Complex variable theory, Lecture notes of Mathematics

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These notes are based on earlier notes by Professor Gareth Jones and Dr Jim Anderson. References in square brackets throughout the notes refer to other prerequisite or parallel Mathematics modules, in particular MATH1051 Calculus II and MATH2039 Analysis. References are also made to the defunct MATH2002, a module which was taught for many years by Keith Hirst, and the MATH2002 notes are posted on the MATH2045 Blackboard site. The MATH2002 notes are particularly useful for proofs which are omitted in these notes; note however that only the material in the main notes is examinable.

In these notes, diagrams illustrate important explanations. Key words and theorems are highlighted in bold font; you should make sure you are completely confident with these for the exam. The most important formulae are highlighted with boxes; again you should ensure that you recall and understand these formulae. The last chapter of the notes summarises, step by step, the methods used to compute contour integrals.

Complex variable theory is covered in many textbooks. No specific book is required for this module as all material needed is included in the notes. The use of textbooks is however recommended as they provide extra examples as well as additional explanations and proofs. Suggested texts (along with their library reference, where available) include

• H A Priestley An introduction to complex analysis Oxford University Press QA
• M R Spiegel Theory and problems of complex variables McGraw HIll QA
• J W Brown and R V Churchill Complex variables and applications McGraw Hill QA
• M J Ablowitz and A S Fokas Complex variables: Introduction and Applications Cam- bridge University Press QA

Note that complex methods are also discussed in most maths for engineering or physical sciences textbooks such as

• C Wylie and L C Barrett Advanced engineering mathematics McGraw Hill QA
• G Stephenson and P M Radmore Advanced mathematical methods for engineering and science students Cambridge University Press QA
• E Kreysig Advanced engineering mathematics Cambridge University Press QA

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15 Evaluation of real integrals 36

16 Problem solving 40

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1 Introduction to complex variable theory

We will study differentiation and integration of complex functions. Much of the theory resembles that of real functions. However, there are two main differences:

(a) Differentiability is a stronger condition in C than in R, e.g. f (x) = x|x| is a differ- entiable function R → R, but g(z) = z|z| is not a differentiable function C → C. In C, differentiable once implies differentiable n times for all n ≥ 1, but this is false in R (e.g. f has no f ′′(0)).

(b) The value of a definite integral

∫ (^) b a f^ (z) dz^ may depend on the chosen path in^ C^ from^ a to b. This can cause complications, but can also make some integrals easier to evaluate: a technique called the Calculus of Residues allows one to evaluate certain definite integrals∫ b a f^ (z) dz^ without needing to find an antiderivative^ F^ (z) for^ f^ (z).

2 Revision of complex numbers

[MATH1051 Calculus II, MATH2002 notes §1]

Complex numbers arise as a natural extension of the number system:

• Within the natural numbers (x ∈ N) one can solve 2 + x = 3, but not 3 + x = 2;
• Within the integer numbers (x ∈ Z) one can solve 3 + x = 2, but not 2x = 3;
• Within the rational numbers (x ∈ Q) one can solve 2x = 3, but not x^2 = 2;
• Within the real numbers (x ∈ R) one can solve x^2 = 2, but not x^2 = −2;
• Within the complex numbers (x ∈ C) one can solve any polynomial equation - see section 10.

A complex number z can be written in Cartesian form z = x + iy where x and y are real numbers and i^2 = −1. We call x the real part of z and denote it by Re z = <(z). Similarly, we call y the imaginary part of z and denote it by Im z = =(z).

A complex number can also be written in polar form as z = reiθ^ = r cos θ + ir sin θ. Here r = |z| is called the modulus or absolute value of z and θ = arg z is called the argument of z. Note that arg z is not uniquely defined: we can add any integer multiple of 2π to arg z without altering z. To avoid confusion it is sometimes helpful to define the principle value of the argument as Arg(z) where −π < Arg(z) ≤ π.

