Finding contour integral of $e^{t(z+z^{-1})}z^{-2}$ I am really struggling with a contour integration question, which I am revising for an exam. I want to show that the contour integral of 
$$\int_\Gamma e^{t(z+z^{-1})}z^{-2} dz=\sum_{m=0}^\infty b_mt^{2m+1}$$ where the $b_m$ are constants to be found, and $\Gamma$ is the positively oriented circle with centre $0$ and radius $R$. 
I was wondering, as I instinctivey noticed that $z+z^{-1}=2cos(\theta)$ where, $z=Re^{i\theta}$ if we should rewrite this integral, so that it becomes $$\int _0^{2\pi}\frac{e^{2tRcos(\theta)}}{(Re^{i\theta})^2(iRe^{i\theta})}d\theta$$
But then, I don't see that this has any poles at all? Any help appreciated, thank you. 
 A: $$e^{tz} = \sum_{n=0}^{+\infty}t^n\frac{z^n}{n!}\,,$$
$$e^{t/z} = \sum_{m=0}^{+\infty}t^m\frac{z^{-m}}{m!}\,.$$
Therefore, the integrand is:
$$e^{tz}e^{t/z}z^{-2}=\sum_{n=0}^{+\infty}\sum_{m=0}^{+\infty}
t^{n+m}\frac{z^{n-m-2}}{n!m!}\,.$$
For each value of $m$, we pick a residue from the singularity at the origin only when $n=m+1$, (with residue $t^{2m+1}/m!(m+1)!$). Therefore the value of the integral is:
$$ \sum_{m=0}^{+\infty}\frac{2\pi i}{m!(m+1)!}t^{2m+1}\,.$$
The value of $b_m$ is:
$$b_m = \frac{2\pi i}{m!(m+1)!}$$
A: Hint:
$$e^{t\left(z+z^{-1}\right)}z^{-2}=z^{-2}\sum_{n=0}^{\infty}\dfrac{t^{n}}{n!}\left(z+z^{-1}\right)^{n}=\\\\=z^{-2}\sum_{n=0}^{\infty}\dfrac{t^{n}}{n!}\sum_{k=0}^{n}\binom{n}{k}z^{-k}z^{n-k}=\sum_{n=0}^{\infty}\dfrac{t^{n}}{n!}\sum_{k=0}^{n}\binom{n}{k}z^{n-2\left(k+1\right)}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\oint_{\Gamma}{\expo{t\pars{z + z^{-1}}} \over z^{2}}\,\dd z & =
\oint_{\Gamma}{\expo{tz}\expo{t/z} \over z^{2}}\,\dd z
\\[5mm] & =
\oint_{\Gamma}{\bracks{\sum_{m = 0}^{\infty}\pars{tz}^{m}/m!} \bracks{\sum_{n = 0}^{\infty}\pars{t/z}^{n}/n!} \over z^{2}}\,\dd z
\\[5mm] & =
\sum_{m = 0}^{\infty}\sum_{n = 0}^{\infty}
{t^{m + n} \over m!\, n!}\
\overbrace{\oint_{\Gamma}{\dd z \over z^{2 - m + n}}}
^{\ds{2\pi\ic\,\delta_{2 - m + n,1}}}
\\[5mm] & =
\sum_{m = 0}^{\infty}
{2\pi\ic \over m!\pars{m - 1}!}\,t^{m + \pars{m - 1}}\bracks{m - 1 \geq 0}
\\[5mm] & =
\bbx{\sum_{m = 0}^{\infty}\
\underbrace{2\pi\ic \over \pars{m + 1}!\,m!}_{\ds{ b_{m}}}\ t^{2m + 1}}
\end{align}
