I try to prove that function function $\tan(z)$ is bounded for all $\epsilon>0$ outside of $\epsilon$-neighborhood of its poles
My attempt is the following:
I have proved that $\tan z=-i\frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}$
Also We know this function is periodic and $\tan(z)$ has finite limit for $|Im z| \to \infty$
$\lim_{z \to \infty} -\frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}=-1$
How does the result follow from this? Can you help me please?
Thank you!
Because of the periodicity it suffices to show that $\tan$ is bounded on the set $$ \{ z \in \Bbb C \mid 0 \le \operatorname{Re} (z) \le \pi, |z - \pi/2| \ge \epsilon \} \, . $$ We split this into three sets: $$ \begin{align} A &= \{ z \in \Bbb C \mid 0 \le \operatorname{Re} (z) \le \pi, |z - \pi/2| \ge \epsilon, |\operatorname{Im}(z)| \le 1 \} \, ,\\ B &= \{ z \in \Bbb C \mid \operatorname{Im}(z) > 1 \} \, ,\\ C &= \{ z \in \Bbb C \mid \operatorname{Im}(z) < -1 \} \, . \end{align} $$
$\tan$ is bounded on $A$ as a continuous function on a compact set.
For $z=x+iy \in B$ with $y > 1$ we have $$ |\tan(z)| = \left| \frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}\right| = \left| \frac{e^{2iz}-1}{e^{2iz}+1}\right| \le \frac{1+|e^{2iz}|}{1-|e^{2iz}|} \\ = \frac{1+e^{-2y}}{1-e^{-2y}} \le \frac{1+e^{-2}}{1-e^{-2}} =: M \, . $$
Similarly it can be shown that also $|\tan(z)| \le M $ for all $z \in C$.
