# 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 follows from the residue theorem.

Title zero as contour integral ZeroAsContourIntegral 2013-03-22 16:46:42 2013-03-22 16:46:42 rspuzio (6075) rspuzio (6075) 5 rspuzio (6075) Corollary msc 30E20