# Is my laurent series expansion for $f(z)=\frac{e^z}{(z-1)^2}$ for $|z-1|>0$ about $z=1$ correct?

Hi could anyone help me with this past exam question (we weren't provided answers)

Find the Laurent series expansion of $$f(z)=\frac{e^z}{(z-1)^2}$$ for $$|z-1|>0$$ about $$z=1$$

I have tried to answer it but unsure if it is right?:
$$f(z)=\frac{1}{(z-1)^2}e^{z+1-1}$$
$$=\frac{e}{(z-1)^2}e^{z-1}$$
$$=\frac{e}{(z-1)^2}\sum^{\infty}_{n=0}\frac{(z-1)^n}{n!}$$
$$=e\sum^{\infty}_{n=0}\frac{(z-1)^{n-2}}{n!}$$

If this is incorrect please could you point me in the right direction? Thanks:)

It's fine. But such a Laurent series is usually expressed as$$\sum_{n=-2}^\infty\frac e{(n+2)!}(z-1)^n,$$so that the exponents of the $$(z-1)$$'s are equal to $$n$$.