Why does $\frac{|f(z)-f(z_0)|}{|f(z)-\overline{f(z_0)}|}\leq\frac{|z-z_0|}{|z-\bar{z}_0|}$ when $\mathrm{Im}z>0\implies\mathrm{Im}f(z)\geq 0$? I'm trying to understand the following inequality. Let $f$ be holomorphic, such that $\mathrm{Im}f(z)\geq 0$ when $\mathrm{Im}(z)>0$. Why is it that
$$
\displaystyle\frac{|f(z)-f(z_0)|}{|f(z)-\overline{f(z_0)}|}\leq\frac{|z-z_0|}{|z-\bar{z}_0|}?
$$
Since $f$ maps the upper half plane to itself, I was thinking of mapping the plane to the unit disk by some linear fractional, and then attempt to use Schwarz' lemma somehow. I haven't been able to execute a good plan. 
Does anyone have any hints and/or solutions to show this inequality? Thank you.

I assume you mean this to hold for $\Im(z) \geq 0$ and also that $\Im(z_0) > 0$ and $\Im(f(z)) > 0$ for all $z \in \mathbb{R}$.  The functions
$$
P(z) = \frac{f(z)-f(z_0)}{f(z)-\overline{f(z_0)}}
$$
and 
$$
Q(z) = \frac{z-z_0}{z-\overline{z_0}}
$$
are holomorphic on the closed upper half plane and satisfy:


*

*$P(z_0)=Q(z_0)=0$.

*If $\Im(z) \geq 0$ then $|P(z)| \leq 1$ and $|Q(z)| \leq 1$.


Moreover, $Q$ has a single simple zero at $z_0$ and if $z \in \mathbb{R}$ then $|Q(z)|=1$.  Therefore
$$
\frac{P(z)}{Q(z)}
$$
is holomorphic on the closed upper half plane and for all $z \in \mathbb{R}$
$$
\left| \frac{P(z)}{Q(z)} \right| = |P(z)| \leq 1.
$$
By the maximum modulus principle this equality must hold on all of the upper half plane.  In other words, $|P(z)| \leq |Q(z)|$ for all $z$ with $\Im(z) \geq 0$.