The two representations z = x + iy = reiθ

3 Subsets of the complex plane

[MATH2039 Analysis, MATH2002 notes §2]

We can generalise the concepts of open and closed intervals from R to C. If z 0 ∈ C and ε > 0, define the neighbourhood of z 0 as

Nε(z 0 ) = {z ∈ C | |z − z 0 | < ε},

a disc, radius ε, centred at z 0. If S ⊆ C, an element z 0 ∈ S is an interior point of S if there exists ε > 0 such that Nε(z 0 ) ⊆ S, i.e. all points close to z 0 are also in S. A subset S ⊆ C is open if every z 0 ∈ S is an interior point.

Example 3.1 S = {z ∈ C | |z| < 1 } is open (take ε = 1 − |z 0 |) but S′^ = {z ∈ C | |z| ≤ 1 } is not (z 0 = 1 is not an interior point).

A subset S ⊆ C is closed if its complement C \ S is open, i.e. if, whenever z 0 6 ∈ S, there exists ε > 0 such that Nε(z 0 ) is disjoint from S.

Example 3.2 S = {z ∈ C | |z| < 1 } is not closed (take z 0 = 1) but S′^ = {z ∈ C | |z| ≤ 1 } is closed (take ε = |z 0 | − 1).

It is useful to illustrate such subsets as follows:

a 2

a 1

a 3

r 1

r 2

r 3

open disk

closed disk

punctured open disk

Im

Re

circle |z−a|=r a (^) r

Figure 2: Examples of discs in the complex plane.

A circle of radius r with centre a in the complex plane has the equation

|z − a| = r,

where a can be complex and r is real and positive. The inside of the circle

|z − a|< r,

is called an open disc. If we include the boundary it becomes a closed disc:

|z − a|≤ r.

If we remove the point a from the disk it is called a punctured disc - such a disc is usually open: 0 < |z − a| < r

but could be closed: 0 < |z − a| ≤ r.

4 Continuity and differentiability

[MATH1051, MATH2039, MATH2002 notes §3, §4]

A mapping f which maps a complex number z to a unique complex number w, that is w = f (z), is a function. There are also mappings such that w is not unique; these multifunctions are not true functions but mappings. For example, log is not a proper function of z, since arg z is not unique. Here we will concentrate however on functions.

Let f be a function defined on an open set S ⊆ C and let z 0 ∈ S. Then f is continuous at z 0 if limz→z 0 f (z) = f (z 0 ), i.e. for each ε > 0 there exists δ > 0 such that

|z − z 0 | < δ ⇒ |f (z) − f (z 0 )| < ε.

If f and g are continuous at z 0 then so are f ± g, f g and (provided g(z 0 ) 6 = 0) f /g. We say that f is continuous on S if it is continuous at every z 0 ∈ S.

We say that f is differentiable at z 0 , with derivative f ′(z 0 ), if

lim z→z 0

f (z) − f (z 0 ) z − z 0

= f ′(z 0 ).

Notice that for the limit to exit, it must be independent of the direction in which h = (z−z 0 ) is taken to zero.

If f and g are differentiable at z 0 then so are f ± g, f g and (provided g(z 0 ) 6 = 0) f /g, with the usual formulae for their derivatives. For example, ( 1 g

(z 0 ) = −

g′(z 0 ) g(z 0 )^2

We say that f is holomorphic on S if it is differentiable at every z 0 ∈ S; it is entire if it is holomorphic on C.

For example, a polynomial function can be written as

f (z) =

∑^ n

k=

akzk

The proof of the Cauchy-Riemann equations is given in MATH1051 Calculus II but let us briefly sketch how the conditions are derived: let f (z) be analytic in a certain domain, and let z = x + iy. It is often convenient to write

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

thereby splitting the complex function f into real and imaginary parts just like z. Then

f (z + h) − f (z) h

u(x + δx, y + δy) + iv(x + δx, y + δy) − u(x, y) − iv(x, y) δx + iδy

where h = δx + iδy.

Take first δy = 0 (so, approach with h tending to zero along the x axis). Then h = δx, so that

limh→ 0 f^ (z+h h)− f^ (z)= limδx→ 0

u(x + δx, y) + iv(x + δx, y) − u(x, y) − iv(x, y) δx

= limδx→ 0

[

u(x + δx, y) − u(x, y) δx

• i

v(x + δx) − v(x, y) δx

]

= ux(x, y) + ivx(x, y).

