2

and a = z(0) = 0, b = z(1) = 2 + i, so∫

C

z dz =

z2

2

∣∣∣∣2+i

0

=

1

2

(2 + i)2 =

3 + 4i

2

2.1 Relation between complex and linear integration

If f(z) = u(x, y) + iv(x, y) = u+ iv, and z = x+ iy, the complex line integral∫

C

f(z) dz

can be expressed in terms of real line integrals as∫

C

f(z) dz =

∫

C

(u+ iv)(dx+ idy) =

∫

C

[udx− vdy] + i

∫

C

[vdx+ udy] (36)

A slightly different approach is needed when the curve is given in parametric

form. Let us calculate the same integral as in the previous example but now

using the above relation. In what follows we denote the derivatives with respect

to t by dot, i.e. dx/dt = x˙. In that example we had x = 2t, y = t (after the

identification of real and imaginary parts), so that dx = x˙dt = 2dt, dy = y˙dt =

dt, u = x, v = y and∫

C

zdz =

∫ 1

0

(2xdt− ydt) + i

∫ 1

0

(2ydt+ xdt) = 3

∫ 1

0

t dt+ 4i

∫ 1

0

t dt =

(3 + 4i)

∫ 1

0

t dt = (3 + 4i)

t2

2

∣∣∣∣1

0

=

3 + 4i

2

In general, if the curve C is given in parametric form as x = x(t), y = y(t) with

a ≤ t ≤ b, then Eq. (36) can be written as∫

C

f(z)dz =

∫ b

a

(ux˙− vy˙)dt+ i

∫ b

a

(vx˙+ uy˙)dt (37)

24

Example 17.: Evaluate∫ (2,4)

(0,3)

(2y + x2)dx+ (3x− y)dy

along (a) the parabola x = 2t, y = t2 + 3; (b) straight lines from (0, 3) to (2, 3)

and then from (2, 3) to (2, 4); (c) the straight line from (0, 3) to (2, 4).

Solution:

25

2.2 Cauchy’s theorem

Let C be simple (i.e. without self-intersection) piece-wise smooth closed contour

and f(z) an analytic function defined in the domain D enclosed by the curve C

and continuous at all points in D and on C. Then∮

C

f(z)dz = 0 (38)

Example 18.: Calculate

∮

C

zdz, where C is the circle of radius R centred on

the coordinate origin.

Solution:

The parametric equations of this circle are x = R cosφ and y = R sinφ. Because

the circle is a closed contour, the variable φ covers the range [0, 2pi]. Then

dx = −R sinφ dφ, dy = R cosφ dφ and∮

C

zdz =

∮

C

(x+ iy)(dx+ idy) =

∮

C

(xdx− ydy) + i

∮

C

(ydx+ xdy) =

−

∫ 2pi

0

(R cosφ×R sinφdφ+R sinφ×R cosφ dφ)+i

∫ 2pi

0

(−R sinφ×R sinφ dφ+R cosφ×R cosφ dφ)

= −2R2

∫ 2pi

0

(cosφ sinφ)dφ+iR2

∫ 2pi

0

(− sin2 φ+cos2 φ)dφ = −R2

∫ 2pi

0

sin 2φ dφ+iR2

∫ 2pi

0

cos 2φdφ

=

1

2

R2 cos 2φ

∣∣2pi

0

+

1

2

iR2 sin 2φ|2pi0 =

1

2

R2(cos 4pi−cos 0)+1

2

iR2(sin 4pi−sin 0) = 0

There is a very important consequence of the Cauchy’s theorem: if f(z) is

an analytic function in a domain D and if a and z are any two points in D then∫ z

a

f(z)dz (39)

is independent of the path in D joining a and z.

Example 19.: If the curve C is joining the points (1, 1) and (2, 3), find the

value of ∫

C

(12z2 − 4iz)dz

Solution:

26

2.3 Cauchy integral formula and Taylor series

Let f(z) be analytic inside a domain D and on a simple closed curve C around

this domain. If a is any point inside C, then∮

C

f(z)

z − a dz =

{

2piif(a), a ∈ D

0, a /∈ D

(40)

where C is traversed in the positive (counterclockwise) sense. Similarly, the nth

derivative of f(z) at z = a is given by

f (n)(a) =

n!

2pii

∮

C

f(z)

(z − a)n+1 dz n = 1, 2, 3 . . . (41)

The above two results are called Cauchy’s integral formulae and are quite re-

markable because they show that if a function f(z) is known on the simple

closed curve C, then the value of the function and all its higher derivatives can

be found at all points inside C.

Example 20.: Evaluate

(a)

∮

C

sinpiz2 + cospiz2

(z − 1)(z − 2) dz and (b)

∮

C

e2z

(z + 1)4

dz

where C is the circle |z| = 3

Solution: Since

1

(z − 1)(z − 2) =

1

z − 2 −

1

z − 1

we have∮

C

sinpiz2 + cospiz2

(z − 1)(z − 2) dz =

∮

C

sinpiz2 + cospiz2

z − 2 dz −

∮

C

sinpiz2 + cospiz2

z − 1 dz

By Cauchy’s integral formula with a = 2 and a = 1, respectively we have∮

C

sinpiz2 + cospiz2

z − 2 dz = 2pii[sinpi2

2 + cospi22] = 2pii

∮

C

sinpiz2 + cospiz2

z − 1 dz = 2pii[sinpi1

2 + cospi12] = −2pii

Since z = 1 and z = 2 are inside C and sinpiz2 + cospiz2 is analytic inside C.

Then the required integral has the value 2pii− (−2pii) = 4pii.

b.

27

One of the consequences of Cauchy’s formulae is that if a function of complex

variable has a first derivative, i.e. analytic in a certain domain, all its higher

derivatives exist in that domain. This means that any analytical function can be

expanded in an infinite series in the vicinity of any point z0 inside that domain,

i.e. it can be expanded into Taylor series with

a0 =

1

2pii

∮

C

f(ζ) dζ

ζ − z0 , a1 =

1

2pii

∮

C

f(ζ) dζ

(ζ − z0)2 ,

a2 =

1

2pii

∮

C

f(ζ) dζ

(ζ − z0)3 , . . . , an =

1

2pii

∮

C

f(ζ) dζ

(ζ − z0)n+1 , . . .

