Is $z=0$ an essential singularity of $z\sin(1/z)$?

Question

I know the Laurent series of $z\sin(1/z)$ has infinite terms of $(1/z^k)$ at $z = 0$. The thing is:

$$\lim_{z\to 0}z\sin\frac 1 z= 0$$

So is it a removable or an essential singularity?

Answer 1

The limit applies for real arguments $x$ as $x$ approaches $0$. But in the complex domain $z$ can approach $0$ along additional paths off the real axis, where the function grows rather than shrinking in size. Put in $z=iy, y \rightarrow 0$, and see what happens.