If, instead, we take δx = 0 (i.e., approach with h to zero along the y axis), then h = iδy and we obtain

limh→ 0 f^ (z+h h)− f^ (z)=

u(x, y + δy) + iv(x, y + δy) − u(x, y) − iv(x, y) iδy

= −iuy(x, y) + vy(x, y).

Now, if f (z) is analytic, then limh→ 0 f^ (z+h h)− f^ (z)= f ′(z) does not depend on the direction, and the two expressions must be equal:

ux + ivx = f ′(z) = vy − iuy

Taking the real and imaginary parts of f ′(x) and equating them we obtain the Cauchy- Riemann equations:

ux = vy, uy = −vx.

Example 5.1 Show that f (z) = z^2 is analytic for all z ∈ C. f (z) = z^2 = (x + iy)^2 = (x^2 − y^2 ) + i 2 xy so u = x^2 − y^2 v = 2xy ux = 2x vy = 2x uy = − 2 y vx = 2y

Hence ux = vy and vx = −uy ⇒ The Cauchy-Riemann equations are satisfied ⇒ f is differentiable.

Example 5.2 Determine whether f (z) = ¯z is analytic. f (z) = ¯z = x − iy, so

u = x v = −y ux = 1 vy = − 1 uy = 0 vx = 0

Hence vx= − uy but ux 6 =vy ⇒ The Cauchy-Riemann equations are not satisfied ⇒ f is not differentiable.

6 Useful functions

[MATH2002 notes §7, §8, §10]

(a) Polynomials: f (z) =

∑n k=0 akz k (^) where each ak ∈ C; the usual rules for algebra,

differentiation and integration apply. We will later see that f has n roots in C, counting multiple roots (Fundamental Theorem of Algebra); this is false in R, e.g. f (x) = x^2 + 1.

(b) Rational functions: f (z) = p(z)/q(z) where p, q are polynomials; defined wherever q(z) 6 = 0; usual rules for algebra, differentiation, integration.

(c) Exponential function:

exp(z) =

∑^ ∞

k=

zk k!

This power series converges for all z ∈ C, so it is differentiable term-by-term [MATH2039, MATH2002 notes §20]:

exp′(z) =

∑^ ∞

k=

kzk−^1 k!

∑^ ∞

k=

zk−^1 (k − 1)!

∑^ ∞

k=

zk k!

= exp(z).

(We will discuss power series and their convergence in section 11.) Basic identities:

exp(z 1 + z 2 ) = exp(z 1 ). exp(z 2 ),

exp(z 1 − z 2 ) = exp(z 1 )/ exp(z 2 ), exp(z)n^ = exp(nz) for n ∈ Z, exp(x + iy) = ex(cos y + i sin y).

Periodicity: exp(z + 2πi) = exp(z),

implying that the map is infinite-to-one C → C \ { 0 }, bijection R × (−π, π] → C \ { 0 }. (The origin of C is excluded as the exponential function never takes the value zero.)

is many-valued (since arg(z) is); the principal value

Log(z) = ln(|z|) + i Arg(z) with − π < Arg(z) ≤ π,

is continuous on S = C \ {z | z ≤ 0 } but not where z ≤ 0 (value changes by ± 2 πi). Holomorphic on S with derivative

Log′(z) =

z

More generally, for any a, b ∈ R with 0 < b − a ≤ 2 π we can define a single-valued branch of log(z) on S = {z = reiθ^ | r > 0 , a < θ < b}, with derivative 1/z.

(g) Powers: define az^ = exp(z log(a)),

which is in general many-valued (since log(a) is). Principal value exp(z Log(a)), e.g. ez^ = exp(z). Holomorphic, with derivative Log(a).az^.

7 Paths and integrals in the complex plane

[MATH2002 notes, §11, §12]

We are used to integrating a real function along the real line: considering the value of f (x) as x ranges between α and β say and finding the area underneath.

For the complex variables z we move in 2D across the complex plane. Because of this it is not only the end points that are important but also the route across the complex plane between them.

A curve γ is the graph of a continuous function z(t) from a real interval a ≤ t ≤ b to the complex plane: γ = {z(t) : a ≤ t ≤ b}.

A path is a finite join of smooth curves. If z(a) = z(b) then γ is closed.

