Why is this an essential singularity?

Question

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?

Answer 1

$z = 0$ 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.

Answer 2

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.

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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.