singularities of the function $f(z)=z^{n}\sin\left ( \frac{1}{z} \right )$ $n\in \mathbb{N}-{0}$

Question

$$\require{cancel}$$ I want to study the singularities of the function

$$f(z)=z^{n}\sin\left ( \frac{1}{z} \right )$$ for $$n\in \mathbb{N}-\{0\}$$ and $$z\in \mathbb{C}$$

I don't understand how to grasp that the function has a essential singularity in $$z=0$$

From real analysis I remember that $$\lim_{x\to0}x^{n}\sin\left ( \frac{1}{x} \right )=0$$

I know that for essential singularities in $$z=z_{k}$$ $$\lim_{z\to z_{z}}\left | f(z) \right |=\cancel{\exists}$$

Where is the mistake or the contradiction? What about the residue?

Thank u so much.

Answer 1

If the singularity would be non-essential, the function $f(z)=z^m\sin \frac{1}{z}$ would be holomorphic for $m$ large enough. But this contradicts the identity theorem, because this function has infinitely many zeros in any neighborhood of $0$, because $f\Bigg(\cfrac{1}{\pi l}\Bigg)=0$ for any $l \in \mathbb Z$.

Answer 2

First Calculate over $\mathbb R$ $$ \lim_{y \to 0 \\ y\in \mathbb R} (iy)^n \sin(\frac{1}{iy}) $$ that is the limit over the imaginary axix, to see why your limit doesn't exist.

The singularity You know that $$\sin(z)= \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}$$

This is true for all $z \in \mathbb C$.

If $x \neq 0$ setting $z=\frac{1}{x}$ (which we can because we know the above relation holds for ALL $z$) you get

$$\sin(\frac{1}{x})= \sum_{k=0}^\infty (-1)^k \frac{1}{x^{2k+1}(2k+1)!}$$

Therefore, $$f(x)= \sum_{k=0}^\infty (-1)^k \frac{x^{n-2k-1}}{(2k+1)!}$$

if you split this series in the Taylor part ($n-2k-1\geq0$) and the negative Laurent part ($n-2k-1 <0$) you see why you have an essential singularity.

The residue is given by the coefficient of $x^{-1}$ which can be read from the series.