Let $p,q\in\mathbb{C}, p\neq q$ and $G = \mathbb{C}\setminus\left\{p,q\right\}$ and $f: G\to \mathbb{C}$ analytic and bounded. Show that $f$ is constant.
Hi. How could I start from this problem? I suspect that with Cauchy's modification ...
actualization.
$\lim_{z\to p}|(z-p)|f(z)|\leq \lim_{z\to p} |(z-p)|M=0$ because $f$ is bounded
therefore $p$ removable singularity, analogous $q$.
thefore exists $g:\mathbb{C}\to\mathbb{C}$ analytic and bounded. By Liouville theorem, $g$ is constant then $f$ is constant. Thanks to José Carlos Santos
By Riemann's theorem on removable singularities, you can extend $f$ to an entire function $F$, which will still be bounded. Therefore, $F$ is constant (by Liouville's theorem) and so, in particular, $f$ itself is constant.