Having trouble finding the Laurent Series Expansion of $f(z)=\frac{ze^{z}}{(z-2)^{2}}$ on the region $0<|z-2|<R$

Question

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 $a_{n\geq 0}$

Answer 1

I finally figured out how to write this as a Laurent Series. I think I sent people down the wrong path focusing on the coefficients using residue theorem and whatnot. Both responses were on the right path apart from the residue stuff but not quite there based on what I found, and I've tested my series against the original function, using Maple symbolic software.

For the problem:

Find the Laurent Series Expansion for $\displaystyle f(z)=\frac{ze^{z}}{(z-2)^{2}}$ on the region $0<|z-2|<R$.

Start by manipulating $f(z)$ into a desirable form, which other answers were close to but it is not clear why one would need $z\to(z+2-2)$ in the numerator, so I left off that manipulation and achieved a correct series expansion. $\displaystyle f(z)=\frac{ze^{z}}{(z-2)^{2}}=\frac{ze^{z+2-2}}{(z-2)^{2}}=\frac{ze^{2}(e^{z-2})}{(z-2)^{2}}$

Then, for $\displaystyle e^{z}=1+\frac{z}{1!}+\frac{z^{2}}{2!}+\frac{z^{k}}{k!}+\cdots=\sum\limits_{k=0}^{\infty}\frac{z^{k}}{k!}$

We can write $\displaystyle e^{z-2}=1+\frac{z-2}{1!}+\frac{(z-2)^{2}}{2!}+\frac{(z-2)^{k}}{k!}+\cdots=\sum\limits_{k=0}^{\infty}\frac{(z-2)^{k}}{k!}$

Then, writing $\displaystyle \frac{ze^{2}}{(z-2)^{2}}(e^{z-2})$, and expanding the exponential:

$$\frac{ze^{2}}{(z-2)^{2}}\left(1+\frac{z-2}{1!}+\frac{(z-2)^{2}}{2!}+\frac{(z-2)^{k}}{k!}+\cdots\right)$$

Allowing us to write the expanded form of this series as:

$$\frac{ze^{2}}{(z-2)^{2}}+\frac{ze^{2}}{(z-2)}+\frac{ze^{2}}{2!}+\frac{ze^{2}(z-2)}{3!}+\frac{ze^{2}(z-2)^{2}}{4!}+\cdots$$

As a side note the Principal part is found to be:

$$\frac{ze^{2}}{(z-2)^{2}}+\frac{ze^{2}}{(z-2)}$$

And the Analytic part to be:

$$\frac{ze^{2}}{2!}+\frac{ze^{2}(z-2)}{3!}+\frac{ze^{2}(z-2)^{2}}{4!}+\cdots$$

Finally, this Laurent Series can be condensed to the following:

$$\sum\limits_{k=-2}^{\infty}\frac{ze^{2}(z-2)^{k}}{(k+2)!}$$

Answer 2

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?

Answer 3

Taking $\displaystyle e^2\left[\frac{1}{z-2}+\frac{2}{(z-2)^2}\right]\cdot \sum_{n=0}^{\infty}\frac{(z-2)^n}{n!}$ from your comment to another answer.

$\displaystyle e^2\left[\sum_{n=0}^{\infty}\frac{(z-2)^{n-1}}{n!}+ \sum_{n=0}^{\infty}\frac{2(z-2)^{n-2}}{n!}\right]$

Re-index
$\displaystyle \displaystyle e^2\left[\sum_{n=-1}^{\infty}\frac{(z-2)^{n}}{(n+1)!}+ \sum_{n=-2}^{\infty}\frac{2(z-2)^{n}}{(n+2)!}\right]$

$\displaystyle 2e^2(z-2)^{-2} + e^2\sum_{n=-1}^{\infty}\frac{(n+4)(z-2)^{n}}{(n+2)!}$

$\displaystyle c_{-2} = 2e^2, c_{-1} = 3e^2$

To follow up on the method in the original post.

Our general Laurent series
$\displaystyle f(z) = \sum_\limits{n=-\infty}^{\infty} c_n (z-a)^n$

Around a contour that includes $a$

$\oint_\gamma f(z)\ dz = (2\pi i) c_{-1}$

We can find the general term of the Laurent series if we multiply the function by the appropriate factor $(z-a)^k$

In this example:

$\displaystyle c_{-1} = \frac {1}{2\pi i} \oint \frac {ze^{z}}{(z-2)^2}$

$\displaystyle c_{n} = \frac {1}{2\pi i} \oint \frac {ze^{z}}{(z-2)^2}(z-2)^{-(n+1)}$

$\displaystyle c_{-2} = \frac {1}{2\pi i} \oint \frac {ze^{z}}{(z-2)} = 2e^2$

From the Cauchy integral formula.
$\displaystyle c_n = \frac {1}{n!}\frac {d}{dz^{n+1}} ze^{z}|_2$

$\displaystyle c_{-1} \frac {d}{dz} ze^{z}|_{2} = (1+z)e^{z}|_2 = 3e^2$

For $n\ge 1$

$\displaystyle c_n = \frac {n+4}{n!} e^{2}$

This approach is more cumbersome but more generalizable.