Prove that $\sum_{n=1}^\infty\frac{\sin(nz)}{2^n}$ is analytic on $\{z\in\mathbb{C}:|\operatorname{Im}(z)|<\log(2)\}$ 
Prove that $f(z)=\sum_{n=1}^\infty\frac{\sin(nz)}{2^n}$ is analytic on $A=\{z\in\mathbb{C}:|\operatorname{Im}(z)|<\log(2)\}$ 

I tried expanding $\sin(nz)$ in terms of $e^{inz}$ but that did not help me unless I am doing something wrong.  I know Weierstrass's M-test comes in to play.

We have
$$
f(z)=\frac{1}{2i}\sum_{n=0}^\infty\frac{e^{inz}-e^{-inz}}{2^n}
=\frac{1}{2i}\sum_{n=0}^\infty\left[\left(\frac{e^{iz}}{2}\right)^n-\left(\frac{e^{-iz}}{2}\right)^n\right]
$$
If $|e^{\pm iz}/2|=\frac{1}{2}e^{\mp\text{Im}(z)}<1$, i.e. $\mp\text{Im}(z)<\log2$ or $|\text{Im}(z)|<\log2$ then the series converges and 
\begin{eqnarray}
f(z)&=&\frac{1}{2i}\left[\frac{e^{iz}}{2}\frac{1}{1-e^{iz}/2}-\frac{e^{-iz}}{2}\frac{1}{1-e^{-iz}/2}\right]\cr
&=&\frac{1}{2i}\left(\frac{e^{iz}}{2-e^{iz}}-\frac{e^{-iz}}{2-e^{-iz}}\right)\cr
&=&\frac{2e^{iz}-1-2e^{-iz}+1}{2i(4-2e^{iz}-2e^{-iz}+1)}\cr
&=&\frac{4i\sin z}{2i(5-4\cos z)}=\frac{2\sin z}{5-4\cos z}.
\end{eqnarray}
This function is obviously analytic!
