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I am given the definition of the residue of a complex function $f$ at an isolated singularity $z_0$ to be the coefficient of $\frac{1}{z-z_0}$ in the Laurent series expansion of $f$ in some punctured ball $B^{\circ}(z_0,r)$ for some $r>0$.

In general, how would I find it for an essential singularity in general then? I can't find the laurent series expansion of $f$ at it s essential singularity and look at the coefficient can I? (as it's not the definition)

I ask because I saw @Daniel Fischer's answer to: What is the residue of this essential singularity?
where he used the Residue theorem but other answers considered the series expansion as well...

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If you have a series $\sum_{n=-\infty}^{\infty}a_n z^n$ that converges in $r < |z| < R$, then you can integrate over $|z|=a$ for any $r < a < R$, and interchange orders of integration and summation in order to get what you want. Every term involving $z^n$ will give an integral of $0$ except for $n=-1$ because $z^n$ has an obvious anti-derivative for all other values of $n$.

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Why can't you? Depending on the function $f$, it may or may not be easy to find the Laurent series expansion. For example, for $f(z) = e^{1/z}$ (which has an essential singularity at $z=0$), the Laurent series is $$ \sum_{n=0}^\infty \frac{z^{-n}}{n!}$$ so the residue is $1$. Other functions may give you more trouble.

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