Cauchy Integral

1. May 5, 2009

luke1001

A. 1. The problem statement, all variables and given/known data

f is analytic inside and on a simple closed contour C and z0 isn't on C. Show:

$$\int f'(z)dz/ (z- zo)$$ = $$\int f(z)dz/ (z- zo)^2$$

3. The attempt at a solution

$$\int f'(z)dz/ (z- zo)$$ = 2$$\pi$$i f'(zo)

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

1. The problem statement, all variables and given/known data

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

$$\int ((e^a)^z) /z) dz$$ = 2$$\pi$$ i (1)

Derive the following (the integral is from 0 to $$\pi$$):

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

3. The attempt at a solution

First, I translated (1) in terms of $$\theta$$ 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!

2. May 5, 2009

Dick

For the first one you have a second order pole at z=z0. How do you find the residue at a second order pole. It's not the way you are trying to do it. For the second, yes, write the integral 1) as a an integral dtheta. Then take the imaginary part and compare with 1). Then argue that the integral from 0 to pi is 1/2 the integral from 0 to 2pi by showing it's symmetrical around theta=pi.