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.