So, we eventually obtain that

f(z) = a0 + a1(z − z0) + a2(z − z0)2 + · · ·+ an(z − z0)n + . . . (42)

This expansion is called Taylor series expansion for function f(z). Let us find

expressions for coefficients an. Substituting z = z0 in (42) we obtain that

a0 = f(z0)

Now we differentiate (42) to obtain

f ′(z) = a1 + 2a2(z − z0) + 3a3(z − z0)2 + · · ·+ nan(z − z0)n−1 + . . .

and once again substitute z = z0 to arrive at

a1 = f

′(z0)

Once again differentiate

f ′′(z) = 2a2 + 3 · 2a3(z − z0) + · · ·+ n(n− 1)an(z − z0)n−2 + . . .

and substitute z = z0:

a2 =

1

2

f ′′(z0)

Once again differentiate

f ′′′(z) = 3 · 2a3 + · · ·+ n(n− 1)(n− 2)an(z − z0)n−3 + . . .

and again substitute z = z0:

a3 =

1

3 · 2 f

′′′(z0)

which can be continued as n→∞. After differentiating n times we obtain

f (n)(z) = n(n− 1)(n− 2) . . . 3 · 2an + . . .

28

Substituting z = z0 yields

an =

1

n!

f (n)(z0) (43)

This is the same series as discussed earlier in (24), so the entire discussion

presented there on convergence of series (Taylor and Laurent) should be ap-

plied here, too. The above discussion shows that there is a connection between

Cauchy’s integral formulae and Taylor series. Now the notion of singularity of

analytical function will have a slightly different connotation and will be essential

in future discussion.

If the Laurent series of a function contains at least one term with negative

power of (z − z0), then the point z0 is called isolated singularity of function

f(z). Isolated singularities are divided in two types. If the Laurent series of

f(z) contains infinitely many terms with negative powers of (z − z0), then z0

is called essential singularity. If the series contains only finite number of terms

with negative powers of (z − z0), then z0 is called a pole. When the Laurent

series expansion has the form

f(z) =

a−m

(z − z0)m + . . .

a−2

(z − z0)2 +

a−1

z − z0 +

+ a0 + a1(z − z0) + a2(z − z0)2 + · · ·+ an(z − z0)n + . . .

i.e. it starts from a term proportional to (z − z0)−m, the pole z0 is called the

mth order pole. In the particular case when m = 1 it is called a simple pole.

Example 21.: Find the Laurent series expansions of the following functions

about the singularity z0 = 0 and determine the type of this singularity.

(i) f(z) =

z2 + 2

z(z2 + 1)

; (ii) f(z) = e1/z

Solution:

(i) We decompose f(z) in partial fractions:

z2 + 2

z(z2 + 1)

=

A

z

+

Bz + C

z2 + 1

=⇒ z2 + 2 = A(z2 + 1) + (Bz + C)z

Take z = 0 =⇒ A = 2. Then we have

z2 = (2 +B)z2 + Cz =⇒ B = −1, C = 0

so finally

z2 + 2

z(z2 + 1)

=

2

z

− z

z2 + 1

=

2

z

− z

1− (−z2) =

2

z

− z[1 + (−z2) + (−z2)2 + · · ·+ (−z2)n + . . . ]

=

2

z

− z + z3 − z5 + · · ·+ (−1)n+1z2n+1 + . . .

29

We see that z = 0 is simple pole.

(ii)

2.4 Residue Theorem

Let z = z0 be a singularity of the function f(z) and (28) its expansion in Laurent

series. Then the coefficient a−1 is called the residue of the function f(z) at the

singularity z0 and we denote

a−1 = res

z=z0

f(z)

Let the simple closed contour C be the boundary of domain D, and f(z) analytic

inside D except isolated singularities z1, z2, . . . , zn, and continuous up to the

domain boundary C. Then∮

C

f(ζ) dζ = 2pii

(

res

z=z1

f(z) + res

z=z2

f(z) + · · ·+ res

z=zn

f(z)

)

(44)

To obtain the residue of a function f(z) at z = z0, it may appear that first the

Laurent expansion of f(z) must be obtained. However, in the case where z = z0

is a pole of order k, there is a simple formula for a−1 given by

a−1 = lim

z→z0

1

(k − 1)!

dk−1

dzk−1

[

(z − z0)kf(z)

]

If k = 1 (z0 is a simple pole) the result is especially simple and is given by

a−1 = lim

z→z0

(z − z0)f(z)

Example 22.: If

f(z) =

z

(z − 1)(z + 1)2

then z = 1 and z = −1 are poles of order one and two, respectively. Therefore

using the standard formula we have:

30

For the residue at z = 1

lim

z→1

(z − 1)

[

z

(z − 1)(z + 1)2

]

=

1

4

For the residue at z = −1

lim

z→−1

1

1!

d

dz

[

(z + 1)2

(

z

(z − 1)(z + 1)2

)]

= −1

4

Example 23.: Evaluate the integral

∮

C

f(z) dz, where C is the circle of radius

1/2 centred at the origin, and f(z) is given by

(i) f(z) =

z2 + 2

z(z2 + 1)

; (ii) f(z) = e1/z

Solution:

(i) This function has three singularities: 0 and ±i, but only 0 is inside the

circle. In previous example we found the Laurent expansion of this function

about 0. In particular, we found that a−1 = 2. Hence∮

z2 + 2

z(z2 + 1)

dz = 2pii res

z=0

f(z) = 4pii

(ii) This function has only one singularity at z = 0. In the previous example we

found the Laurent expansion of this function about 0. In particular, we found

that a−1 = 1. Hence ∮

e1/z dz = 2pii res

z=0

f(z) = 2pii

Example 24.: Evaluate ∮

C

ezt

z2(z2 + 2z + 2)

dz

around the circle C with equation |z| = 3.

31

2.5 Using the residue theorem for calculating real inte-

grals

The residue theorem can be used to calculate real integrals. We consider this

topic using examples.

Example 25.: Calculate ∫ ∞

−∞

dx

x2 − 6x+ 10 .

