Complex Integration of DTFT Question
A discrete-time signal $u \in \mathcal{l}^2(\mathcal{Z})$ has DTFT
\begin{equation}
\hat{u}(\omega) = \frac{5+3\cos(\omega)}{17+8\cos(\omega)}
\end{equation}
Use complex integration to find $u(k)$ for $k\in\mathcal{Z}$

My Attempt
If I'm not mistaken, to find $u[k]$, I should find the inverse Fourier Transform
\begin{equation}
u(k) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{5+3\cos(\omega)}{17+8\cos(\omega)} e^{j\omega k} d\omega
\end{equation}
Is this where I use complex integration?  In order to do so, doesn't my equation need poles?  I can't think of when $17+8\cos(\omega)$ would equal zero for this to not be analytic.  

Using the residue theorem (complex integration) on integrals of rational functions of sines and cosines is commonplace.  The trick is to let $z=e^{i \omega}$.  Then $d\omega = -i dz/z$ and the integral may be expressed as a complex integral over the unit circle:
$$\begin{align}u(k) &= -\frac{i}{2 \pi} \oint_{|z|=1} \frac{dz}{z} \frac{5+\frac{3}{2} (z+z^{-1})}{17+4 (z+z^{-1})} z^k\\ &= -\frac{i}{4 \pi} \oint_{|z|=1} dz \, z^{k-1} \frac{3 z^2+10 z+3}{4 z^2+17 z+4}\end{align}$$
Because $k \in \mathbb{Z}$, we have no branch points.  Consider $k \ge 1$.  Then the only poles in the integrand are where
$$4 z^2+17 z+4=0 \implies z_{\pm} = \frac{-17 \pm 15}{8} $$
i.e., $-1/4$ or $-4$, respectively.  As only $z_+=-1/4$ is the only pole within the unit circle, this is the only pole contributing to the integral.  
By the residue theorem, the integral is $i 2 \pi$ times the residue at the pole $z=-1/4$.    Thus
$$u(k) = \frac12 \left (-\frac14 \right)^{k-1} \frac{11}{16 \cdot 15} = -\frac{11}{120} \left (-\frac14 \right)^{k} $$ 
when $k \ge 1$.  When $k=0$, there is an additional pole at $z=0$, which contributes $3/8$ to add to the above piece, or
$$u(0) = \frac{3}{8}-\frac{11}{120} = \frac{17}{60} $$
For $k \lt 0$, we can use the fact that
$$u(-k) = u(k)^*$$
