Solution:

(a) By using Euler's identity,

the signal x(n) can be expressed as,

$ x(n)=\left(\cos \omega_0 n\right) \mu(n)=\frac{1}{2} e^{j \operatorname{man} n} u(n)+\frac{1}{2} e^{-j \omega p n} u(n)\\ $

that,

$ X(z)=\frac{1}{2} Z\left\{e^{j \operatorname{mov} n} u(n)\right\}+\frac{1}{2} Z\left\{e^{-j \operatorname{mov} n} u(n)\right\}\\ $

If we set $\alpha=e^{\pm j \omega_0}\left(|\alpha|=\left|e^{\pm j a_0}\right|=1\right)$ in (3.2.2). …