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I need to show that \begin{align} \exp \left(a z+b z^{-1}\right)&=\sum_{n=-\infty}^{\infty} a_{n} z^{n} \;\;\; a, b \in \mathbb{C} \\\ a_{n}&=\frac{1}{2 \pi} \int_{0}^{2 \pi} e^{(a+b) \cos \theta} \cos \Big[(a-b) \sin \theta-n \theta\Big] d \theta \end{align} I am stuck at one of my arguments, which I find shaky. What I have so far is: I note that f(z) is a composition of a polynomial, with the exponential function, it is holomorphic on the punctured disk K'(a,r) centered at the point a = 0 which is a singularity. This is analogous to the annulus $$A(a, R_1, R_2) , R_1 \rightarrow 0, R_1 \rightarrow \infty$$There exists a Laurent series for the function at a. The general term a_n is given by

\begin{align} a_n &= \frac{1}{2\pi i} \int_{\partial K(a,r)}\frac{f(z)}{(z-a)^{n+1}}dz \\\ &= \frac{1}{2\pi i} \int_{\partial K(0,r)}\frac{\exp \left(a z+b z^{-1}\right)}{z^{n+1}}dz\\\ &= \frac{1}{2\pi i} \int_{0}^{2\pi}\frac{\exp \left(a r e^{i\theta}+b (re^{i\theta})^{-1}\right)}{(re^{i\theta})^{n+1}}r i e^{i\theta}d\theta \\\ &= \frac{1}{2\pi} \int_{0}^{2\pi} e^{\left(a r e^{i\theta}+b (re^{i\theta})^{-1}\right)}(re^{i\theta})^{-n}d\theta \\\ &= \frac{1}{2\pi} \int_{0}^{2\pi} e^{\big( a r [cos(\theta)+i\sin(\theta)]+b [cos(\theta)-i\sin(\theta)]r^{-1} \big)} (re^{i\theta})^{-n} d\theta \\\ &= \frac{1}{2\pi} \int_{0}^{2\pi} e^{\big( (a r+b r^{-1})cos(\theta)+i(a r-b r^{-1})\sin(\theta)\big)} (re^{i\theta})^{-n} d\theta \\\ &= \frac{1}{2\pi} \int_{0}^{2\pi} e^{(a r+b r^{-1})\cos\theta}e^{i(a r-b r^{-1})\sin\theta}e^{-i\theta n} r^{-n} d\theta \\\ &= \frac{1}{2\pi} \int_{0}^{2\pi} e^{(a r+b r^{-1})\cos\theta}e^{i\big[(a r-b r^{-1})\sin\theta -\theta n\big]} r^{-n} d\theta \\\ &= \frac{1}{2\pi} \int_{0}^{2\pi} e^{(a r+b r^{-1})\cos\theta} \Big(\cos\left[(a r-b r^{-1})\sin\theta-\theta n\right] + i\sin\left[(a r-b r^{-1})\sin\theta - \theta n\right] \Big) r^{-n} d\theta \end{align} $f(z)$ has a primitive (as it is the composition of the exponential with polynomials),so the last term of the integral is zero, since it is the evaluation of its primitives at the upper and lower boundaries of the integral. The primitive of $i\sin z$ is $i \cos z$, so evaluating the last term at $F(0)$ and $F(2\pi)$ yields $F(2\pi) - F(0) = i\cos(2\pi n)-i\cos(0) = i - i = 0$, and thus \begin{align} a_n &= \frac{1}{2\pi} \int_{0}^{2\pi} e^{(a r+b r^{-1})\cos\theta} \cos\left[(a r-b r^{-1})\sin\theta-\theta n\right] r^{-n} d\theta \end{align} Now i need to get rid of the r's... I somehow want to just say they are equal to one, since the closed path integral is independent of the path. We are integrating around the boundary of the disc, a closed circle, and the integral must be independent of r. This means it must hold for any r. Is this reasoning correct?

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  • $\begingroup$ Please see DLMF equation 10.12.1 about the g.f. of Bessel functions. $\endgroup$
    – Somos
    Mar 17, 2020 at 19:34
  • $\begingroup$ Sory, I am unsure how I am supposed to use that. I didn't learn about bessel functions yet, as far as i am aware. $\endgroup$
    – Achnos
    Mar 18, 2020 at 9:53

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I am unsure how you get rid of the last term. How do you find the primitive when the last term also contains $e^{(ar+br^{-1})\cos(\theta)}$ from outside the parenthesis?

Your reasoning for getting rid of the r's is definitely correct

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  • $\begingroup$ Hmm, i didn't actually think about that, so maybe that part is not correct $\endgroup$
    – Achnos
    Mar 21, 2020 at 14:15

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