Solution

3+j

−b b

C

y

x

Figure 2: Representation for Example 33

Consider contour Γ that consists of half-circle C of radius b centred at the

origin and the interval on the real axis [−b, b], i.e. Γ = C ∪ [−b, b]. The roots of

x2 − 6x+ 10 are

x1,2 = 3±

√

9− 10 = 3±√−1 = 3± i

32

Let us consider the complex function 1/(z2 − 6z + 10). This function has two

simple poles at z = 3± i. However, only 3 + i is inside contour Γ. Hence,∮

Γ

dz

z2 − 6z + 10 = 2pii resz=3+i

1

z2 − 6z + 10 = 2pii limz→3+i

z − 3− i

(z − 3− i)(z − 3 + i)

= 2pii lim

z→3+i

1

z − 3 + i =

2pii

3 + i− 3 + i =

2pii

2i

= pi

On the other hand,∮

Γ

dz

z2 − 6z + 10 =

∫ b

−b

dx

x2 − 6x+ 10 +

∫

C

dz

z2 − 6z + 10

So that ∫ b

−b

dx

x2 − 6x+ 10 +

∫

C

dz

z2 − 6z + 10 = pi

Let us take the limit b→∞. In the second integral we take z = beiθ, so∫

C

dz

z2 − 6z + 10 =

∫ pi

0

ibeiθ dθ

b2e2iθ − 6beiθ + 10 → 0 as b→∞∫ b

−b

dx

x2 − 6x+ 10 →

∫ ∞

−∞

dx

x2 − 6x+ 10 as b→∞

As a result we obtain ∫ ∞

−∞

dx

x2 − 6x+ 10 = pi

Of course, this integral can be easily calculated by the traditional method:∫ ∞

−∞

dx

x2 − 6x+ 10 =

∫ ∞

−∞

dx

1 + (x− 3)2 =

∫ ∞

−∞

du

1 + u2

= tan−1 u

∣∣∣∞

−∞

=

pi

2

−

(

−pi

2

)

= pi

Example 26.: Calculate ∫ ∞

−∞

cosx

x2 + 1

dx.

Solution

33

3 Advanced vector calculus: double operators,

polar coordinates

In many applications, we do not consider individual vectors or scalars, but func-

tions that give a vector or scalar at every point. Such functions are called vector

fields or scalar fields. For example:

(a) Suppose we want to model the flow of air around an aeroplane. The velocity

of the air flow at any given point is a vector. These vectors will be different at

different points, so they are functions of position (and also of time). Thus, the

air velocity is a vector field. Similarly, the pressure and temperature are scalar

quantities that depend on position, or in other words, they are scalar fields.

(b) The magnetic field inside an electrical machine is a vector that depends on

position, or in other words a vector field. The electric potential is a scalar field.

34

Although we will mainly be concerned with scalar and vector fields in three-

dimensional space, we will sometimes use two-dimensional examples because

they are easier to visualise.

3.1 The gradient of a scalar field

If f is a scalar field, then we define ∇(f) = (fx, fy, fz) =

(

∂f

∂x ,

∂f

∂y ,

∂f

∂z

)

. The

result of this operator is a vector field. Often the gradient of f is written as

grad(f) rather than ∇(f).

Example 27.:

(a) for the function f = x3 + y4 + z5, we have ∇(f) = (3x2, 4y3, 5z4).

(b) for the function f = sinx sin y sin z we have

∇(f) = (cosx sin y sin z, sinx cos y sin z, sinx sin y cos z)

(c) for the function r =

√

x2 + y2 + z2 we have

rx =

1

2

2x√

x2 + y2 + z2

=

x√

x2 + y2 + z2

=

x

r

and similarly ry = y/r and rz = z/r. This means that

∇(r) =

(x

r

,

y

r

,

z

r

)

(a) We write E for the electric field (which is a vector field) and φ for the

electric potential (which is a scalar field). These are related by the equation

E = ∇(φ). (All this is valid only when there are no significant time-varying

magnetic fields.)

(b) Similarly, there is a gravitational potential function, Ψ, and the gravita-

tional force field is proportional to ∇(Ψ).

(c) The net force on a particle of air involves ∇(p), where p is the pressure.

If we have a single charge at the origin, then the resulting electric potential

function is φ = Ar−1 for some constant A, where r = (x2 +y2 +z2)1/2, as usual.

Note that

(r−1)x = −1

2

2x

(x2 + y2 + z2)3/2

= − x

r3

and similarly (r−1)y = −y/r3 and (r−1)z = −z/r3. This gives the electric field

as

E = ∇(φ) = −A (x, y, z)

r3

= −A r

r3

3.2 The div and curl operators

Now suppose we have a vector field u = (f, g, h), so that f , g, and h are all

functions of the coordinates. We can think of ∇ as itself being a strange kind

35

of vector, in which the components are differential operators

∇ =

(

∂

∂x

,

∂

∂y

,

∂

∂z

)

.

This means that in the case of the div and curl operators we can make use of

the dot and cross operators as follows:

∇ · u =

(

∂

∂x

,

∂

∂y

,

∂

∂z

)

· (f, g, h) = ∂f

∂x

+

∂g

∂y

+

∂h

∂z

= fx + gy + hz

∇× u =

∣∣∣∣∣∣

i j k

∂

∂x

∂

∂y

∂

∂z

f g h

∣∣∣∣∣∣ = (hy − gz, fz − hx, gx − fy),

Let u and v be vector fields, let f be a scalar field, and let p be a function of

one variable. Then:

∇(f + g) = ∇(f) +∇(g), ∇(fg) = f∇(g) + g∇(f)

∇ · (u + v) = ∇ · u +∇ · v, ∇ · (fu) = f∇ · (u) +∇(f) · u

∇× (u + v) = ∇× u +∇× v, ∇× (fu) = f∇× u +∇(f)× u

∇(p(f)) = p′(f)∇(f), ∇ · (u× v) = v · (∇× u)− u · (∇× v)

3.3 Second-order operators

There are several ways to combine the three operators

scalar field ∗ grad→ vector field ∗ div → scalar field

scalar field ∗ grad→ vector field ∗ curl→ vector field

vector field ∗ div → scalar field ∗ grad→ vector field

vector field ∗ curl→ vector field ∗ div → scalar field

vector field ∗ curl→ vector field ∗ curl→ vector field

No other combinations make sense. For example we cannot define curl(div(u)),

because div(u) is a scalar field, and we can only take the curl of a vector field.

It is important that two of the above combinations are automatically zero, e.g.

• For any scalar field f we have curl(grad(f)) = ∇× (∇(f)) = 0

• For any vector field u we have div(curl(u)) = ∇ · (∇× u) = 0

36

These can be checked directly. For a scalar field f , we have ∇(f) = (fx, fy, fz).

