MomodromyTheorem 2004-10-17 03:37:09 2006-01-09 11:36:07 Theorem % 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 Let $C(t)$ be a one-parameter family of smooth paths in the complex plane with common endpoints $z_0$ and $z_1$. (For definiteness, one may suppose that the parameter $t$ takes values in the interval $[0,1]$.) Suppose that an analytic function $f$ is defined in a neighborhood of $z_0$ and that it is possible to analytically continue $f$ along every path in the family. Then the result of analytic continuation does not depend on the choice of path. Note that it is \emph{crucial} that it be possible to continue $f$ along all paths of the family. As the following example shows, the result will no longer hold if it is impossible to analytically continue $f$ along even a single path. Let the family of paths be the set of circular arcs (for the present purpose, the straight line is to be considered as a degenerate case of a circular arc) with endpoints $+1$ and $-1$ and let $f(z) = \sqrt{z}$. It is possible to analytically continue $f$ along every arc in the family except the line segment passing through $0$. The conclusion of the theorem does not hold in this case because continuing along arcs which lie above $0$ leads to $f(z_1) = +i$ whilst continuing along arcs which lie below $0$ leads to $f(z_1) = -i$.