If z(t 1 ) = z(t 2 ) only if t 1 = t 2 for all t ∈ (a, b) then γ is simple, which means that it does not intersect itself. Note that this does not include end points.

When we integrate complex functions, we usually do so over special paths: a contour is a finite join of straight lines or arcs forming a simple closed path.

Since a closed path starts where it finishes we need to also define its orientation. By convention we take the integral in an anticlockwise direction.

Curve Path

Re

Im

α β z(b)

z(a) γ

Re

Im

Closed path Simple path

Re

Im

Re

Im

Im

Re

Contour Not simple

Not closed

Contour

Not just arcs and straight lines

circular

Figure 4: Curves in the complex plane.

Im

Re

a r

γ

Figure 5: The circle C(a : r) = {z : |z − a| = r} = {z(t) = a + reit^ : 0 ≤ t < 2 π}.

In particular, if C goes once around S^1 in the positive direction then the integral is 2πi. Note that the positive direction is anticlockwise around the origin of the complex plane.

Example 7.2 Integrate the function (z − a)n^ around a circle with radius r and center a, when n is an integer, i.e., find In =

C(a:r)(z^ −^ a)

n (^) dz.

The integration contour is γ = {z : |z − a| = r}: see Figure 5. This can be parameterised as γ = z(t) = a + reit^ : 0 ≤ t < 2 π.

We have z(t) = a + reit^ ⇒ z ′ (t) = ireit, and so:

In =

∫ (^2) π

0

a + reit

− a

)n ireitdt = irn+

∫ (^2) π

0

ei(n+1)tdt

rn+ n+

[

ei(n+1)t

] 2 π 0 = 0,^ for^ n^6 =^ −^1 i [ t ]^20 π = 2πi, for n = − 1.

This result is important for Cauchy’s Residue Theorem later on:

∫

C(a:r)

(z − a)n^ dz =

0 , n 6 = − 1 , 2 πi, n = − 1 for any r. (2)

Example 7.3∫ Let C be the unit square, with corners at 0, 1 , 1+i and i, and let us evaluate

C z

(^2) dz. We divide C into four subcontours C 1 ,... , C 4 along the four sides, starting with the side C 1 from 0 to 1. Again note that the direction of the contour will be anticlockwise.

Along C 1 we have z = g(t) = t with t ∈ [0, 1], so g′(t) = 1 and

∫

C 1

z^2 dz =

0

t^2 dt =

Along C 2 we have z = g(t) = 1 + it with t ∈ [0, 1], so g′(t) = i and

∫

C 2

z^2 dz =

0

(1 + 2it − t^2 )i dt = i − 1 −

i 3

2 i 3

Along C 3 we have z = g(t) = 1 + i − t with t ∈ [0, 1], so g′(t) = −1 and

∫

C 3

z^2 dz =

0

(2i − 2(1 + i)t + t^2 ). − 1 dt = − 2 i + (1 + i) −

− i.

Along C 4 we have z = g(t) = i(1 − t) with t ∈ [0, 1], so g′(t) = −i and

∫

C 1

z^2 dz =

0

(1 − 2 t + t^2 )i dt = (1 − 1 +

)i =

i 3

Summing these, we get ∫

C

z^2 dz =

∑^4

k=

Ck

z^2 dz = 0.

All that work for nothing! We will see shortly that this was entirely predictable.

Theorem 1 Let −C denote the contour C in the reverse direction, i.e. parametrized by g(a + b − t) for t ∈ [a, b]. Then

∫

−C

f (z) dz = −

C

f (z) dz.

This is proved in the MATH2002 notes, §12.

Theorem 2 The Fundamental Theorem of Calculus: Let f have an antiderivative F (i.e. F ′^ = f ) in a region D containing a contour C from g(a) = z 0 to g(b) = z 1. Then

∫

C

f (z) dz = F (z 1 ) − F (z 0 ).

Proof. ∫

C

f (z) dz =

∫ (^) b

a

F ′(g(t))g′(t) dt =

∫ (^) b

a

d dt

F (g(t)) dt = F (g(b)) − F (g(a)) = F (z 1 ) − F (z 0 ).

By construction

R =

W

∫ (^) b

a

g(t)dt =

∫ (^) b

a

W −^1 g(t)dt.

