Contour Integral: $\int_{c} e^{(z+1/z)} dz$ where $c$ is a unit circle.

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From Cauchy's Residue theorem, I know that the value of integral is equal to: $$2\pi*i*\sum{R}$$ where $R$=Residues of poles inside the contour. And $z_0=0$ being the only residue inside the unit circle, I wrote the Laurent series of the function $f(z)= e^{(z+1/z)}$ which has coefficient of $1/z$ as a sum of series $(1+3+5+7+..=\infty)$. So the given integral blows up.

Need to confirm if my answer is correct or not.

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$\textbf{Hint:}$ Instead of residue theorem consider that on the unit circle

$$\oint_C e^{z+\frac{1}{z}}dz = \oint_C e^{z+\bar{z}}dz$$

Then use the theorem

$$\oint_{\partial \Omega} f\:dz = 2i\iint\limits_\Omega \frac{\partial f}{\partial \bar{z}}\:dx dy$$

(where $z=x+iy$) and integrate over the disk in Cartesian coordinates. You will immediately get an integral representation of a Bessel function.