Let f(z,w) be holomorphic in $\mathbb{C}^{n}$ and not identically zero on the w-axis. Let {$b_{j}$} be the set of singularities of f(z,w) in some disk of radius $|w| < r$. Why does the residue theorem imply that $b_{1}^{m} + ... + b_{k}^{m} = \frac{1}{2\pi i}\int_{|w| = r} \frac{w^{m}\frac{\partial f}{\partial w}(z,w)}{f(z,w)} dz$? I'm reading the proof of the Weierstrass preparation theorem and I can't see how the formula $\sum Res = \frac{1}{2\pi i} \int f(z) dz$ implies the above.


  • $\begingroup$ You are integrating over a circle in the $w$-plane with a $dz$-integral? Also, when you are talking about singularities $b_j$, I assume one of the variables is kept fixed and the other one varies? Which one is which? $\endgroup$ – Lukas Geyer May 20 '15 at 19:56

Assuming they are all nonremovable simple poles, you see that $\text{Res}(\frac{w^m\frac{df}{dz}f(z,w)}{f(z,w)},b_k)=\frac{w^m\frac{df}{dz}f(z,w)}{\frac{df}{dz}f(z,w)}\vert_{w=b_k}=b_k^m$. Thus, plug right into to Cauchy's Residue Theorem.

  • $\begingroup$ Oh okay, I see how the first equality follows now. Thanks $\endgroup$ – user238194 May 20 '15 at 20:02

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