proof of argument principle

Since $f$ is meromorphic, $f^{\prime}$ is meromorphic, and hence $f^{\prime}/f$ is meromorphic. The singularities of $f^{\prime}/f$ can only occur at the zeros and the poles of $f$.

I claim that all singularities of $f^{\prime}/f$ are simple poles. Furthermore, 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

 ${f^{\prime}(x)\over f(x)}={n\over x\!-\!p}+{g^{\prime}(x)\over g(x)}$

Since $g(p)\neq 0$, the only singularity of $f^{\prime}\!/\!f$ at $p$ comes from the first summand. 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 $p$ if $f$ has a pole at $p$, the assertion is proven.

By the Cauchy residue theorem, the integral

 ${1\over 2\pi i}\int_{C}{f^{\prime}(z)\over f(z)}dz$

equals the sum of the residues of $f^{\prime}/f$. Combining this fact with the characterization of the poles of $f^{\prime}/f$ 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.

Title proof of argument principle ProofOfArgumentPrinciple 2013-03-22 14:34:32 2013-03-22 14:34:32 rspuzio (6075) rspuzio (6075) 11 rspuzio (6075) Proof msc 30E20 Cauchy’s argument principle