Solve Cauchy Integral on Unit Circle: f'(z)/(z-zo) = f(z)/(z-zo)^2

[luke1001]

A.

Homework Statement

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

The Attempt at a Solution

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

Homework Statement

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)

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!

 

[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.