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For the function

$$f(z) = \frac{1}{z} + b$$

what type of singularity is $z = 0$? I would say that it is a pole singularity, as $|f(z)| \rightarrow \infty$ as $z \rightarrow z_0$, but a problem that I am looking at hints to apply Casorati-Weierstrass to $f$, which implies that $z = 0$ is in fact an essential singularity.

How do I tell which one it is?

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If $\;b\;$ is a constant, there you have your Laurent series:

$$\frac1z+b$$

so by definition $\;z=0\;$ is a simple pole.

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Wikipedia says:

If $U$ is an open subset of the complex plane $C$, $a$ is an element of $U$ and $f : U $ \ $\ \{a\} → C$ is a function which is holomorphic over its domain. If there exists a holomorphic function $g : U → C$ and a positive integer $n$, such that for all $z$ in $U$ \ $\{a\}$: $$f(z)=\frac{g(z)}{(z-a)^n}$$

Substituiting $g(z) = 1 + bz, a = 0, n = 1$ we will get the initial expression for $f(z)$, so it's a simple pole.

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