Mathematical Methods for Physics and Engineering : A Comprehensive Guide

COMPLEX VARIABLES

contourCandz 0 is a point withinCthen

f(z 0 )=

1

2 πi

∮

C

f(z)

z−z 0

dz. (24.46)

This formula is saying that the value of an analytic function anywhere inside a

closed contour is uniquely determined by its values on the contour§and that the

specific expression (24.46) can be given for the value at the interior point.

We may prove Cauchy’s integral formula by using (24.43) and takingγto be a

circle centred on the pointz=z 0 , of small enough radiusρthat it all lies inside

C. Then, sincef(z) is analytic insideC, the integrandf(z)/(z−z 0 ) is analytic

in the space betweenCandγ. Thus, from (24.43), the integral aroundγhas the

same value as that aroundC.

We then use the fact that any pointzonγis given byz=z 0 +ρexpiθ(and

sodz=iρexpiθ dθ). Thus the value of the integral aroundγis given by

I=

∮

γ

f(z)

z−z 0

dz=

∫ 2 π

0

f(z 0 +ρexpiθ)

ρexpiθ

iρexpiθ dθ

=i

∫ 2 π

0

f(z 0 +ρexpiθ)dθ.

If the radius of the circleγis now shrunk to zero, i.e.ρ→0, thenI→ 2 πif(z 0 ),

thus establishing the result (24.46).

An extension to Cauchy’s integral formula can be made, yielding an integral

expression forf′(z 0 ):

f′(z 0 )=

1

2 πi

∫

C

f(z)

(z−z 0 )^2

dz, (24.47)

under the same conditions as previously stated.

Prove Cauchy’s integral formula forf′(z 0 )given in (24.47).

To show this, we use the definition of a derivative and (24.46) itself to evaluate

f′(z 0 ) = lim

h→ 0

f(z 0 +h)−f(z 0 )

h

= lim

h→ 0

[

1

2 πi

∮

C

f(z)

h

(

1

z−z 0 −h

−

1

z−z 0

)

dz

]

= lim

h→ 0

[

1

2 πi

∮

C

f(z)

(z−z 0 −h)(z−z 0 )

dz

]

=

1

2 πi

∮

C

f(z)

(z−z 0 )^2

dz,

which establishes result (24.47).

§The similarity between this and the uniqueness theorem for the Laplace equation with Dirichlet

boundary conditions (see chapter 20) is apparent.