Laurent Series Expansion of $e^{(\frac{w}{2})(z-z^{-1})}$

Question

I'm trying to expand the following in the region $|z|>0$ $$f(z)=e^{(\frac{w}{2})(z-z^{-1})}$$ to get its Laurent series expansion. If $a_n$ is the coefficient of the $n^{th}$ term in the expansion, I should be able to get, $$a_n = \sum_{k=0}^{\infty}\frac{(-1)^k(w/2)^{n+2k}}{k!(n+k)!}$$ My Approach so far:

By Maclurin expansion we have,
$$f(z) = \sum_{n=0}^{\infty}\left(\frac{wz}{2}-\frac{w}{2z}\right)^n\frac{1}{n!} $$
Now by Binomial theorem, $$f(z) = \sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{n!}{(n-k)!k!}\left(\frac{wz}{2}\right)^k\left(\frac{-w}{2z}\right)^{n-k}\frac{1}{n!} $$
$$f(z) = \sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{(-1)^{n-k}}{(n-k)!k!}\left(\frac{w}{2}\right)^nz^{2k-n} $$
How can I proceed after this step? Or am I doing something completely wrong?

Answer 1

You can also use the product:

$$f(z)=e^{wz/2}e^{-wz^{-1}/2}=\\\left(\sum_{k=0}^\infty\frac1{k!}\left(\frac{w}2\right)^kz^k\right)\left(\sum_{j=0}^{\infty}\frac1{j!}\left(-\frac w2\right)^j z^{-j}\right)$$

Combining:

$$\sum_{j,k} \frac{1}{k!j!}(-1)^j\left(\frac w2\right)^{j+k}z^{k-j} $$

Letting $n=k-j,$ you get:

$$\sum_{n=-\infty}^{\infty}z^n\sum_{k=\max(n,0)}^{\infty}\frac1{k!(k-n)!}(-1)^{k-n}\left(\frac w2\right)^{2k-n}$$

Everything converges absolutely here, so all these steps are allowed.

I don’t think you are going to get a closed formula for the coefficient of $z^n:$

$$a_n(w)=\sum_{k=\max(n,0)}^{\infty}\frac1{k!(k-n)!}(-1)^{k-n}\left(\frac w2\right)^{2k-n}$$ I can’t even find a closed form for the case $n=0.$ When $n=0,$

But it is an entire function in $w.$ It is odd when $n$ is odd and even when $n$ is even.

$$a_n(w)=\sum_{p=0}^\infty\frac{(-1)^{p}}{2^{n+2p}}\binom{n+2p}p \frac{w^{n+2p}}{(n+2p)!}$$

Since $f(z)f(z^{-1})=1,$ we get some conditions on the $a_n.$

When $n=-m<0,$ then:

$$a_n(w)=\sum_{k=0}^{\infty}\frac{(-1)^{m+k}}{2^{m+2k}}\binom{2m+k}{k}\frac{ w^{2k+m}}{(2k+m)!}$$

This shows that for $m\geq 0,$ $$a_{-m}(w)=(-1)^m a_{m}(w)=a_{m}(-w).$$

That’s because, if we write the function as $F(w,z),$ $F(w,z^{-1})=F(-w,z).$