Finding the z-transform of $h[n] = a^n\cos(2\pi \frac{n}{F_s}f_0)$ for $n ≥ 0$ and zero for $n < 0$ So I'm trying to decide whether the cosine part is intended to be plugged in for $z$ or whether it is strictly part of $h[n]$. (the number a lies in the open unit disk) 

I mean I was pretty sure it was all part of $h[n]$ but then upon performing the z-transform I get this rational function 

$$\frac{1 - a\cos(2\pi\frac{f_0}{F_s})z^{-1}}{1-2a\cos(2\pi\frac{f_0}{F_s})z^{-1} + a^2z^{-2}}$$

The thing is then I'm supposed to evaluate the poles and zeros and if you just ignore the cosine parts you get this really nice rational expression which factors and simplifies down to $\displaystyle\frac{z}{z-a}$.  

So that has gotten me thinking that maybe I'm not understanding things correctly and the cosine portion is supposed to be plugged in for $z$ or something.  Can anyone clarify this for me?
The time domain signal (or impulse response)

$$h(n)=a^{n}\cos n\theta_0,\quad 
\theta_0=2\pi\frac{f_0}{f_s},\; n\ge 0$$

is very common: it is a damped sinusoidal function (assuming $|a|&lt;1$) which occurs frequently, because it is one possible response of a second order linear time-invariant system. So concerning your doubt, the cosine part is definitely meant to be part of the time domain signal. The poles and zeros can be found by rewriting $H(z)$:

$$\begin{align}H(z)&amp;=\frac{1-az^{-1}\cos \theta_0}{1-2az^{-1}\cos\theta_0+a^2z^{-2}}\\&amp;=
\frac{z(z-a\cos \theta_0)}{z^2-2az\cos\theta_0+a^2}\end{align}\tag{1}$$

From (1) it is easy to determine the zeros of $H(z)$:
$$z_{0,0}=0\quad z_{0,1}=a\cos\theta_0$$

For determing the poles, we write $H(z)$ as a partial fraction expansion:

$$H(z)=\frac{1}{2}\left [ \frac{1}{1-ae^{j\theta_0}z^{-1}}
+ \frac{1}{1-ae^{-j\theta_0}z^{-1}} \right ]\tag{2}$$

From (2) we see that the poles are given by
$$z_{\infty,0}=ae^{j\theta_0}\quad z_{\infty,1}=ae^{-j\theta_0}$$
We have complex conjugate poles because $h(n)$ is real-valued. Assuming that $h(n)$ is an impulse response, we can see from the poles that the system is stable if $|a|&lt;1$ because then the poles are inside the unit circle of the complex plane.