ProofOfArgumentPrinciple 2004-09-04 01:55:33 2006-09-18 07:19:11 Proof argument principle 0 complex variables complex analysis complex integrals contour integration residues % 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 Since $f$ is meromorphic, $f'$ is meromorphic, and hence $f'/f$ is meromorphic. The singularities of $f'/f$ can only occur at the zeros and the poles of $f$. I claim that all singularities of $f'/f$ are simple poles. Furthemore, if $f$ has a zero at some point $p$, then the residue of the pole at $p$ is positive and equals the multiplicity of the zero of $f$ at $p$. If $f$ has a pole at some point $p$, then the residue of the pole at $p$ is negative and equals minus the multiplicity of the pole of $f$ at $p$. To prove these assertions, write $f(x) = (x-p)^n g(x)$ with $g(p) \neq 0$. Then $${f'(x) \over f(x)} = {n \over x-p} + {g'(x) \over g(x)}$$ Since $g(p) \neq 0$, the second term on the right hand side is not singular at $p$. The only singularity at $p$ comes from the first term. Since $n$ is either the order of the zero of $f$ at $p$ if $f$ has a zero at $p$ or minus the order of the pole of $f$ at $p$ is $f$ has a pole at $p$, the assertion is proven. By the Cauchy residue theorem, the integral $${1 \over 2 \pi i} \int_C {f'(z) \over f(z)} dz$$ equals the sum of the residues of $f'/f$. Combining this fact with the characterization of the poles of $f'/f$ and their residues given above, one deduces that this integral equals the number of zeros of $f$ minus the number of poles of $f$, counted with multiplicity.