Finding the z-transform of $h[n] = a^n\cos(2\pi \frac{n}{F_s}f_0)$ for $n ≥ 0$ and zero for $n < 0$
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.