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  $g(p)\ne 0$. Then

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

Since $g(p)\ne 0$, 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

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

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.