# How do I show with Laurent Series Expansion that $1/z$ has a simple pole for $z=z_0=0$?

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

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.

• If you want to deduce the nature of the singularity at zero, you should center your Laurent series at zero. – D_S Jan 19 '18 at 5:01

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.