zero as contour integral

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

$$p=\frac{1}{2\pi i}{\oint }_C\frac{z{f}^{\prime }(z)}{f(z)}𝑑z$$

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 $z{f}^{\prime }(z)/f(z)$ at $p$ is $p$, so the above follows from the residue theorem.

Title	zero as contour integral
Canonical name	ZeroAsContourIntegral
Date of creation	2013-03-22 16:46:42
Last modified on	2013-03-22 16:46:42
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	5
Author	rspuzio (6075)
Entry type	Corollary
Classification	msc 30E20