Cauchy Integral

  • Thread starter luke1001
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A.

Homework Statement



f is analytic inside and on a simple closed contour C and z0 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!
 

Answers and Replies

  • #2
Dick
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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.
 

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