TITLE: Computing the Laurent series of $e^{z+\frac{1}{z}}$
QUESTION [7 upvotes]: I'm really stuck on this, and I have no idea how to start. Writing it at $f(z)=e^z e^{\frac{1}{z}}$ and their expansions didn't really give any insight. I am aware it is possible to multiply the series of these functions, but this was not covered in this course. I am allowed to assume $f$ is holomorphic on $\mathbb{C}\setminus\{0\}$.

REPLY [8 votes]: Let $C_{n}$ ($n \in \mathbb{Z}$) be a coefficient of the Laurent expansion of $e^{z+\frac{1}{z}}$ at the origin.
By definition,
$$ C_{n} = \frac{1}{2 \pi i} \int_{|z|=1} \frac{e^{z+\frac{1}{z}}}{z^{n+1}} \ dz$$
since the unit circle lies in an annulus in which $e^{z+\frac{1}{z}}$ is analytic.
Let $z= e^{i \theta}$.
Then $$C_{n} = \frac{1}{2 \pi i} \int_{-\pi}^{\pi} \frac{e^{e^{i \theta}+e^{- i \theta}}}{e^{i(n+1) \theta}} i e^{i \theta} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{2 \cos \theta} e^{-in \theta}$$
$$ = \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{2 \cos \theta} \cos(n \theta) \ d \theta =  \frac{1}{\pi} \int_{0}^{\pi} e^{2 \cos \theta} \cos (n \theta) \ d \theta$$
But for an integer $n$, the modified Bessel function of the first kind has the integral representation 
$$I_{n}(x) = \frac{1}{\pi} \int_{0}^{\pi} e^{x \cos \theta} \cos(n \theta) d \theta$$
http://dlmf.nist.gov/10.32
Therefore,
$$ C_{n} = I_{n}(2)$$
Also notice from the integral representation that $I_{n}(x) = I_{-n}(x)$.