2Contour integration and Cauchy's theorem

IB Complex Methods

2.4 Cauchy’s integral formula

Theorem

(Cauchy’s integral formula)

.

Suppose that

f

(

z

) is analytic in a domain

D and that z

0

∈ D. Then

f(z

0

) =

1

2πi

I

γ

f(z)

z −z

0

dz

for any simple closed contour γ in D encircling z

0

anticlockwise.

This result is going to be very important in a brief moment, for proving one

thing. Afterwards, it will be mostly useless.

Proof. (non-examinable) We let γ

ε

be a circle of radius ε about z

0

, within γ.

z

0

γ

ε

γ

Since

f(z)

z−z

0

is analytic except when z = z

0

, we know

I

γ

f(z)

z −z

0

dz =

I

γ

ε

f(z)

z −z

0

dz.

We now evaluate the right integral directly. Substituting z = z

0

+ εe

iθ

, we get

I

γ

ε

f(z)

z −z

0

dz =

Z

2π

0

f(z

0

+ εe

iθ

)

εe

iθ

iεe

iθ

dθ

= i

Z

2

0

π(f(z

0

) + O(ε)) dθ

→ 2πif(z

0

)

as we take the limit ε → 0. The result then follows.

So, if we know

f

on

γ

, then we know it at all points within

γ

. While this

seems magical, it is less surprising if we look at it in another way. We can write

f

=

u

+

iv

, where

u

and

v

are harmonic functions, i.e. they satisfy Laplace’s

equation. Then if we know the values of

u

and

v

on

γ

, then what we essentially

have is Laplace’s equation with Dirichlet boundary conditions! Then the fact

that this tells us everything about

f

within the boundary is just the statement

that Laplace’s equation with Dirichlet boundary conditions has a unique solution!

The difference between this and what we’ve got in IA Vector Calculus is that

Cauchy’s integral formula gives an explicit formula for the value of

f

(

z

0

), while

in IA Vector Calculus, we just know there is one solution, whatever that might

be.

Note that this does not hold if

z

0

does not lie on or inside

γ

, since Cauchy’s

theorem just gives

1

2πi

I

γ

f(z)

z −z

0

dz = 0.

Now, we can differentiate Cauchy’s integral formula with respect to

z

0

, and

obtain

f

0

(z

0

) =

1

2πi

I

γ

f(z)

(z −z

0

)

2

dz.

We have just taken the differentiation inside the integral sign. This is valid since

it’s Complex Methods and we don’t care

the integrand, both before and after,

is a continuous function of both z and z

0

.

We see that the integrand is still differentiable. So we can differentiate it

again, and obtain

f

(n)

(z

0

) =

n!

2πi

I

γ

f(z)

(z −z

0

)

n+1

dz.

Hence at any point

z

0

where

f

is analytic, all its derivatives exist, and we have

just found a formula for them. So it is differentiable infinitely many times as

advertised.

A classic example of Cauchy’s integral formula is Liouville’s theorem.

Theorem (Liouville’s theorem*). Any bounded entire function is a constant.

Proof.

(non-examinable) Suppose that

|f

(

z

)

| ≤ M

for all

z

, and consider a circle

of radius r centered at an arbitrary point z

0

∈ C. Then

f

0

(z

0

) =

1

2πi

I

|z−z

0

|=r

f(z)

(z −z

0

)

2

dz.

Hence we know

1

2πi

≤

1

2π

· 2πr ·

M

r

2

→ 0

as r → ∞. So f

0

(z

0

) = 0 for all z

0

∈ C. So f is constant.