How to find the Laurent Serie of the function $f(z)=e^{z+1/z}$ around zero. Then, show that:
$$\dfrac{1}{2\pi}\int_0^{2\pi}{e^{2\cos\theta}\cos{n\theta}}=\sum_{j=0}^{\infty}{\dfrac{1}{(n+j)!j!}}$$
What i have done: i first started working on $e^{z+1/z}$ $$z=x+iy=e^{i\theta}=\cos\theta+i\sin\theta$$ $$e^{z+1/z}=e^z\cdot e^{1/z}$$ $$e^{z}=e^{\cos(\theta)+i\sin(\theta)}$$ $$e^{1/z}=e^{\cos(-\theta)+i\sin(-\theta)}$$ $$e^{z+1/z}=e^z\cdot e^{1/z}=(e^{\cos(\theta)+i\sin(\theta)})(e^{\cos(-\theta)+i\sin(-\theta)})=$$ $$=e^{2\cos{\theta}}$$
But i have seen also that: $e^z = e^{x+iy} = e^x e^{iy} = e^x(\cos y + i\sin(y))$ so $e^{1/z} = e^{-x-iy} = e^{(-x)} e^{-iy} = e^{(-x)}(\cos (-y) + i\sin(-y))$, which gives: $$e^{z+1/z}=(\cos{y}-i\cdot \sin{y})(\cos{y}+i\cdot \sin{y})$$
I can´t go further.
Laurent's theorem: Let $f$ be analytic in the annulus $r<\left | z-z_0 \right |<R$, then $f$ can be expressed there as the sum of two series $$f(z)=\sum_{n=-\infty }^{\infty }c_nz^n=\sum_{n=1 }^{\infty }c_{-n}\frac{1}{z^n}+c_0+\sum_{n=1 }^{\infty }c_nz^n$$ both series convering in the annulus, and converging uniformly in any closed subannulus. The coefficients are given by $$c_n=\frac{1}{2\pi i}\oint_{C}^{ }\frac{f(z)}{(z-z_0)^{n+1}}dz,\;n=0,\pm 1,\pm 2, \dots$$ where $C$ is any positively oriented simple closed contour lying in the annulus and containing $z_0$ in its interior.
Solution:
Using the fact that $$e^{z}=\sum_{n=0}^{\infty}\frac{1}{n!}z^n=1+z+\frac{1}{2!}z^2+\frac{1}{3!}z^3+\frac{1}{4!}z^4+\dots$$ we can wrtie $$e^{z+\frac{1}{z}}=\sum_{n=0}^{\infty}\frac{1}{n!}\left ( z+\frac{1}{z} \right )^n=1+\left (\frac{1}{z}+z \right )+\frac{1}{2!}\left ( \frac{1}{z^2}+2+z^2 \right )+\frac{1}{3!}\left ( \frac{1}{z^3}+\frac{3}{z}+3z+z^3 \right )+\frac{1}{4!}\left ( \frac{1}{z^4}+\frac{4}{z^2}+6+4z^2+z^4 \right )+\dots$$
Compare the above expansion with the equation mentioned in the first theorem.
For example, let us find $c_0$ (coefficient of terms without any $z$): $$c_0 = 1+ \frac1{2!} 2 + \frac1{4!} 6 + \cdots = \sum_{j=0}^{\infty} \frac1{(2 j)!} \binom{2 j}{j} =\sum_{j=0}^{\infty} \frac1{(j!)^2} = I_0(2)$$ where $I_0(2)$ is the modified Bessel function of the first kind (of zero order.)
Similarly, you can find that for any $n$: $$c_n= \sum_{j=0}^{\infty} \frac1{(n+2 j)!} \binom{n+2 j}{j} = \sum_{j=0}^{\infty} \frac1{(n+j)! j!} = I_n(2)$$ where $I_n(2)$ is the modified Bessel function of the first kind (of $n$th order.)
Therefore the Laurent series becomes $$f(z)=\sum_{n=-\infty }^{\infty }c_nz^n=\sum_{n=1 }^{\infty }I_n(2)\frac{1}{z^n}+I_0(2)+\sum_{n=1 }^{\infty }I_n(2)z^n= I_0(2)+ \sum_{n=1 }^{\infty }I_n(2)\left ( \frac{1}{z^n}+z^n \right )$$
We now use the integral equation (see Laurent's theorem): \begin{align} \require{cancel} c_n &=\frac{1}{2\pi i}\oint_{C}^{ }\frac{f(z)}{(z-z_0)^{n+1}}dz \\ & =\frac{1}{2\pi i}\oint_{\left | z \right |=1}^{ }\frac{e^{z+\frac{1}{z}}}{z^{n+1}}dz \\ & =\frac{1}{2\pi i}\int_{0}^{2\pi}\frac{e^{e^{i\theta}+e^{-i\theta}}}{e^{i\left (n+1 \right )\theta}}ie^{i\theta}d\theta \\ & =\frac{1}{2\pi }\int_{0}^{2\pi}e^{2cos\theta}e^{-in\theta}d\theta \\ & = \frac{1}{2\pi }\int_{0}^{2\pi}e^{2cos\theta}\cos n\theta \;d\theta-\cancelto{0}{\frac{1}{2\pi }\int_{0}^{2\pi}e^{2cos\theta}\sin n\theta\;d\theta} \\ & =\frac{1}{2\pi }\int_{0}^{2\pi}e^{2cos\theta}\cos n\theta \;d\theta \\ & =\sum_{j=0}^{\infty} \frac1{(n+j)! j!} \end{align}