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I'm confused about how we are supposed to use Laurent expansion of a function to tell types of its singularity.

For example, the book says if $f(z)$'s Laurent expansion about $z_0$ has only finite terms in the principal part, then $z_0$ is a pole.

However, if we expand $\frac{1}{1-z}$ around $z=0$ on the annulus $|z|>1$, we will have infinite terms: $\sum_{n=0}^{\infty}-\frac{1}{z^n}$. The book then moves on to say $z=1$ is an essential singularity which is apparently wrong!

So does the typical "Laurent expansion and singularity relation" only holds when we expand function around an isolated singularity?

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Your Laurent series expansion of $\displaystyle \frac{1}{1-z}$ around $z=0$ is correct which you have written.

$\bullet$ But , at $z=0$ , $\displaystyle \frac{1}{1-z}$ has NO singularity, it is analytic at $z=0$ ; so the question of any type singularity does not arise.

$\bullet$ $z=1$ is a pole if $\displaystyle \frac{1}{1-z}$ , NOT an essential singularity ; because Laurent series expansion of $\displaystyle \frac{1}{1-z}$ around $z=1$ is itself $\displaystyle \frac{1}{1-z}=-\frac{1}{z-1}$. Clearly it has only one term in principal part , around z=1$.

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