Why is this an essential singularity? I have the following complex function: $$f(z)=\frac{\cos\left(\frac{\pi z}{2}\right)\sin\left(\frac{1}{z}\right)}{(z^2-1)(z-2)}$$
I kind of shown that $z=\pm 1$ and $z =2$ are simple poles. However I don't know how to show that $z=0$ is an essential singularity. How do I even find the Laurent series of $sin\left(\frac{1}{z}\right)$ around $0$?
I know started from the Taylor expansion $$\sin(z)=\sum_{n=0}^{\infty}\frac{(-1)^kz^{2k+1}}{(2k+1)!}$$ and then I wrote it for $\sin\left(\frac{1}{z}\right)$ $$\sin\left(\frac{1}{z}\right)=\sum_{n=0}^{\infty}\frac{(-1)^k}{(2k+1)!z^{2k+1}}$$ but then I don't know how to find the Laurent series of it, shouldn't it be from $-\infty$ to $\infty$? All I can do here seems to be $$\sin\left(\frac{1}{z}\right)=\sum_{n=0}^{\infty}\frac{(-1)^kz^{-(2k+1)}}{(2k+1)!}$$ which has negative exponents only?
Can you help me?
 A: $z = 0$ is an essential singularity because if you develop the function
$$\sin \frac{1}{z}$$
In powers of $z$, then ALL the terms show a pole at $z = 0$, and there is no way to remove them.
Hence it's indeed essential.
P.s. The Lurent series of $\sin$ is the actual Taylor series. 
A: Since
$$
g(z)=\frac{\cos(\pi z/2)}{(z^2-1)(z-2)}
$$
is holomorphic in a neighborhood of $0$ and $g(0)=1/2$, this factor can be removed for discussing the singularity at $0$ of $f$. Thus we remain with
$$
h(z)=\sin\frac{1}{z}
$$
which is certainly holomorphic in $0<|z|<1$, so $0$ is an isolated singularity. It is not removable, so it is either essential or a pole. If the latter, there should exist a positive integer $n$ such that
$$
\lim_{z\to0}z^n\sin\frac{1}{z}
$$
exists (finite) and is not $0$.
However, if we approach $0$ along the real axis, we have
$$
\lim_{x\to0}x^n\sin\frac{1}{x}=0
$$
for every positive integer $n$. Hence the singularity is essential.

For the singularity at $1$, you should consider
$$
\lim_{z\to1}\frac{\cos(\pi z/2)}{z-1}=
\lim_{w\to0}\frac{\cos(\frac{\pi}{2}-w)}{-\frac{2w}{\pi}}=
\lim_{w\to0}-\frac{\pi}{2}\frac{\sin w}{w}=-\frac{\pi}{2}
$$
so $1$ is a removable singularity for $f$ and the same for $-1$.
If you want to see a different approach, let's look at $z=-1$. Granted that
$$
\frac{1}{(z-1)(z-2)}\sin\frac{1}{z}
$$
is holomorphic in a neighborhood of $-1$, we need to consider
$$
g(z)=\frac{\cos(\pi z/2)}{z+1}
$$
of which we want to find the Laurent series at $-1$; setting $z+1=w$, we have
$$
\cos\left(\frac{\pi}{2}w+\frac{\pi}{2}\right)=-\sin\frac{\pi w}{2}
$$
so
$$
-\frac{1}{w}\sin\frac{\pi w}{2}=
-\frac{1}{w}\left(\frac{\pi}{2}w-
  \frac{\pi^3}{2^3\cdot 3!}w^3+\dotsb\right)=
-\frac{\pi}{2}+\frac{\pi^3}{48}w^2+\dotsb
$$
so the singularity is removable.
