Show that the singular point of the following functions is a pole and verify using Laurent’s expansion

Question

Given that I have $\frac{1-e^{2z}}{z^3}$ and $\frac{e^{2z}}{(1-z)^2}$, how would I go about verifying it in each case using Laurent's expansion? I can find the residues and I know the singular points are poles, just unsure of how to show it.

Answer 1

Remember that $z_0$ is a pole with order $k$ for $f$ if and only if $$f(z)=\sum\limits_{n\in\Bbb{Z}}a_n(z-z_0)^n$$ has $a_n=0$ for all $n<k$ and $a_k\neq 0$.

Answer 2

In the first case you have$$\frac{1-e^{2z}}{z^3}=-\frac{2z+\frac{2^2}{2!}z^2+\frac{2^3}{3!}z^3+\cdots}{z^3}=-\frac2{z^2}+\frac1z+\frac83+\cdots$$In the second case, use the fact that$$\frac{e^{2z}}{(1-z)^2}=e^2\frac{e^{2(z-1)}}{(z-1)^2}$$and do the same thing as above.