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Suppose that $f$ is holomorphic in a disk $D$ without zeroes on the boundary $\partial D$. Denote $S_k(f) = \frac{1}{2\pi i}\int\limits_{\partial D} \frac{f'}{f}z^k dz$. How can I show that $S_k(f) = \sum\limits_i \alpha_i^k$, where $\alpha_i$ are zeroes of $f$ counted with multiplicites?

It looks like a corollary from Cauchy's integral formula.

Hasek
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    It's a generalized argument principle. See https://math.stackexchange.com/questions/1803460/help-to-understand-the-generalization-of-the-argument-principle and https://math.stackexchange.com/questions/1183864/generalization-of-the-argument-principle. – lhf Apr 14 '17 at 12:03
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    Just calculus of residues: what are the poles and residues of $z^k f'/f $ ? – Angina Seng Apr 14 '17 at 12:07
  • @LordShark, poles are zeroes $\alpha_i$ of $f$ and value of a residue in a pole of order $k$ is $\alpha_i^k$. – Hasek Apr 14 '17 at 14:00

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