Taking into account that fxy = fyx and so on we find that

∇× (∇(f)) =

∣∣∣∣∣∣

i j k

∂

∂x

∂

∂y

∂

∂z

fx fy fz

∣∣∣∣∣∣ = (fzy − fyz, fxz − fzx, fyx − fxy) = (0, 0, 0)

Now consider instead a vector field u = (p, q, r) We have

∇× u =

∣∣∣∣∣∣

i j k

∂

∂x

∂

∂y

∂

∂z

p q r

∣∣∣∣∣∣ = (ry − qz, pz − rx, qx − py)

so

∇·(∇×u) = (ry−qz)x+(pz−rx)y+(qx−py)z = ryx−qzx+pzy−rxy+qxz−pyz

= pzy − pyz + qxz − qzx + ryx − rxy = 0

There are three more possible combinations

• For a scalar field f we have div(grad(f)) = ∇ · (∇(f)) = fxx + fyy + fzz.

This is usually written as ∇2(f), and called the Laplacian of f . Note

that the Laplacian of a scalar field is a scalar field. We can also define the

Laplacian of a vector field by the rule

∇2(p, q, r) = (∇2(p),∇2(q),∇2(r)) = (pxx+pyy+pzz, qxx+qyy+qzz, rxx+ryy+rzz)

with the Laplacian of a vector field being again a vector field

• For a vector field u = (p, q, r) we have

grad(div(u)) = ∇(∇ · u) = ∇(px + qy + rz) =

= (pxx + qyx + rzx, pxy + qyy + rzy, pxz + qyz + rzz)

• The last remaining combination can be expressed in terms of the above

two by the equation

curl(curl(u)) = ∇× (∇× u) = ∇(∇ · u)−∇2(u)

In two dimensions the situation is similar but simpler.

• For any scalar field f we have

div(grad(f)) = ∇ · (∇(f)) = fxx + fyy

which is again called the Laplacian and denoted by ∇2(f)

• We also have

curl(grad(f)) = curl(fx, fy) = fyx − fxy = 0

37

A vector field u is it incompressible (or solenoidal) if div(u) = 0, and that it is

irrotational (or conservative) if curl(u) = 0.

- For any scalar field f (in two or three dimensions) we have a vector field

∇(f) = grad(f). The rule curl(grad(f)) = 0 tells us that grad(f) is irrota-

tional

- For any vector field v in three dimensions we have another vector field curl(v).

The rule div(curl(v)) = ∇ · (∇× v) = 0 tells us that curl(v) is incompressible.

Example 28: For the two-dimensional vector field u = (x2 − y2 + 2xy, x2 −

y2 − 2xy) calculate div(u) and curl(u).

Solution:

38

3.4 Potential functions

If u is an irrotational vector field, a potential function for u is a scalar field , p

such that∇(p) = u. (Because curl(grad(p)) = 0, only irrotational field can have

a potential.) Potential functions always exist (but they may be multi-valued),

and it is often useful to find them.

Consider the vector field u = (y + z, z + x, x+ y). This has

∇× u =

∣∣∣∣∣∣

i j k

∂

∂x

∂

∂y

∂

∂z

y + z z + x x+ y

∣∣∣∣∣∣ = (1− 1, 1− 1, 1− 1) = 0

so it is irrotational. It therefore makes sense to look for a potential function, or

in other words a function p(x, y, z) with (px, py, pz) = (y + z, z + x, x + y). As

we want px = y + z, we must have

p =

∫

(y + z)dx = xy + xz + C1 = xy + xz + q(x, y)

where the constant C1 is independent of x, i.e. q(y, z). We thus have qx = 0,

and the equation py = z + x becomes x+ qy = z + x, or in other words qy = z.

Integrating this we get

q =

∫

z dy = yz + C2 = yz + r(z)

where now the constant C2 is independent of x and y, say C2 = r(z). We now

have p = xy + xz + q = xy + xz + yz + r, so the equation pz = x+ y becomes

x + y + rz = x + y, so rz = 0. As r can only depend on z and we have rz = 0

we see that r is a genuine constant. We can choose it to be zero, and we find

that the function p = xy + xz + yz is a potential function for u.

Now consider the vector field u = (0, 0, x2). This has

∇× u =

∣∣∣∣∣∣

i j k

∂

∂x

∂

∂y

∂

∂z

0 0 x2

∣∣∣∣∣∣ = (0,−2x, 0) 6= 0

so it is not irrotational, so it cannot have a potential function. We will nonethe-

less try to find one, and see what goes wrong. A potential function p would

have to have (px, py, pz) = (0, 0, x

2). As px = py = 0, we see that p can only

depend on z. That means that the derivative pz also depends only on z, so we

cannot have pz = x

2. Thus, there is no potential function.

39

4 Line integrals, surface integrals, Stokes’ theo-

rem, volume integrals, divergence theorem

Engineering applications often have to deal with curves in 3-dimensional space.

For example

• A wire in an electrical machine is a curve. To calculate the magnetic field

created by a current in the wire, or the force exerted on the wire by an

externally applied magnetic field, we need equations for the curve.

• The path of a moving particle over time defines a curve. If the particle

is charged then it will feel a force from any electric or magnetic fields;

to understand the effect of this, we need various equations relating the

position, velocity, force and acceleration to the fields.

We can describe a curve by giving the x, y and z coordinates (or equivalently,

the position vector r = (x, y, z)) in terms of another parameter t. (In the case

of a moving particle we often take t to be time, but that is not compulsory.)

For example the equation

r = (x, y, z) = (at, b cos(t), b sin(t))

describes a helix winding around the x-axis.

This is the path followed by an electron moving in a uniform magnetic field, but

it could also describe a wire wound round a cylinder.

4.1 Integration along curves

To integrate along a curve C, we divide C into many small pieces, each running

from some position r to a nearby position r + δr. Each such piece will give a

contribution to the integral, and we add up the contributions to get an approx-

imation to the required value. For the exact value, we pass to the limit where

the length of the small pieces tends to zero.

• The length of the curve is approximately the sum of the lengths |δr| over

all the small pieces. The exact length is denoted by

∫

C

|dr|.

• If a particle moves along a curve C through a force field F, then the work

done against the force is − ∫

C

F · dr.

40

For the last integral it makes a difference which direction we follow when travers-

ing the curve: the answer we get when traversing the curve backwards will be

the negative of the answer we get when traversing the curve in the forward

direction.

In practice, we calculate these integrals as follows. We parametrise the curve

as r = (x(t), y(t), z(t)) for some range of values of t (say a ≤ t ≤ b), and we

write x˙ = dx/dt and so on. We then have

dr =

dr

dt

dt = r˙dt = (x˙ dt, y˙ dt, z˙ dt)

|dr| =

√

x˙2 + y˙2 + z˙2 dt,

so

length(C) =

∫

C

|dr| =

∫ b

t=a

√

x˙2 + y˙2 + z˙2 dt

work =

∫

C

F · dr =

∫ b

t=a

F · r˙ dt

and so on.

