ArgumentPrinciple 2004-09-04 01:33:32 2007-04-07 17:29:23 Algorithm argument principle argument complex anaysis contour integration % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \PMlinkescapeword{origin} \PMlinkescapeword{side} \PMlinkescapeword{sides} \PMlinkescapeword{formula} \PMlinkescapeword{relation} If a function $f$ is meromorphic on the interior of a rectifiable simple closed curve $C$, then \begin{align}\label{eq:arg-princ} {1 \over 2 \pi i} \oint_C {f'(z) \over f(z)} dz \end{align} equals the difference between the number of zeros and the number of poles of $f$ counted with multiplicity. (For example, a zero of order two counts as two zeros; a pole of order three counts as three poles.) This fact is known as the \emph{argument principle}. The principle may be stated in another form which makes the origin of the name apparent: If a function $f$ is meromorphic on the interior of a rectifiable simple closed curve $C$ and has $m$ poles and $n$ zeros on the interior of $C$, then the argument of $f$ increases by $2 \pi (n - m)$ upon traversing $C$. The relation of this statement to the previous statement is easy to see. Note that $f'/f = (\log f)'$ and that $\log (z) = \log |z| + i \arg z$. Substituting this into formula \eqref{eq:arg-princ}, we find \[ 2 \pi i (n - m) = \oint_C {f'(z) \over f(z)} dz = \oint_C d \log |f(z)| + i \oint_C d \arg (f(z))\,. \] The first integral on the rightmost side of this equation equals zero because $\log|f|$ is single-valued. The second integral on the rightmost side equals the change in the argument as one traverses $C$. Cancelling the $i$ from both sides, we conclude that the change in the argument equals $2 \pi (n - m)$. Note also that the integral \eqref{eq:arg-princ} is the winding number, about zero, of the image curve $f \circ C$.