How to find the set $E$ of convergence of the series $\sum_{n=1}^{\infty}\frac{\sin nz}{n^{2}}$ . Is the convergence uniform on $E$? I tried to find the set $E$ of convergence of the series 
$$\sum_{n=1}^{\infty}\frac{\sin nz}{n^{2}}$$ 
where $z$ is complex number but get stuck. I know that 
$$\sin(x+iy) = \sin x\cosh y+i\cos x\sinh y$$ 
but not very clear how to apply it. Can anyone help me out?
And also, is the convergence uniform on $E$ ? 
Thank you guys a lot.
 A: Note that we can write
$$\begin{align}
\sum_{n=1}^N \frac{\sin(nz)}{n^2}&=\frac1{2i}\sum_{n=1}^N \frac{e^{inz}-e^{-inz}}{n^2}\\\\
&=\frac1{2i}\sum_{n=1}^N \frac{e^{-ny}e^{inx}}{n^2}-\frac1{2i}\sum_{n=1}^N \frac{e^{ny}e^{-inx}}{n^2}\tag1
\end{align}$$
The first sum on the right-hand side of $(1)$ converges absolutely for all $x$ and all $y\ge0$, converges uniformly for all $x$ and all $y\ge y_0^+>0$, and diverges for $y<0$.  
Analogously, the second sum on the right-hand side of $(1)$ converges absolutely for all $x$ and all $y\le0$, converges uniformly for all $x$ and all $y\le y_0^-<0$, and diverges for $y>0$.
The only common domain on which both sums converge is $\{z| x\in \mathbb{R}$, $y=0\}$.  On this domain, the series of interest converges absolutely.
If we restrict the domain to real numbers only, the series $\sum_{n=1}^\infty\frac{\sin(nx)}{n^2}$ as a function of $x$ uniformly converges for all $x$ as guaranteed by the Weierstrass M-Test.
But, for $z\in \mathbb{C}$, the series fails to converge uniformly for any $z$ since in every neighborhood of any point for which the series converges contains a point for which the series diverges. 
