Show that a function has essential singularity I had to find all singularities of a function $$ f(z)=\frac{(e^{iz}-1) e^{1/z}}{\sin z}$$
and determine their type.

Clearly, $f$ has singularities in $z=k\pi$, $k\in\mathbb{Z}$.
I've checked that if $k\ne0$, then $f$ has simple poles at such $z$. But what if $z=0$? I know that $\exp(1/z)$ has an essential singularity in $0$, but what about my function $f$? How can i show it...? Thanks for any help.
 A: As you already know that $z=0$ is an isolated singularity, you just have to examine the behavior of the function nearby. The standard method would be to analyze the Laurent series expansion of $f$ in some small annulus around $z=0$ (if there are infinitely many terms with negative exponent, then it is an essential singularity); but if you know something about essential singularities, you can take shorter paths.
Consider the sequenze $z_n=1/n$, you have:
$$f(z_n)=e^n\dfrac{e^{i/n}-1}{\sin 1/n}=e^n\dfrac{\cos(1/n)-1 + i\sin(1/n)}{\sin(1/n)}$$
so
$$\lim_{n\to\infty}|f(z_n)|=\lim_{n\to\infty}e^n\left|\dfrac{\cos(1/n)-1 + i\sin(1/n)}{\sin(1/n)}\right|=\lim_{x\to 0^+}e^{1/x}\left|\dfrac{\cos(x)-1+i\sin(x)}{\sin (x)}\right|=+\infty$$
as $e^{1/x}\to+\infty$, $(\cos(x)-1)/\sin(x)\to 0$ (you could say this is L'Hopital rule, or Taylor or multiply and divide by $1+\cos x$ or whatever you like) and $i\sin(x)/\sin(x)=i$.
However, for $w_n=i/n$, we have
$$f(w_n)=e^{-in}\dfrac{e^{-1/n}-1}{i\sinh(1/n)}$$
so
$$\lim_{n\to\infty}|f(w_n)|=1\;.$$
So, $z=0$ has to be an essential singularity, as we have two sequences of points, going to $z=0$, such that $|f|$ has different limits along them.