Since R is real, R =

∫ (^) b a Re(W^

− (^1) g(t))dt. However,

R =

∫ (^) b

a

Re(W −^1 g(t))dt ≤

∫ (^) b

a

|W −^1 g(t)|dt =

∫ (^) b

a

|g(t)|dt,

where we first use Re(z) ≤ |z| and then use |W | = 1.

Theorem 5 Estimation Lemma: If |f (z)| ≤ M for all z on a contour C of length L then (^) ∣ ∣ ∣

C f^ (z) dz

∣ ≤ M L.

Proof. Let C be parametrized by g : [a, b] → C. Then

∣ ∣ ∣

C

f (z) dz

∫ (^) b

a

f (g(t))g′(t) dt

∫ (^) b

a

|f (g(t))||g′(t)| dt ≤

∫ (^) b

a

M |g′(t)| dt = M L.

The Estimation Lemma is sometimes referred to as ML estimation.

Example 7.6 Let

IR =

CR

z + 1 z^3 − 1

dz,

where CR is the circle |z| = R of radius R > 1 centred at 0. This contour has length L = 2πR. On this contour, |z| = R, so the triangle inequality gives |z+1| ≤ |z|+| 1 | = R+1, and the reverse triangle inequality gives |z^3 − 1 | ≥ ||z^3 | − | 1 || = |R^3 − 1 | = R^3 − 1, so that

∣ ∣ ∣

z + 1 z^3 − 1

∣ ≤^ M^ =^

R + 1

R^3 − 1

It follows from the Estimation Lemma that

|IR| ≤ M L = 2π

R(R + 1)

R^3 − 1

In particular, we see that IR → 0 as R → +∞. (See the end of §14 for more on this example.) We will see similar applications of the Estimation Lemma in §15(b).

8 Cauchy’s Theorem.

[MATH2002 notes, §14, §15]

Cauchy’s theorem is one of the most important theorems in mathematics:

Theorem 6 Cauchy’s theorem If f is holomorphic inside and on a closed contour C, then (^) ∫

C f^ (z) dz^ = 0.

Proof. Let f (z) = u + iv with u, v ∈ R, and let C be parametrized by g(t) = x(t) + iy(t) for a ≤ t ≤ b, with x(t), y(t) ∈ R. Then

∫

C

f (z) dz =

∫ (^) b

a

(u + iv)(x′^ + iy′) dt

∫ (^) b

a

((ux′^ − vy′) + i(uy′^ + vx′)) dt

C

(u dx − v dy) + i

C

(u dy + v dx)

D

( (^) ∂v

∂x

∂u ∂y

dx dy + i

D

(∂u

∂x

∂v ∂y

dx dy,

with the last step justified by applying Green’s Theorem (Vector Calculus notes) to the vector fields (u, −v) = u − iv = f and (v, u) = v + iu = if , where D is the region enclosed by C, so that C is the boundary ∂D of D. But both integrands are identically zero in D, since f satisfies the Cauchy-Riemann equations there (see §5), so

C f^ (z) dz^ = 0.

[Simple case of Green’s Theorem: imagine C consisting of an upper contour C+ and a lower contour C−, containing points x + iy+ and x + iy− for x 0 ≤ x ≤ x 1 , with y+ ≥ y−. Then ∫

C

u dx = −

C+

u dx+

C−

u dx = −

∫ (^) x 1

x 0

[

u(x+iy)

]y+ y− dx^ =^ −

∫ (^) x 1

x 0

∫ (^) y+

y−

∂u ∂y

dy dx = −

D

∂u ∂y

dx dy

and similarly (with changes of sign to take account of orientation) ∫

C

v dy =

D

∂v ∂x

dx dy ,

C

u dy =

D

∂u ∂x

dx dy ,

C

v dx = −

D

∂v ∂y

dx dy .]

There is an alternative proof in §14 of the MATH2002 notes, avoiding Green’s Theorem, for the special case where C is the boundary of a triangle.

Example 8.1 If f (z) is an entire function, such as a polynomial, or exp(z), sin(z), cos(z), sinh(z) or cosh(z), then

C f^ (z) dz^ = 0 for every closed contour^ C, since^ f^ is holomorphic every- where.