Expand $$f(z)=\frac {e^{2z}}{(z-1)^3}$$ about $z=1$ as a Laurent;s series. Also indcate the region of convergence.
My attempt :
Let the given complex function be analytic in an annulus $r<|z-1|<R$ Then $f(z)$ can be expanded into a laurent's series about $z=1$
Let $z-1=u$ then $z=u+1$ Putting in $f(z)$ and expanding, i get
$$f(z)=\frac {1}{u^3}[1+2(u+1)+2(u+1)^2+...]$$
Does it look good ?
A better approach is to factor out an $e^{2}$ before using the series expansion for $e^{2z}$
So you get: $$f(u)=\frac{e^{2u+2}}{u^{3}}=\frac{e^{2}e^{2u}}{u^{3}}=\frac{e^{2}}{u^{3}}[1+2u+2u^{2}+\frac{4}{3}u^{3}+...]$$
This way you don't have to worry about all the $u+1$ terms.