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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.

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  • $\begingroup$ Formatting tip: When entering trig functions, notice that adding \ before the functions looks much better see $sin x$ gives $sin x$ whereas $\sin x$ gives $\sin x$. The same goes in the case of log functions, see $\log x$ and $\ln x$ give $\log x$ and $\ln x$ and look better, as opposed to $log x$ and $ln x$ which give $log x$ and $ln x$. $\endgroup$ – user409521 Apr 5 '17 at 16:24
  • $\begingroup$ @InfiniteMonkey I can't understand. Could you explain more about it? $\endgroup$ – Parting Apr 5 '17 at 16:51
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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.

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