Laurent expansion of ##ze^{1/z}##

[Terrell]

Homework Statement

Find a Laurent series of $f(z)=ze^{1/z}$ in powers of $z-1$. Is there an easier way to go about this as this is not a typical expansion I see on textbooks. It seems that my incomplete solution is too complicated. Please help, exam is in two days and I am working on past exams. Worked out solutions are welcome, also. P.S. it may seem like I'm cramming, but it's not under my control -

Homework Equations

The Attempt at a Solution

Since $f(z)$ is entire, then it is analytic in any annular domain. So let's pick the domain $\frac{1}{2} \lt \vert z-1\vert \lt 3$. By Laurent's theorem the coefficients
\begin{align}a_k=\frac{1}{2\pi i}\int_{c}\frac{(1+e^{it})e^{\frac{1}{1+e^{it}}}}{e^{it\cdot k}}\end{align}
where $C$ is the contour $z(t)=1+e^{it}$ and hence, $dz=ie^{it}dt$. Note that the contour $C$ is within our selected annular domain as it should be according to the theorem.

 

[Charles Link]

I think I see a trick to this one: $f'(z)=(z-1)f(z)$, and if you do the algebra, and I think you can show $f^n(z)=(z-1)f^{n-1}+(n-1) f^{n-2}(z)$. $\\$That would make it suitable for a Taylor expansion about $z=1$. Note that $f^n(1)=(n-1)f^{n-2}(1)$. $\\$ It looks like the odd number derivatives are zero and the even ones are $(n-1)! =(n-1)(n-3)(n-5)...(1)$. (I think I have the correct definition of n!=Edit: yes I googled it: see http://mathworld.wolfram.com/DoubleFactorial.html) And hopefully I don't get dinged for supplying a solution to this, because other than this approach, I wouldn't know what to do with this one.

 

[Terrell]

f′(z)=(z−1)f(z)

I don't see why this is true. I got \begin{align}f'(z)=e^{1/z}-\frac{e^{1/z}}{z}\end{align}
But $(z-1)f(z)=(z-1)ze^{1/z}=z^2e^{1/z}-ze^{1/z}$

 

[Charles Link]

I made an algebraic error here: $f'(z)=\frac{(z-1)}{z^2} f(z)$. Oh well, back to the drawing board.

 

[Charles Link]

I think a similar method might still be possible: Using $f'(z)=f(z)g(z)$ where $g(z)=\frac{1}{z}-\frac{1}{z^2}$, so that $f'=fg$, $f''=f'g+fg'$, etc, I think you should be able to get some kind of expression for $f^n(z)$.

 

[Terrell]

I think a similar method might still be possible: Using f′(z)=f(z)g(z)f′(z)=f(z)g(z) f'(z)=f(z)g(z) where g(z)=1z−1z2g(z)=1z−1z2 g(z)=\frac{1}{z}-\frac{1}{z^2} , so that f′=fgf′=fg f'=fg , f′′=f′g+fg′f″=f′g+fg′ f''=f'g+fg' , etc, I think you should be able to get some kind of expression for fn(z)fn(z) f^n(z) .

Is there a standard general approach you are trying to follow? Thanks!

 

[Charles Link]

Is there a standard general approach you are trying to follow? Thanks!

I am just trying to find something that works. Had I not made a simple algebraic error, the first solution that I had would have been a good one. Upon further inspection, I'm not having any luck with anything that works...

 

[Terrell]

I am just trying to find something that works. Had I not made a simple algebraic error, the first solution that I had would have been a good one. Upon further inspection, I'm not having any luck with anything that works...

I find it hard to believe they put this on a past exam. It's suppose to be a test of understanding, not puzzles. I'll try my hand at this again later, I'm working on other problems for the moment. Thank you!

 

[Charles Link]

I find it hard to believe they put this on a past exam. It's suppose to be a test of understanding, not puzzles. I'll try my hand at this again later, I'm working on other problems for the moment. Thank you!

I hope the people who designed the problem didn't make the same mistake that I did in thinking there was a simple solution. It's unlikely, but certainly quite possible.

 

[Terrell]

I hope the people who designed the problem didn't make the same mistake that I did in thinking there was a simple solution. It's unlikely, but certainly quite possible.

I would be outrage! These are comprehensive exams and they don't get returned to us.

 

[Charles Link]

@Orodruin You have much more math expertise. Does this one have a simple solution that we are missing?

 

[stevendaryl]

Here's my short-cut way to an answer:

$f(z) = z e^\frac{1}{z}$ has a singularity only at $z=0$. Which means that it doesn't have any singularity at $z=1$. If it doesn't have a pole at $z=1$, that means that it can be expanded in an ordinary Taylor series, using just positive powers of $z-1$.

Given that, I think I would just let $z = u+1$ and then our function becomes $f(u) = (1+u) e^{\frac{1}{1+u}}$. You can find the Taylor series by just taking derivatives, rather than doing complex integrals. I guess it depends on whether the assignment is intended to give you practice at doing complex integration.

 

[Charles Link]

With a Taylor expansion, and taking derivatives, I was unable to succeed at getting a simple formula for the series. This only worked when I made an algebraic error that was clearly wrong. Perhaps the OP or someone else can find such a formula.