Example 30.: Let C be the curve given by

r = (x, y, z) = (6t, 3

√

2t2, 2t3)

for 0 ≤ t ≤ 1. Calculate the length of this curve.

Solution: According to the definition we have

dr = (6, 6

√

2t, 6t2) dt =⇒ |dr| =

√

36 + 72t2 + 36t4 dt = 6

√

1 + 2t2 + t4 dt = 6(1+t2) dt,

so

length =

∫

C

|dr| =

∫ 1

t=0

6(1 + t2) dt =

[

6t+ 2t3

]1

t=0

= 8.

Example 31.: Consider a particle moving along a path r = (x, y, z) =

(t, 0, t/2) (for 0 ≤ t ≤ 1) against a force field F = (y2 + z2 − 1, 0, 0) (This could

reasonably model the wind force in a wind tunnel of radius one centred on the

x-axis). Find the work done against the force.

Solution:

41

If f is a function of one variable, then

∫ b

x=a

f ′(x) dx = f(b) − f(a) in the

virtue of the Fundamental Theorem of Calculus. For any curve C from a to b,

and any scalar field p, we have∫

C

∇(p) · dr = p(b)− p(a).

Let us suppose we have a curve C from a to b, and we want to calculate the

integral I =

∫

C

F · dr for some vector field F. Suppose that F is conservative

(i.e. curl(F) = 0). We can then find a potential function p with ∇(p) = F and

it will follow that

∫

C

F · dr = p(b)− p(a).

Note that in this method, we do not need to know anything about C except

where it starts and ends. This often makes calculations much easier.

If we have trouble finding a potential function, it may be better to use the

following approach: Suppose we have a curve C from a to b, and we want to

calculate the integral I =

∫

C

F · dr for some vector field F. Suppose that F is

conservative. We can then find a different curve C ′ from a to b for which the

calculation is easier, and then I will be equal to

∫

C′ F · dr. The reason why this

method works is that both

∫

C

F · dr and ∫

C′ F · dr are equal to p(b) − p(a),

where p is the potential function. For this to be valid, we need to know that p

exists (so we must check that F is conservative) but we do not actually need to

find p.

Example 32.: Let C be the helical path given by r = (t, cos(10pit), sin(10pit))

for 0 ≤ t ≤ 1, which runs from a = (0, 1, 0) to b = (1, 1, 0). Let F be the vector

field (yz, xz, xy). Calculate

∫

C

F · dr.

Solution:

42

Now let us discuss a case where the vector space is not conservative. Now

for the same helical path as before (see above) consider the vector field G =

(0,−z, y).

43

Let F be an conservative vector field. We can then define a potential function

p for F by the rule

p(a, b, c) = the integral

∫

C

F · dr, for any curve C from (0, 0, 0) to (a, b, c) .

The answer will not depend on the choice of curve, so we can choose whichever

curve makes the integral easiest. A straight line is often good, but sometimes

a broken line (from (0, 0, 0) to (a, 0, 0) to (a, b, 0) to (a, b, c), for example) is

better. Note again that this is only valid for conservative fields. Fields that are

not conservative do not have a potential function.

Example 33. The vector field F = (yz, xz, xy) was earlier shown to be conser-

vative, i.e. curl(F) = 0. Find a potential function p that satisfies grad(p) = F.

Solution: To find p(a, b, c), we evaluate

∫

L

F · dr, where L is the straight line

from (0, 0, 0) to (a, b, c). This line can be parametrised by r = (x, y, z) =

(ta, tb, tc) for 0 ≤ t ≤ 1, which gives

dr = (a, b, c)dt

F = ((tb)(tc), (ta)(tc), (ta)(tb)) = (t2bc, t2ac, t2ab)

F · dr = 3t2abc dt =⇒ p(a, b, c) =

∫

L

F · dr =

∫ 1

t=0

3t2abc dt =

[

t3abc

]1

t=0

= abc.

It is convenient to write this calculation in terms of a, b and c, to avoid confusion

between the end of the path (where (x, y, z) = (a, b, c)) and the points along the

path (where (x, y, z) = (ta, tb, tc)). However, we can restate the final answer as

p(x, y, z) = xyz, which is more convenient for later use.

Let us consider a vector field F and a curve C, with the vector dr pointing

along the curve. We can construct another vector dn of the same length per-

pendicular to dr. The integral

∫

C

F · dr measures the extent to which F points

along the curve. For some purposes, however, we want to measure the flow of

F across the curve, in which case we want to evaluate

∫

C

F · dn rather than∫

C

F · dr.

Note that dr = (dx, dy) = (x˙, y˙)dt, and dn is obtained by rotating this a

quarter turn clockwise, so dn = (dy,−dx) = (y˙,−x˙)dt (after all the two vectors

are perpendicular).

Example 34.: Let L be the straight line from (1, 0) to (0, 1), so r = (x, y) =

(1 − t, t) for 0 ≤ t ≤ 1. If F be the vector field (x2 − y2, 2xy), find the flux of

the vector F through the line L.

Solution: Since dr = (−1, 1)dt, it means that dn = (1, 1)dt. On L we have

F = ((1− t)2 − t2, 2t(1− t)) = (1− 2t+ t2 − t2, 2t− 2t2) = (1− 2t, 2t− 2t2),

44

so

F · dn = ((1− 2t) + (2t− 2t2))dt = (1− 2t2)dt

so ∫

C

F · dn =

∫ 1

t=0

(1− 2t2) dt =

[

t− 23 t3

]1

t=0

= 1− 2

3

=

1

3

Example 35.: Let us calculate the flow of the field

F = (x+ 2y, 3x+ 4y)

out of the unit circle C.

Solution:

4.2 Surface Integrals

As well as considering curved paths, we also need to consider curved sur-

faces in three-dimensional space. Such a surface can be parametrised as r =

(x(s, t), y(s, t), z(s, t)) for some pair of parameters s and t.

As typical surface we can talk about is a hemisphere, i.e. the surface that

covers half of a sphere. For instance the upper half of a spherical shell of radius

2 can be described in terms of parameters φ and θ by

(x, y, z) = (2 sin(φ) cos(θ), 2 sin(φ) sin(θ), 2 cos(φ))

45

(for 0 ≤ θ ≤ 2pi and 0 ≤ φ ≤ pi/2).

