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I'd appreciate some help with this problem.

(a) Show that $$\frac{1}{(1-e^z)^n} - \frac{1}{(1-e^z)^{n+1}} \quad (n=1,2,3,\ldots)$$ has an anti-derivative whenever $e^z \neq 1$.

(b) Explain why $$\oint_{C} \frac{1}{(1-e^z)^n}dz = \oint_{C} \frac{1}{(1-e^z)^{n+1}}dz$$ for every natural number $n$ where $C$ is the circle $|z|=1$.

For (a) the term reduces to $$\frac{-e^z}{(1-e^z)^{n+1}} = \frac{d}{dz}\left[\frac{1}{(1-e^z)^n}\right]$$ for every natural number $n$ and is valid when $e^z \neq 1$. But how to prove it?

For (b) both integrands are discontinuous at $z=0$, so how do I proceed?

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up vote 0 down vote accepted

I don't see what there is left to prove in (a); the function $1\over (1-e^z)^n$ is defined on $\{z:e^z\neq 1\}$, and the derivative of this function is the function in your problem.

In (b), you are supposed to use the result in (a). Instead of showing that the integrals are equal, you can show that the difference between them is $=0$. Note that the difference of the integrals is just the integral of the function in (a), which you have shown has an anti-derivative.

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Ok, that clears it up. Thank you very much, Mr. Manne. – Pomegranate Sep 13 '12 at 12:29

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