- #1

luke1001

- 3

- 0

## Homework Statement

f is analytic inside and on a simple closed contour C and z

_{0}isn't on C. Show:

[tex]\int f'(z)dz/ (z- zo)[/tex] = [tex]\int f(z)dz/ (z- zo)^2 [/tex]

## The Attempt at a Solution

[tex]\int f'(z)dz/ (z- zo)[/tex] = 2[tex]\pi[/tex]i f'(zo)

[tex]\int f(z)dz/ (z- zo)^2 [/tex] = [tex]\int [f(z)dz/(z- zo)]/(z-zo) [/tex] = 2[tex]\pi[/tex]i [f(zo)/(zo-zo)] (I got stuck here!?)

## Homework Statement

B. C is the unit circle z = e^(i[tex]\theta[/tex]). For any real constant a:

[tex]\int ((e^a)^z) /z) dz[/tex] = 2[tex]\pi[/tex] i (1)

Derive the following (the integral is

*from 0 to [tex]\pi[/tex]*):

[tex]\int [(e^a)^(cos\theta)) * cos (a sin\theta)] d\theta [/tex] = [tex]\pi[/tex] (2)

## The Attempt at a Solution

First, I translated (1) in terms of [tex]\theta[/tex] and got a very long term. Then I substituted the term to (2). But what I got was a huge and complicated mess. Did I miss out something obvious? Thanks!