Another example is the off-centre cylinder. Let S be a cylindrical surface of ra-

dius 1, centred on the line joining (1, 1,−1) to (1, 1, 1). Then S can be described

in terms of parameters s and t by

(x, y, z) = (1 + cos(s), 1 + sin(s), t)

(for 0 ≤ s ≤ 2pi and −1 ≤ t ≤ 1).

46

For any function f(x, y), the equation z = f(x, y) defines a surface.

We can use the variables x and y themselves as parameters, and then the full

parametrisation is

(x, y, z) = (x, y, f(x, y)).

To integrate over S, we need a formula for the area of a small piece of S

in terms of a parametrisation r = (x(s, t), y(s, t), z(s, t)). If s and t vary by δs

and δt, then the corresponding part of the surface will be a small parallelogram

spanned by the vectors rs δs = (xsδs, ys δs) and rt δt = (xtδt, yt δt).

We write δA for the area of this parallelogram. We also write δA for the vector

(rs × rt)δs δt. This is perpendicular to rs and rt which means that it is normal

to the surface), and |δA| = δA. In the limit we get dA = (rs × rt)ds dt and

dA = |dA| = |rs × rt|ds dt. Also dA = n dA, where n is the unit normal to S.

Consider again a hemispherical shell of radius a. We have

r = (a sin(φ) cos(θ), a sin(φ) sin(θ), a cos(φ))

47

rφ = (a cos(φ) cos(θ), a cos(φ) sin(θ), −a sin(φ))

rθ = (−a sin(φ) sin(θ), a sin(φ) cos(θ), 0)

rφ × rθ =

∣∣∣∣∣∣

i j k

a cos(φ) cos(θ) a cos(φ) sin(θ) −a sin(φ)

−a sin(φ) sin(θ) a sin(φ) cos(θ) 0

∣∣∣∣∣∣ =

= (a2 sin2(φ) cos(θ), a2 sin2(φ) sin(θ), a2 sin(φ) cos(φ)) = a2 sin(φ)er

dA = a2 sin(φ)er dφ dθ, dA = |dA| = a2 sin(φ)dθ dφ.

It follows that the area of the surface is

A =

∫∫

S

1 dA =

∫ 2pi

θ=0

∫ pi

2

φ=0

a2 sin(φ)dθ dφ =

= 2a2pi

∫ pi

2

φ=0

sin(φ) dφ = 2a2pi

[

− cos(φ)

]pi

2

φ=0

= 2a2pi.

Now let us consider the case of a cylinder. In this case we have

r = (1 + cos(s), 1 + sin(s), t) (0 ≤ s ≤ 2pi, −1 ≤ t ≤ 1)

rs = (− sin(s), cos(s), 0), rt = (0, 0, 1)

rs × rt =

∣∣∣∣∣∣

i j k

− sin(s) cos(s) 0

0 0 1

∣∣∣∣∣∣ = (cos(s), sin(s), 0)

dA = (cos(s), sin(s), 0) ds dt

|rs × rt| = |(cos(s), sin(s), 0)| =

√

cos2(s) + sin2(s) = 1

dA = |rs × rt| ds dt = ds dt.

It follows that the area of the surface is∫∫

S

1 dA =

∫ 2pi

s=0

∫ 1

t=−1

1 ds dt = 2pi(1− (−1)) = 4pi.

For a general surface given by z = f(x, y) we have

r = (x, y, f(x, y)), rx = (1, 0, fx), ry = (0, 1, fy)

rx × ry =

∣∣∣∣∣∣

i j k

1 0 fx

0 1 fy

∣∣∣∣∣∣ = (−fx,−fy, 1)

dA = (rx × ry) dx dy = (−fx,−fy, 1) dx dy

|rx × ry| =

√

f2x + f

2

y + 1

48

and

dA =

√

f2x + f

2

y + 1 dx dy

Now let us apply these relations to a particular case when we consider that

z = f(x, y) = cosh(x+ y)/

√

2 for 0 ≤ x, y ≤ 1. The task is to calculate the area

of this surface for the values of x and y given earlier. It is easy to see that

fx = sinh(x+ y)/

√

2 fy = sinh(x+ y)/

√

2

So now we can apply the formulae just derived to obtain√

1 + f2x + f

2

y =

√

1 +

1

2

sinh2(x+ y) +

1

2

sinh2(x+ y) =

√

1 + sinh2(x+ y) =

=

√

cosh2(x+ y) = cosh(x+ y)

With this relation we can calculate the area element as

dA =

√

1 + f2x + f

2

y dx dy = cosh(x+ y) dx dy.

It follows that the area of the surface is

A =

∫∫

S

1 dA =

∫ 1

x=0

∫ 1

y=0

cosh(x+ y)dy dx =

∫ 1

x=0

[

sinh(x+ y)

]1

y=0

dx =

=

∫ 1

x=0

sinh(x+ 1)− sinh(x) dx =

[

cosh(x+ 1)− cosh(x)

]1

x=0

= (cosh(2)− cosh(1))− (cosh(1)− cosh(0)) = cosh(2)− 2 cosh(1) + 1 ' 1.676

Let us now look into the problem of the flow across a surface. Consider a

surface S in a region where there is a vector field F.

49

We want to calculate the flow of F across S. At each point on S there is a

normal vector n. The flow across S only involves the component of F in the

normal direction, or in other words F · n. We need to integrate this with re-

spect to area, giving

∫∫

S

F · ndA. However, ndA is the same as the vector dA

considered earlier, which can be calculated from a parametrisation by the rule

dA = (rs × rt)ds dt.

Example 36.: Let S be the surface given by z = f(x, y) = xy for 0 ≤ x, y ≤ 1,

and let F be the vector field (x+ y+ z, x+ y+ z, x+ y+ z). Calculate the flow

of the vector field through the surface S.

Solution

Now let us concentrate on the surface integration. This topic has a special

role in electromagnetism since it is closely connected to the Maxwell’s equations.

• For any three-dimensional region, the total electric field crossing the bound-

ary of the region is −10 times the total charge in the region

50

• On the other hand, the magnetic field crossing the boundary always can-

cels out to give a total of zero

• Now suppose we have a surface S in three-dimensional space. Suppose that

has a boundary that is a closed curve C (so the surface could be a disk or

a hemispherical bowl, but not a complete sphere). Then the circulation of

E around C is minus the rate of change of the total magnetic field passing

through S.

• Similarly, the circulation of B around C is µ0 times the rate of change

of the current passing through S (including the ’”displacement current”

0E˙).

