A meromorphic function $f$ is called meromorphic herglotz function if $\mathrm{Im}(z)>0$ implies $\mathrm{Im}(f(z))>0$
I need to prove that all the poles and zeros of $f$ are in $\mathbb{R}$. Morover, each pole and zero is simple and the poles and zeros alterante.
There is a proof here, but I can't understand why $\operatorname{arg}(f)$ takes all the values in $[0,2\pi)$ and why that implies that all the zeros and poles are in $\mathbb{R}$.
If a meromorphic function $f$ has a pole or zero at $z_{0}$, then, for a unique non-zero integer $n$, $$ f(z) = (z-z_{0})^{n}g(z) $$ where $g$ is holomorphic near $z_{0}$ with $g(z_{0})\ne 0$. Then $$ f(z) = (z-z_{0})^{n}g(z_{0})+(z-z_{0})^{n+1}\left[\frac{g(z)-g(z_{0})}{z-z_{0}}\right]. $$ By choosing $z=re^{i\theta}+z_{0}$ with $r$ small enough, you can arrange for the first term to dominate so that the image of $C_{r}=\{ re^{i\theta}+z_{0} : 0 \le \theta \le 2\pi \}$ is very nearly circular.
