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Note: This material is contained in Kreyszig, Chapter 13.

Complex integration

We will define integrals of complex functions along curves in C. (This is a bit similar to [real-

valued] line integrals ∫

γ

P dx + Qdy in R

2

.)

A curve is most conveniently defined by a parametrisation. So a curve is a function γ: [a,b] →

C (from a finite closed real intervale [a,b] to the plane). We can imagine the point γ(t) being

traced out by a pen which is at position γ(t) at time t. We can write γ(t) = x(t) + iy(t) in terms

of its real an imaginary parts.

Then we define γ (t) = x (t) + iy (t) (can be viewed as the tangent vector or velocity vector

to the curve) and we will only be dealing with curves where γ (t) is defined and continuous.

A curve is called closed if γ(a) = γ(b) (start and end point coincide).

A curve is called simple if it never goes though the same point twice (with the possible

exception that γ(a) = γ(b) is allowed — apart from this all γ(t) have to be different points).

Example. A simple example to keep in mind is a circle, say the circle of radius r > 0 about

the origin where we travel once around it anticlockwise starting and ending at the point r on the

positive axis.

Then γ

r

: [0,2π] → C,

γ

r

(t) = re

it

= r cost + ir sint

is one obvious parametrisation.

Definition If γ: [a,b] → C is a curve in C and f(z) is a complex-valued function defined at

least for all z = γ(t), then we define

∫

γ

f(z)dz = ∫

b

a

f(γ(t))γ (t)dt.

(This last is a fairly ordinary integral, except that f(γ(t))γ (t) will have complex values. Say

f(γ(t))γ (t) = p(t) + iq(t) (in terms of the real part p(t) and imaginary part q(t)). Then the

complex integral means simply

∫

b

a

p(t) + iq(t)dt = ∫

b

a

p(t)dt + i∫

b

a

q(t)dt

Technically we will require that these ordinary integrals of p and q should exist, but that will

be ok in all our examples. Continuity of f and of γ (t) is enough to make the integral ok.

Example. Take γ

r

as in the example above and f(z) = 1/z. Then we can explicitly compute

∫

γ

r

1

z

dz = ∫

2π

0

1

re

it

ire

it

dt = ∫

2π

0

idt = 2πi

(In practice, we can rarely do the calculation so directly. Typically we will use theorems to

simplify the curve first.)

Elementary properties (of complex integrals). The basic properties are reminiscent of those

for line integrals in R

2

(except that we now have complex values).

Complex integrals

2

1. The exact parametrisation of the curve γ is not important, although the direction is. So for

example, if we take the circle |z| = r but parametrise it in a different way, while still going

once around anticlockwise — say by σ

r

: [0,1] → C with σ

r

(t) = e

2πit

, then the integral

will not change. So

∫

γ

r

f(z)dz = ∫

σ

r

f(z)dz

Changing the direction of the curve changes the the integral by a factor −1. For exam-

ple in the case of the circle, µ

r

: [0,2π] → C with µ

r

(t) = e

−2πit

has ∫

µ

r

f(z)dz =

−∫

γ

r

f(z)dz. These fact follow by ordinary substitution (or change of variables).

2. If f(z) = F (z) for some analytic F(z) and γ: [a,b] → C is a curve with all points γ(t)

in the set where F(z) is analytic, then

∫

γ

f(z)dz = ∫

b

a

F (γ(t))γ (t)dt = ∫

b

a

d

dt

F(γ(t))dt = [F(γ(t))]

b

t=a

= F(γ(b))−F(γ(a))

is the difference of the values F(end) − F(start).

3. In particular, if the integrand f(z) has an analytic antiderivative F(z) that works all along

γ, then the exact path γ does not enter in to the value of ∫

γ

f(z)dz (as long as γ stays in

the set where F is analytic) and the integral will be 0 if the path is closed (start = end).

If you look back at the last example you will see that the integral of f(z) = 1/z around

the closed curve γ

r

was not zero. Thus there is no antiderivative of 1/z that works all the

way around γ

r

. [Recall that log z is an antiderivative of 1/z except on the negative axis.

The jump we make in the argument arg(z) at the negative axis actually corresponds to the

value 2πi of the integral.]

4. We can use Green’s theorem for complex valued P(x,y) and Q(x,y). That is

∫

γ

P dx + Qdy = ∫∫

R

(∂Q

∂x

−

∂P

∂y

) dxdy

is true for complex-valued P, Q if γ is a simple closed curve in R

2

, R is the interior of γ

and both P and Q are well-behaved inside and on γ.

Here we interpret the integrals of complex things as the integral of the real part +i times

the integral of the imaginary part.

