$f$ is holomorphic at $B(z_0,r)\setminus\{z_0\}$ and $f$ doesn't except real values – i.e $f(z)\notin\mathbb{R}$ for all $z\in \mathbb{R}$. Then $z_0$ is a removable singularity point ($f$ can be extended holomorphically in $z_0$).
Well I tried to use Riemann theorem, and show that $\exists 0<r'\le r$ such that $f$ is bounded in $B(z_0,r)$, but didn't succeed to do so. Formerly I solved a similar question which demand that $\Re(f)>0$ and then by defining $e^{-f(z)}$ which is holomorphic and bounded, which by taking $log$ holomorphic branch promises $f$ is holomorphic. Is there any manipulation or composition I may make to $f$, to get a similar results?
I also tried to assume that $f$ is not bounded, so in $B(z_0,r)$ one may find $z$ such that $|f(z)|$ is arbitrary big. However, is there any kind of intermediate value principle which assures that $f$ must "cross" the real line in case $|f(z)|$ is not bounded in $B(z_0,r)$?
Best Answer
EDITED: If $f$ has a pole at $z_0$, $1/f$ has a zero there, and by the Open Mapping Theorem $1/f$ would take all values in some interval near $0$.
If $f$ has an essential singularity at $z_0$, Picard says it can omit at most one value near $z_0$.
Removable is all that's left.
EDIT If you don't want to use the heavy artillery of Picard, note that if $f$ takes no real values, since $B(z_0,r)$ is connected the values it does take are either in the upper or lower half plane.