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$\gamma_1$ is the circle |z| = 4 and $\gamma_2$ is the boundary of the square with sides along x = plus and minus 1, and y = plus and minus 1.

Show that $$\int_{\gamma_1} f(z)dz = \int_{\gamma_2} f(z)dz$$

when $$f(z) = \frac{z}{1-e^z}$$

So there is a singularity at z = 0. But I cant see how to get the denominator into a form that will let me apply Cauchy's Integral Formula.

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The singularity is removable. (Recall the Taylor expansion of the exponential function.) Also note the multiples of $2\pi i$ are outside of both $\gamma_1$ and $\gamma_2$. –  anon Apr 24 '12 at 0:11
    
In case this is homework, can you add the homework tag please? –  Gerry Myerson Apr 24 '12 at 0:40
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By the way, you have been posting an alarming number of complex analysis and linear algebra questions here lately. I'm not sure that's an appropriate use of m.se - it might be better to go see a teacher or hire a tutor instead. –  Gerry Myerson Apr 24 '12 at 0:45
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@GerryMyerson Its not homework, its exam preparation. I have end of year exams in less than a week, you wont have to read anymore complex analysis questions from me after that as i am planning to study other topics over the summer. If people dont want to read my questions no one is forcing them to, but I thought the whole point of m.se was for helping less knowledgeable people out and facilitating faster learning. After the exams I was planning on thanking some specific people out here who have helped me over the last few weeks. like anon above, he/she has been a really big help. Cheers anon! –  Jim_CS Apr 24 '12 at 10:26
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Good luck on your exams. –  Gerry Myerson Apr 24 '12 at 13:15
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