We have to show that$\displaystyle\sum_{n=1}^\infty\frac{sin( nz)}{n}$ diverges if $Im\,(z)\neq0$
My try:$$\sin (nz) = \mathfrak{Im}(e^{nzi})$$
$$\sum_{n=1}^{\infty} \frac{\sin nz}{n} = \mathfrak{Im} \sum_{n=1}^{\infty} \frac{(e^{zi})^n}{n} = - \mathfrak{Im} \ln\left(1-e^{zi}\right). $$ Then how to proceed?
The statement $\sin(nz)=\text{Im}\{e^{inz}\}$ is true if $z\in \mathbb{R}$, but is not true if $\text{Im}(z)\ne 0$ since then $\sin(nz)\in \mathbb{C}$ with non-zero imaginary part.
Let $z=x+iy$. Then, note that $\sin(nz)=\sin(nx)\cosh(ny)+i\cos(nx)\sinh(ny)$.
If $y\ne 0$, then the general term $a_n=\frac{\sin(nx)\cosh(ny)+i\cos(nx)\sinh(ny)}{n}$ of the series does not approach $0$ as $n\to \infty$.
Hence, the series diverges.
Based on the same idea than Dr. MV.
$$\sin z = \frac{e^{iz}-e^{-iz}}{2i}$$ hence for $z=x+iy$
$$\vert \sin z \vert \ge \frac{1}{2} \left(\vert e^{y} \vert -\vert e^{-y} \vert\right).$$ Consequently $$\frac{\sin nz}{n} \ge \frac{1}{2} \frac{e^{ny}}{n}$$ diverges for $y >0$. And similar for $y <0$.