Why z=0 is a simple pole of $\frac{z}{1-e^{z^2}}$

Question

Why $z=0$ is a simple pole of $\frac{z}{1-e^{z^2}}$
I understood why for $z\neq 0 $ it is a simple pole but for the other part I didn't an explanation will be nice :)

Answer 1

It suffices to show that $\lim_{z \to 0} z\left(\frac{z}{1 - e^{z^2}}\right) = L \neq 0$ (i.e. the limit exists, is finite, and is non-zero). We can simply apply L'Hopital Rule: $$ \lim_{z \to 0} \frac{z^2}{1 - e^{z^2}} = \lim_{z \to 0} \frac{2z}{ - 2ze^{z^2}} = \lim_{z \to 0} -e^{-z^2} = -1 $$ Hence it's a pole of order $1$.

Answer 2

By L'Hopital's Rule or series expansion you can see that $\frac {e^{z}-1} z \to 1$ as $ z \to 0$. If $f$ is your function it follows that $zf(z) \to -1$ as $z \to 0$. This implies that $f$ has a pole of order $1$at $0$.

Answer 3

Take the Laurent expansion

$$-z^{-1}\left(1+\frac{1}{2}z+\frac{1}{6}z^2+\cdots\right)^{-1} = -z^{-1}\left(1-\frac{1}{2}z-\frac{1}{6}z^2+\frac{1}{4}z^2+\cdots\right)$$

$$=-\frac{1}{z} + \frac{1}{2}-\frac{1}{12}z+O(z^2)$$

by geometric series. It is clearly a simple pole at $z=0$ with residue $-1$

Answer 4

Because

$$\lim_{x\to0}z\cdot\frac z{1-e^{z^2}}=\lim_{x\to0}\frac{z^2}{1-e^{z^2}}=\lim_{x\to0}\frac{2z}{-2ze^{z^2}}=\lim_{x\to0}-\frac1{e^{z^2}}=-1\neq0$$