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(a) Let $f$ be a holomorphic function in $|z| < 2$ such that (*) $\int_{|z| = 1} \frac{f(z)}{nz-1} = 0$ for each $n = 2, 3, ....$ Prove that $f(z) \equiv 0$ in $B(0; 2)$.

(b) If we only assume that $f$ is a holomorphic function defined in $0 < |z| < 2$ such that (*) holds, can we still draw the conclusion that $f \equiv 0$ ?


I have managed to solve part (a) by using Cauchy's Integral Formula as well as the Identity Theorem.

However, how to go about solving part (b)? I have tried to use the Laurent series expansion but I wouldn't know whether I am in the right direction.

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(a) The integral is essentially the residue of $c$ at $\frac1n$ if $f$ is holomorphic in the open disc. If $f(\frac1n)$ were nonzero for some $n$, we'd obtain nonzero residue. Hence $f(\frac1n)=0$ and by the identity theorem, $f\equiv 0$, as you correctly found out.

(b) Now the integral computes the residue at $\frac1n$ plus the residue at $0$. We conclude that for all $n$ we have $$\operatorname{Res}(\frac{f(z)}{nz-1}, \frac1n)+ \operatorname{Res}(\frac{f(z)}{nz-1}, 0)=0$$ The first summand is simply $\frac1nf(\frac1n)$, the second is $-\operatorname{Res}(f(z), 0)$ if $f$ has a simple pole at 0. Would $f(z)=\frac1z$ make a fine example?

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Alright thanks for your example. :) –  Markeur Oct 23 '12 at 10:42

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