I was asked to determine the type of the isolated singularity of the function $f(z)$ at $z=0$: $$f\left(z\right)=\frac{z\left(e^{\frac{1}{z}}-1\right)}{e^z-1}$$ I know that if the function's limit doesn't exist at $z=0$, then we can say that the function has an essential singular point at $z=0$. I wasn't able to show that the limit doesn't exist. Also, I was wondering if there is a way to find the full Laurent expansion.
2026-03-30 07:28:19.1774855699
How can I find the limit of the complex function: $f\left(z\right)=\frac{z\left(e^{\frac{1}{z}}-1\right)}{e^z-1}$?
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