Maxwell’s equations told us about the values of scalar and vector fields and their

derivatives at every point in space.The above statements are about various kinds

of integrals of such scalar and vector fields over curves, surfaces and three-

dimensional regions. The main point of this final section of the course is to

understand why these integral statements are the same as the earlier differential

statements.

Let D be a region in the plane. The edge of the region will be a curve,

which we call C. For any vector field u, we can consider the integral

∫

C

u · dn

measuring the flux of u across C.

4.3 The two-dimensional divergence theorem

Let D be a region in the plane whose boundary is a closed curve C. The two-

dimensional divergence theorem says that for any vector field u, we have∫∫

D

div(u) dA =

∫

C

u · dn.

Functions like 1/(x2 + y2) (which blows up to infinity at the origin) are allowed

if the origin lies outside D, but disallowed if the origin is inside D.

Let us consider the above divergence theorem, i.e.∫∫

D

div(u) dA =

∫

C

u · dn.

where C is the boundary of D and the contour is traversed in the anticlockwise

direction. Then let u be (p, q). We then have

div(u) = px + qy

so

u · dn = (p, q) · (dy,−dx) = p dy − q dx.

Therefore ∫∫

D

(px + qy) dA =

∫

C

(p dy − q dx).

51

Example 37. Let D be the disc where x2 + y2 ≤ m2, so C is a circle of radius

m. Take u = (ax + by, cx + dy) for some constants a, b, c and d. Prove the

divergence theorem.

Solution:

Given the expression of u we have

div(u) = (ax+ by)x + (cx+ dy)y = a+ d,

so ∫∫

D

div(u)dA = (a+ d)area(D) = pim2(a+ d)

On the other hand, we can parametrise C by r = (x, y) = (m cos(t),m sin(t)),

so dn = (y˙,−x˙)dt = (m cos(t),m sin(t)) dt. On C we also have

u = (ax+ by, cx+ dy) = (am cos(t) + bm sin(t), cm cos(t) + dm sin(t))

so

u·dn = (am cos(t)+bm sin(t))(m cos(t))dt+(cm cos(t)+dm sin(t))(m sin(t))dt =

= m2(a cos2(t)+(b+c) sin(t) cos(t)+d sin2(t))dt =

m2

2

(a+a cos(2t)+(b+c) sin(2t)+d−d cos(2t)) =

=

m2

2

((a+ d) + (a− d) cos(2t) + (b+ c) sin(2t))

As a result we have∫

C

u · dn = m

2

2

[

(a+ d)t+

1

2

(a− d) sin(2t)− 1

2

(b+ c) cos(2t)

]2pi

t=0

=

=

m2

2

2pi(a+ d) = pim2(a+ d).

Example 38. Let D be the rectangle as shown below. The boundary of the

domain consists of C1, . . . , C4. If u is the horizontal vector field u = (e

−x−y, 0),

prove the divergence theorem.

Solution:

52

Example 39.: The figure shows the deltoid curve C The curve C can be written

Figure 3: The deltoid curve used for Example 39

in parametric form as

x = 2 cos(t) + cos(2t) y = 2 sin(t)− sin(2t).

Calculate the area of this object.

Solution: Obviously it is hard to find the area of D directly. However, we can

evaluate it by a trick using the divergence theorem. Consider the vector field

53

F = (x, 0), so

div(F) =

∂x

∂x

+

∂0

∂y

= 1

As a consequence we have that

∫∫

D

div(F)dA = area(D).

The Divergence Theorem tells us that this is the same as

∫

C

F · dn, where

dn = (y˙,−x˙) dt = (2 cos(t)− 2 cos(2t), 2 sin(t) + 2 sin(2t)) dt

Furthermore

F = (x, 0) = (2 cos(t) + cos(2t), 0)

so

F·dn = (2 cos(t)−2 cos(2t))(2 cos(t)+cos(2t)) = 4 cos2(t)−2 cos(t) cos(2t)−2 cos2(2t) =

= (2+2 cos(2t))−(cos(3t)+cos(t))−(1+cos(4t)) = 1−cos(t)+2 cos(2t)−cos(3t)−cos(4t)

Therefore

area =

∫ 2pi

t=0

F · dn = 2pi

4.4 Green’s theorem

Let D be a region in the plane whose boundary is a closed curve C. Green’s

theorem says that for any vector field u that is well-behaved (i.e. is not discon-

tinuous) everywhere in D, we have∫∫

D

curl(u) · dA =

∫

C

u · dr.

Since D is plane dA = ndA, where n is unit normal vector to D. Accordingly,

if u = (P,Q), then∫∫

D

curl(u) · dA =

∫∫

D

(

∂Q

∂x

− ∂P

∂y

)

dA

Example 40.: Let D be the unit disc, so the boundary curve C is the unit

circle. Let u be the vector field (x3, x3). Verify the Green’s Theorem.

Solution:

54

4.5 The (three-dimensional) Divergence Theorem

The two dimensional knowledge can be expanded to study three-dimensional

cases. Let E be the three-dimensional solid region enclosed by a surface S. Let

u be a vector field that is well-behaved everywhere in E. Then∫∫∫

E

div(u) dV =

∫∫

S

u · dA

This can be proved by an argument similar to that used for the two-dimensional

version. The physical interpretation is also similar: in a steady state, the rate

of flow of particles escaping through S must balance the rate of creation of par-

ticles in E.

Example 41.: Let S be the unit sphere, and let E be the solid ball enclosed

by S. Consider the vector field u = (x, 0, 0). Verify the divergence theorem in

3D.

Solution: The given vector has div(u) = ∂x/∂x+ ∂0/∂y + ∂0/∂z = 1, so∫∫∫

E

div(u)dV =

∫∫∫

E

dV = volume of E = 4pi/3.

On S we have

r = (x, y, z) = (sin(φ) cos(θ), sin(φ) sin(θ), cos(φ))

55

and

u = (x, 0, 0) = (sin(φ) cos(θ), 0, 0).

The unit normal vector is n = er = r, so u · n = x2 = sin2(φ) cos2(θ). We have

also seen before that dA = sin(φ) dφ dθ, so∫∫

S

u · dA =

∫∫

S

u · n dA =

∫ 2pi

θ=0

∫ pi

φ=0

sin3(φ) cos2(θ) dφ dθ =

=

(∫ 2pi

θ=0

cos2(θ) dθ

)(∫ pi

φ=0

sin3(φ) dφ

)

= pi

∫ pi

φ=0

1

4

(3 sin(φ)− sin(3φ))dφ =

=

pi

4

[

− 3 cos(φ) + 1

3

cos(3φ)

]pi

φ=0

=

pi

4

((3− 1

3

)− (−3 + 1

3

)) =

4pi

3

As expected, this is the same as

∫∫∫

E

div(u)dV .

