Laurent series for coshz

Question

I'm having a difficulty approaching this problem. Any help would be highly appreciated. So I want to show $cosh(z+\frac{1}{z})=\sum b_n(z^n+\frac{1}{z^n})$, where $b_n=\frac{1}{2\pi} \int cosn\theta cosh(2cos \theta)d\theta$.

How can I approach this problem? I have tried writing out $cosh(z+\frac{1}{z})$ as its Laurent series, but not sure how do get the coefficients..

Answer 1

First, recall that

$$\cosh(a+b)=\cosh(a)\cosh(b)+\sinh(a)\sinh(b)$$

And apply Cauchy products, using the well known

$$\cosh(x)=\sum_{n=0}^\infty\frac{x^{2n}}{(2n)!}$$

$$\sinh(x)=\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}$$