a must be an interior point of
$\displaystyle \gamma$
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The problem is that there is a pole on the countour. Also something else is fishy you sure that it is an open disk?
Ah I see, I screwed that one up. Regarding gamma being an open disk, I thought that it was open since I read in my notes somewhere that the interior of gamma was open and therefore presumed that the whole disk was open as well. What can I do if the singularity is ON the contour? Will residues work (I haven't gone over this section yet).
Okay lets rephrase the question to gamma(0,2). I would be able to use Cauchy's Integral Formula then right, since a is inside the contour.
Yes, this would give
$\displaystyle f(1)=\frac{1}{2\pi i}\int_{\gamma (0,2)}\frac{f(z)}{z-1}dz, \quad f(z)=z$ so we get
$\displaystyle 1=\frac{1}{2\pi i}\int_{\gamma (0,2)}\frac{z}{z-1}dz \iff \int_{\gamma (0,2)}\frac{z}{z-1}dz=2\pi i$
So finaly
$\displaystyle \frac{1}{2i}\int_{\gamma (0,2)}\frac{z}{z-1}dz=\frac{1}{2i}(2\pi i)=\pi$