2Contour integration and Cauchy's theorem

IB Complex Methods

2.3 Contour deformation

One useful consequence of Cauchy’s theorem is that we can freely deform contours

along regions where f is defined without changing the value of the integral.

Proposition.

Suppose that

γ

1

and

γ

2

are contours from

a

to

b

, and that

f

is

analytic on the contours and between the contours. Then

Z

γ

1

f(z) dz =

Z

γ

2

f(z) dz.

a

b

γ

2

γ

1

Proof.

Suppose first that

γ

1

and

γ

2

do not cross. Then

γ

1

−γ

2

is a simple closed

contour. So

I

γ

1

−γ

2

f(z) dz = 0

by Cauchy’s theorem. Then the result follows.

If

γ

1

and

γ

2

do cross, then dissect them at each crossing point, and apply

the previous result to each section.

So we conclude that if

f

has no singularities, then

R

b

a

f

(

z

) d

z

does not depend

on the chosen contour.

This result of path independence, and indeed Cauchy’s theorem itself, becomes

less surprising if we think of

R

f(z) dz as a path integral in R

2

, because

f(z) dz = (u + iv)(dz + i dy) = (u + iv) dx + (−v + iu) dy

is an exact differential, since

∂

∂y

(u + iv) =

∂

∂x

(−v + iu)

from the Cauchy-Riemann equations.

The same idea of “moving the contour” applies to closed contours. Suppose

that

γ

1

is a closed contour that can be continuously deformed to another one,

γ

2

, inside it; and suppose f has no singularities in the region between them.

γ

2

γ

1

××

×

We can instead consider the following contour γ:

γ

××

×

By Cauchy’s theorem, we know

H

γ

f

(

z

) d

z

= 0 since

f

(

z

) is analytic throughout

the region enclosed by

γ

. Now we let the distance between the two “cross-cuts”

tend to zero: those contributions cancel and, in the limit, we have

I

γ

1

−γ

2

f(z) dz = 0.

hence we know

I

γ

1

f(z) dz =

I

γ

2

f(z) dz = 0.