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Thread: contour integral around $\infty$

  1. #1
    Nov 2012

    contour integral around $\infty$

    Consider the function

    $\displaystyle f(z)=Log(\frac{z-a}{z-b})$

    where $\displaystyle a,b\in D(0,r)$ , the disc of radius $\displaystyle r$ centered at the origin, open, and $\displaystyle r>0$. Show that $\displaystyle f$ is holomophic in an open subset of complex plane, containing the complement of the disc:$\displaystyle \mathbb C-D(0,r)$ and compute the integral

    $\displaystyle \oint_{|z|=r}z^nf(z)dz$

    where $\displaystyle n\geq 0$ is an integer.

    I'm quite confused about this problem. Because, the $Log$ in the RHS should be the principal branch, i.e. $\displaystyle Log(z)=log|z|+iArg(z)$, with $\displaystyle Arg(z)\in (-\pi,\pi) and Arg(0)=0$. Then i should prove that $\displaystyle \frac{z-a}{z-b}$ does not intersect the real negative axis, but how to do this? However, suppose i could prove the holomorphicity, then i'm left to compute the integral, for which i tought to use the residue formula applied to $\displaystyle \infty$ point, so calling $\displaystyle I$ the integral,

    $\displaystyle I=2\pi i Ind(\gamma,\infty)Res(f,\infty)$

    where $\displaystyle \gamma$ is the contour $\displaystyle |z|=1$, so that $\displaystyle Ind(\gamma,\infty)=-1$. But then the computation of residue is very hard for me. Some help?
    Last edited by tenderline; Nov 29th 2012 at 06:46 AM.
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