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.  
 A: 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
$$u(-k) = u(k)^*$$
