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# zero as contour integral

Suppose that $f$ is a complex function which is defined in some open set $D\subseteq\mathbb{C}$ which has a simple zero at some point $p\in D$. Then we have

$p={1\over 2\pi i}\oint_{C}{zf^{{\prime}}(z)\over f(z)}\,dz$ |

where $C$ is a closed path in $D$ which encloses $p$ but does not enclose or pass through any other zeros of $f$.

This follows from the Cauchy residue theorem. We have that the poles of $f^{{\prime}}/f$ occur at the zeros of $f$ and that the residue of a pole of $f^{{\prime}}/f$ is $1$ at a simple zero of $f$. Hence, the residue of $zf^{{\prime}}(z)/f(z)$ at $p$ is $p$, so the above formula follows from the residue theorem.

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## Mathematics Subject Classification

30E20*no label found*

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