Singularities of exp(z/sin z) I want to know all the sigularities of the complex valued function f(z)=exp(z/sin z) and their nature?
 A: You could start with the singularities of $1/\sin z$ or the zero's of $\sin z
$. The latter are given by $z=0,\pm \pi ,\pm 2\pi \cdots $. Taking $z=x$ real
\begin{equation*}
\exp [z/\sin z]\rightarrow \exp [\frac{x+k\pi }{\sin (x+k\pi )}]
\end{equation*}
For $k=0$ this remains well-behaved but for $k\neq 0$ this is not the case.
\begin{equation*}
\frac{1}{\sin (x+k\pi )}=\frac{1}{\sin x\cos (k\pi )+\cos x\sin (k\pi )}
\overset{x\rightarrow k\pi }{\sim }\frac{1}{\sin x\cos (k\pi )}=(-1)^{k}
\frac{1}{\sin x}
\end{equation*}
Thus if $x\downarrow 0$ and $k$ is even this object becomes large positive
whereas it becomes large negative for odd $k$. On the other hand, if $
x\uparrow 0$ the situation is reversed and we conclude that $\exp [z/\sin z]$
\ becomes singular in all $z=\pm \pi ,\pm 2\pi \cdots $.
A: Polynomial functions of $z$, the sine function of $z$, and the exponential function of $z$ are holomorphic. 
The inverse function $z^{-1}$ is not, and as shown by @Urgje, trouble arises when when $z = \pm k \pi,$ $k=1,2,3,\dots$.
The conformal map shows 
$$
 f(z) = \exp\left( {\frac{z}{\sin z}} \right)
$$
with Re $z$ in blue, and Im $z$ in red,

The singularity at $z=0$ is removable by defining $f(0) = e$. The remaining singularities are essential. 
