Why does the pole of $\frac{1-e^{2z}}{z^3}$ have order $2$?

Question

Why does pole at $z=0$ have order $2$, I ask this because as far as I can see it would have order $3$ as the denominator is of order $3$.

But when I input it into wolfram alpha it says it has order $2$. Any help would be much appreciated.

Thanks.

Answer 1

Because $$\frac{1-e^{2z}}{z^3} = 2\frac{1-e^{2z}}{2z}\frac{1}{z^2} $$ and $\frac{1-e^{2z}}{2z}$ goes to $-1$ when $z$ goes to $0$.

Answer 2

$$ e^{2z} =1 +2z+\frac{(2z)^2}{2!}+\cdots $$ So $$ \frac{1-e^{2z}}{z^3}=-\frac{\left(2+\frac{4z}{2}+\cdots \right)}{z^2} $$ revealing the pole of order 2 at $z=0$.