Theorem 1 (Cauchy’s theorem) If γ is a simple closed anticlockwise curve in the complex

plane and f(z) is analytic on some open set that includes all of the curve γ and all points

inside γ, then

∫

γ

f(z)dz = 0

Complex integrals

3

Proof. We write dz = dx + idy and use Green’s theorem on

∫

γ

f(z)dz = ∫

γ

f(z)dx + (if(z))dy = ∫∫

R

(∂(if(z))

∂x

−

∂f(z)

∂y

) dxdy

(with R denoting the interior of γ). If you recall the proof of the CR equations you will remember

that (because the limit defining f (z) = lim

h→0

f(z+h)−f(z)

h

can be taken in any direction or in all

directions at once

f (z) =

∂f(z)

∂x

=

1

i

∂f(z)

∂y

It follows that the integrand of the double integral we got from Green’s theorem

∂(if(z))

∂x

−

∂f(z)

∂y

= i(

∂f(z)

∂x

−

1

i

∂f(z)

∂y

) = 0

and so we get ∫

γ

f(z)dz = 0. [Another way to do this is to write f(z) = u + iv and use the CR

equations to get the integrand of the double integral to be zero.]

Remark. It is vital that there are no bad points of f(z) inside or on γ. Look again at the last

example and see that f(z) = 1/z is fine everywhere except at z = 0. This can (and does in that

case) make the integral nonzero.

Corollary 2 Suppose we have two anticlockwise simple closed curves γ

1

and γ

2

with one entirely

contained in the interior of the other. Suppose f(z) is analytic on some open set that includes

both γ

1

and γ

2

and the region between the two curves. Then

∫

γ

1

f(z)dz = ∫

γ

2

f(z)dz

(The proof involves making a ‘narrow bridge’ between the two curves and a simple closed

curve Γ that goes almost once around the outer curve, in across one side of the bridge, the wrong

way around the inner curve and back across the bridge.

f(z) will be analytic on and in-

side Γ and then ∫

Γ

f(z)dz = 0 by

Cauchy’s theorem. Let the width of

the bridge tend to zero and we find

that we get the result we want be-

cause the integral along the bridge

in different directions cancel.)

Example. Looking back at the example ∫

γ

r

1/z dz we saw they all turned out to be 2πi no

matter what the radius r > 0 was. We can now see that this independence of r follows because

of (the Corollary to) Cauchy’s theorem. Also we can see that we will also get the same 2πi

Complex integrals

4

for integrals around more complicated curves that go once anticlockwise around the origin. For

example the ellipse σ: [0,2π] → C with

σ(t) = 4cost + 3isint

contains γ

r

if r < 3. So ∫

σ

1/z dz = ∫

γ

2

1/z dz = 2πi.

Theorem 3 (Cauchy’s integral formula) Suppose that γ is a simple closed anticlockwise curve

in the complex plane and f(z) is analytic on some open set that includes all of the curve γ and

all points inside γ. Then for any point z

0

inside γ we have

∫

γ

f(z)

z − z

0

dz = 2πif(z

0

)

(We will not proved this but the idea is that the integral will remain the same if we replace γ

by a small circle around z

0

. If we let the radius of the small circle → 0 then we can show that

the integral must be very close to 2πif(z

0

). As the integral is idependent of the radius, it must

actually be 2πif(z

0

).)

Example. Consider

∫

γ

2

z

2

+ z + 1

z

2

+ 2z − 3

dz

The integrand is bad at two points because the denominator is zero at two places.

z

2

+ z + 1

z

2

+ 2z − 3

=

z

2

+ z + 1

(z + 3)(z − 1)

(The bad points are z = 1 and z = −3.) Only one of these bad points z = 1 is inside γ

2

. We can

in fact write the integral as

∫

γ

2

z

2

+ z + 1

z

2

+ 2z − 3

dz = ∫

γ

2

f(z)

z − 1

dz where f(z) =

z

2

+ z + 1

z + 3

and then the Cauchy integral formula gives the answer 2πif(1) = 2πi

3

4

= 3πi/2 for the integral

we started with.

It did not matter that the curve was exactly the circle of radius 2, only that it went around 1

but not around −3.

Power series. One can make use of Cauchy’s integral formula to prove that every analytic

function f(z) can be represented by a power series in any disc where it is analytic.

If f(z) is analytic in an open set that includes the disc {z ∈ C : |z −z

0

| < r} of radius r > 0

about z

0

, then

f(z) =

∞

∑

n=0

a

n

(z − z

0

)

n

for all z with |z − z

0

| < r.

The coefficients a

n

can be represented as integrals

a

n

=

1

2πi

∫

|z−z

0

|=s

f(z)

(z − z

0

)

n+1

dz for any 0 < s < r, or as a

n

=

f

(n)

(z

0

)

n!

Complex integrals

5

Example. Take f(z) = 1/z and z

0

= 2. Then f(z) is analytic for |z − z

0

| < 2 and so there

is a power series for f(z) there.

A more complicated example is f(z) =

e

z

(z − 1)(z − 2)

. For any z

0

different from 1 and 2,

there is power series for f(z) in the largest disc |z − z

0

| < r that misses 1 and 2. Specifically r

is the shorter of the two distances |z − 1| and |z − 2|.