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$$ f: \mathbb{C} \backslash \{ z_0 \} \rightarrow \mathbb{C} \; , \; z \mapsto \frac{1}{z} \; , \; z_0 = 0 $$

To show that $f$ has a simple pole for $z=z_0=0$ I'm doing the Laurent Series Expansion. That's what I have thus far:

$$ \frac{1}{z} = \frac{1}{1+(z-1)} = \frac{1}{1-(-(z-1))} = \frac{1}{1-(1-z)} = \frac{a_0}{1-q} $$

$$ \text{with} \;\;\; a_{0} = 1 \;,\;\;\; q = 1-z $$

$$ \overset{|1-z|<1}{\Rightarrow} \;\; \frac{1}{z} = \sum^{\infty}_{k=0}a_0 q^{k} = \sum^{\infty}_{k=0}(1-z)^{k} $$

So how do I show that $\frac{1}{z}$ has neither a removable singularity nor an essential singularity, but a non-removable singularity pole of order 1 for $z=z_0=0$? This should be very easy now... but I don't get it.

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  • $\begingroup$ If you want to deduce the nature of the singularity at zero, you should center your Laurent series at zero. $\endgroup$ – D_S Jan 19 '18 at 5:01
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In a punctured neighborhood of zero, $$\frac{1}{z} = \sum\limits_{n \in \mathbb{Z}} a_nz^n$$ for uniquely determined coefficients $a_n$. By uniqueness, $a_{-1} = 1$, and all the other terms are zero.

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  • $\begingroup$ Of course... makes sense! Thank you. $\endgroup$ – integralette Jan 19 '18 at 9:13

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