I'm just not sure how to finish out this problem.
This is how the problem was given:
Find the Laurent Series Expansion for $\displaystyle f(z)=\frac{ze^{z}}{(z-2)^{2}}$ on the region $0<|z-2|<R$.
I can see it has a pole at $z_{0}=2$ of order $n=2$.
I understand we can write the Laurent Series for $0<|z-z_{0}|<R$ of a pole of order $n$ as the following:
$\displaystyle \frac{a_{-n}}{(z-z_{0})^{n}} + \frac{a_{-(n-1)}}{(z-z_{0})^{n-1}} + \cdots +\frac{a_{-1}}{(z-z_{0})} + a_{0} + a_{1}(z-z_{0}) + a_{2}(z-z_{0})^{2} + a_{3}(z-z_{0})^{3} +\cdots$
With $z_{0}=2\text{, and }n=2$ we would find the following Laurent Series:
$\displaystyle \frac{a_{-2}}{(z-2)^{2}} + \frac{a_{-1}}{(z-2)} + a_{0} + a_{1}(z-2) + a_{2}(z-2)^{2} + a_{3}(z-2)^{3}\cdots$
If I understand correctly we stop at $-2$ on the negative side because all $n<-2$ returns a zero for that coefficient. However, on the positive side we take $n\to \infty$ so we can rewrite the series with $z_{0}=2\text{, and }n=2$ plugged in and get the following:
$$\displaystyle \frac{a_{-2}}{(z-2)^{2}} + \frac{a_{-1}}{(z-2)} + a_{0} + a_{1}(z-2) + a_{2}(z-2)^{2} + a_{3}(z-2)^{3}\cdots$$
$$=\sum_{n=-2}^{-1}a_{n}(z-2)^{n} + \sum_{n=0}^{\infty}a_{n}(z-2)^{n}$$
This is where I get confused. I see in my text that for a simple pole we can find:
using $\displaystyle f(z)=\frac{ze^{z}}{(z-2)^{2}}$
(I included the simple pole because the book explicitly states this is $a_{-1}$, but I know this does not apply to a pole of order $n$)
$\displaystyle a_{-1}=\text{Res}(f(z),z_{0})=\lim\limits_{z\to z_{0}}(z-z_{0})f(z)$
and for the Residue at a Pole of order $n$ we have
$\displaystyle \text{Res}(f(z),z_{0})=\frac{1}{(n-1)!}\lim\limits_{z\to z_{0}}\left(\frac{d^{n-1}}{dz^{n-1}}\left[(z-z_{0})f(z)\right]\right)$
But what coefficient is the Residue of a Pole of order $n$? $a_{-2}\text{ or }a_{-1}$? and what if the pole had a greater order, $n>2$, resulting in many more $a_{-n}$ terms?
Then, how do I find $a_{n}\text{ for }n\geq 0$?
EDIT:
The textbook I reference throughout this post is
Complex Analysis A First Course with Applications by Dennis Zill & Patrick Shanahan, 3ed
EDIT 2 using information from @user317176's response:
I found $a_{-1}$ using the residue theorem
$$a_{-1}=\text{Res}\left(\frac{ze^{z}}{(z-2)^{2}},z_{0}=2\right)$$
$$a_{-1}=\frac{1}{(2-1)!}\lim\limits_{z\to 2}\left(\frac{d^{2-1}}{dz^{2-1}}\left[(z-2)^{2}\frac{ze^{z}}{(z-2)^{2}}\right]\right)$$
$$a_{-1}=\frac{1}{1!}\lim\limits_{z\to 2}\left(\frac{d^{1}}{dz^{1}}\left[(z-2)^{2}\frac{ze^{z}}{(z-2)^{2}}\right]\right)$$
$$a_{-1}=\lim\limits_{z\to 2}\left(\frac{d}{dz}\left[ze^{z}\right]\right)$$
$$a_{-1}=\lim\limits_{z\to 2}\left(e^{z}+ze^{z}\right)$$
$$a_{-1}=e^{2}+2e^{2}$$
$$a_{-1}=3e^{2}$$
still trying to figure out how to work out $a_{-2}$ and $a_{n\geq 0}$
Best Answer
Note that we can write
$$\begin{align} \frac{ze^z}{(z-2)^2}&=\frac{(z-2+2)e^{(z-2+2)}}{(z-2)^2}\\\\ &=e^2\left[\left(\frac{ e^{z-2}}{z-2}\right)+2\left(\frac{e^{z-2}}{(z-2)^2}\right)\right] \end{align}$$
Now use the Taylor series $e^{z-2}=\sum_{n=0}^\infty \frac{(z-2)^n}{n!}$.
Can you finish now?