Example 42.: Let E be the solid vertical cylinder of radius a and height 2b

centred at the origin, and let S be the surface of E. Consider the vector field

u = (−y, x, z3). Verify the divergence theorem in 3D.

Solution:

56

4.6 Stokes’ Theorem

Stokes’ Theorem is analogous to Green’s Theorem, but it applies to curved

surfaces as well as to flat regions in the plane. Suppose we have a surface S

whose boundary is a closed curve C, and a well-behaved vector field u. Then∫∫

S

curl(u) · dA = ±

∫

C

u · dr.

We need a little more discussion to eliminate the ambiguity in the sign. To

make sense of the right hand side, we need to specify the direction in which

we move around C. The integral in one direction will be the negative of the

integral in the opposite direction. Similarly, on the left hand side we have the

integral of curl(u) ·n dA, where n is a unit vector normal to the surface. There

are two possible directions for n (each opposite to the other) and there is no

natural rule to choose between them. However, the choice of n can be linked to

the choice of direction around the curve as follows: if you walk in the specified

direction with your feet on C and your head pointing in the direction of n, then

the surface S should be on your left. Provided that we follow this convention,

we will have ∫∫

S

curl(u) · dA =

∫∫

S

curl(u) · n dA = +

∫

C

u · dr.

Example 43.: Consider the surface S given by z = x2 − y2 with x2 + y2 ≤ 1.

Consider the vector field given as u = (−y, x, 0). Check Stokes’ Theorem.

Solution:

First we parametrise S as

r = (x, y, z) = (r cos(s), r sin(s), r2 cos2(s)− r2 sin2(s))

with 0 ≤ r ≤ 1 and 0 ≤ s ≤ 2pi. Using cos2(s)− sin2(s) = cos(2s):

r = (x, y, z) = (r cos(s), r sin(s), r2 cos(2s))

which gives

rr = (cos(s), sin(s), 2r cos(2s)) rs = (−r sin(s), r cos(s), −2r2 sin(2s))

Now, we can calculate

rr × rs =

∣∣∣∣∣∣

i j k

cos(s) sin(s) 2r cos(2s)

−r sin(s) r cos(r) −2r2 sin(2s)

∣∣∣∣∣∣

= (−2r2 sin(s) sin(2s)−2r2 cos(s) cos(2s), 2r2 cos(s) sin(2s)−2r2 sin(s) cos(2s), r cos2(s)+r sin2(s))

Using the identities

sin(a) sin(b) + cos(a) cos(b) = cos(a− b) = cos(b− a)

57

and

sin(a) cos(b)− cos(a) sin(b) = sin(a− b) = − sin(b− a)

the result becomes rr × rs = (−2r2 cos(s), 2r2 sin(s), r), so

dA = (rr × rs) dr ds = (−2r2 cos(s), 2r2 sin(s), r) dr ds.

Next, we have

curl(u) =

∣∣∣∣∣∣

i j k‘

∂

∂x

∂

∂y

∂

∂z

−y x 0

∣∣∣∣∣∣

so ∫∫

S

curl(u) · dA =

∫ 2pi

s=0

∫ 1

r=0

2r dr ds =

∫ 2pi

s=0

1 ds = 2pi.

On the other hand, we can parametrise the boundary curve C (where r = 1) as

r = (x, y, z) = (cos(s), sin(s), cos(2s)).

On this curve we have

u = (−y, x, 0) = (− sin(s), cos(s), 0), dr = (− sin(s), cos(s),−2 sin(2s)) ds

therefore

u · dr = (sin2(s) + cos2(s)) ds = ds

and ∫

C

u · dr =

∫ 2pi

s=0

ds = 2pi.

As expected, this is the same as

∫∫

S

curl(u) · dA.

Example 44. Let us consider the cylinder with r = a and −b ≤ z ≤ b with

0 ≤ θ ≤ 2pi.

Check Stokes’s Theorem for the vector field u = (−zy, zx, z2).

58

Solution:

Let us denote by S the lateral surface of the cylinder. We can parametrise S as

r = (x, y, z) = (a cos(θ), a sin(θ), z)

so

rθ = (−a sin(θ), a cos(θ), 0), rz = (0, 0, 1)

Now we can calculate

rθ × rz =

∣∣∣∣∣∣

i j k

−a sin(θ) a cos(θ) 0

0 0 1

∣∣∣∣∣∣ = (a cos(θ), a sin(θ), 0)

so

dA = (rθ × rz)dθdz = a(cos(θ), sin(θ), 0) dθ dz.

Note that dA points outwards, away from the z-axis. Also

curl(u) =

∣∣∣∣∣∣

i j k

∂

∂x

∂

∂y

∂

∂z

−zy zx z2

∣∣∣∣∣∣ = (0− x, −y − 0, z − (−z)) = (−x,−y, 2z).

On the surface S this becomes

curl(u) = (−a cos(θ), −a sin(θ), 2z)

so

curl(u) · dA = (−a2 cos2(θ)− a2 sin2(θ)) dθ dz = −a2 dθ dz

meaning that∫∫

S

curl(u) · dA = −a2

∫ 2pi

θ=0

∫ b

z=−b

dθ dz = −a2 × 2pi × 2b = −4pia2b.

The boundaries of S are C1 and C2. Directions as shown keep S on the left when

walking with head in the direction of dA, away from the z-axis. Compatible

parametrisations of the two boundaries are

C1 (x, y, z) = (a cos(t),−a sin(t), b)

and

C2 (x, y, z) = (a cos(t), a sin(t),−b).

On C1 we have

dr = (−a sin(t),−a cos(t), 0) dt u = (−zy, zx, z2) = (ab sin(t), ab cos(t), b2)

so

u · dr = −a2b sin2(t)− a2b cos2(t) = −a2b

59

and ∫

C1

u · dr =

∫ 2pi

t=0

−a2b dt = −2pia2b

The discussion of C2 is similar:

dr = (−a sin(t), a cos(t), 0) dt u = (ab sin(t),−ab cos(t), b2),

therefore

u · dr = −a2b sin2(t)− a2b cos2(t) = −a2b

and ∫

C2

u · dr = −2pia2b.

Putting these together, we get

∫

C

u · dr = −4pia2b, which is the same as∫∫

S

curl(u).dA, as expected.

60