How do I prove that this limit does not exist over complex field? $lim_{zrightarrow 0} z^2 sin{left(frac{1}{z}right)}$

So, here's the problem. I'm trying to figure out the type of singularity at $z=0$ to find a residue at the given point.
One of the possible approaches is find a limit of the function as $z$ tends to the singularity point, and based on the outcome, you could decide, whether it's a pole, an essential singularity or a removable singularity.

Based on Laurent's series, I expect this to be an essential singularity:
$f\left(z\right)=z-\frac{1}{6 z}+\frac{1}{120 z^3}-\frac{1}{5040 z^5}+\frac{1}{362880 z^7} + \frac{\left(-1\right)^{n}\left(\frac{1}{z}\right)^{2n-1}}{\left(2n+1\right)!}$,
since the series has infinitely many negative degree terms (i.e., the principal part of the Laurent series is an infinite sum). If that's so, the aforementioned limit has to be non-existent at $z=0$.

I tried to get an approach to the limit like this:
since $\left|\sin{\left(\frac{1}{z}\right)}\right|\le 1$, we have $-1\le\sin{\left(\frac{1}{z}\right)}\le1$. Multiplying by $z^2$ we get the following: $-z^2\le z^2\sin{\left(\frac{1}{z}\right)}\le z^2$. Since both $z^2$ and $-z^2$ are equal to zero as z tends to $0$, so is the $z^2 \sin{\left(\frac{1}{z}\right)}$ according to the squeeze theorem.
So, actually the limit exists.

However, shortly after this I realised that this approach is of no use over complex field since there's no such thing as order for complex numbers, which is why my approach has nothing to do with the actual problem, and the whole thing was solved over reals.

So, I'm puzzled. How do I prove that this limit does not exist at all?

Answer

Let $z=x\in \mathbb{R}$. Then, clearly

$$\lim_{z\to 0}z^2\sin(1/z)=\lim_{x\to 0}x^2\sin(1/x)=0$$

Now, let $z=iy$, $y\in \mathbb{R}$, $y>0$. Then, clearly

$$\lim_{z\to 0}z^2\sin(1/z)=\lim_{y\to 0^+}y^2\sin(i/y)=i\lim_{y\to 0^+}y^2\sinh(1/y)=i\infty$$

Inasmuch as the two limits are unequal, the limit $\lim_{z\to 0}z^2\sin(1/z)$ fails to exist.

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The Laurent series for $z^2\sin(1/z)$ is given by

$$z^2\sin(1/z)=\sum_{n=0}^\infty \frac{(-1)^n z^{-(2n-1)}}{(2n+1)!}\tag1$$

The residue is equal to the coefficient on the term $z^{-1}$, which occurs in $(1)$ when $n=1$. Hence, we see that

$$\text{Res}\left(z^2\sin\left(\frac1z\right),z=0\right)=-\frac16$$