Let $f(z)=\cfrac{e^{-3z}}{z^2(z-2)^2}$, find the Laurent series about $z=0$.
On the region $0<|z|<2$, I get
$\cfrac{1}{(z-2)^2}=\displaystyle\sum_{n=1}^{\infty}\cfrac{nz^{n-1}}{2^{n+1}}$,
then $\cfrac{1}{z^2(z-2)^2}=\displaystyle\sum_{n=1}^{\infty}\cfrac{nz^{n-3}}{2^{n+1}}$
And I get confused here, is it ok if I let
$f(z)=\cfrac{e^{-3z}}{z^2(z-2)^2}=e^{-3z}\displaystyle\sum_{n=1}^{\infty}\cfrac{nz^{n-3}}{2^{n+1}}$
Or I have to express $e^{-3z}$ as its Taylor series and the multiplicate?
Thank you
You have $$f(z)=\cfrac{e^{-3z}}{z^2(z-2)^2}=\Big(\sum_{m=0}^\infty \frac{(-3)^m }{m!}z^m\Big)\times \Big(\sum_{n=0}^\infty \frac{n}{2^{n+1}}z^{n-3}\Big)=\sum_{i=-2}^\infty c_i z^i$$ and you look for the general expression of the $c_i$'s.
For better legibility, we shall write $$\sum_{i=-2}^\infty c_i z^i=\Big(\sum_{m=0}^\infty a_m z^m\Big)\times \Big(\sum_{n=0}^\infty b_n z^{n-3}\Big)$$
In order to have a degree $(k-3)$, you must add several terms the product of whicb making $z^k$ and then you should arrive to something like $$c_{k+3}=\sum_{m=0}^k a_m \,b_{k-m}$$