Sinyalimiz ve $\mathcal{Z}$ dönüşümünün tanımı :
$$x[n]=\alpha^n\cos(\Omega_0n)u[n]$$
$$X(n)=\mathcal{Z}\{x[n]\}=\sum_{n=-\infty}^\infty\,x[n]\,z^{-n}$$
Bizden istenen :
$$\alpha^n\cos(\Omega_0n)u[n]\xrightarrow{\:\:\:\:\:\:\:\:\mathcal{Z}\:\:\:\:\:\:\:\:}\sum_{n=-\infty}^\infty\,\alpha^n\cos(\Omega_0n)u[n]\,z^{-n}$$
Basamak fonksiyonunun özelliğini kullanarak sadeleştirelim.
$$\alpha^n\cos(\Omega_0n)u[n]\xrightarrow{\:\:\:\:\:\:\:\:\mathcal{Z}\:\:\:\:\:\:\:\:}\sum_{n=0}^\infty\,\alpha^n\cos(\Omega_0n)\,z^{-n}$$
Kosinüsü açalım.
$$\alpha^n\cos(\Omega_0n)u[n]\xrightarrow{\:\:\:\:\:\:\:\:\mathcal{Z}\:\:\:\:\:\:\:\:}\sum_{n=0}^\infty\,\alpha^n\frac{e^{i\Omega_0n}+e^{-i\Omega_0n}}{2}\,z^{-n}$$
Sadeleştirelim.
$$\alpha^n\cos(\Omega_0n)u[n]\xrightarrow{\:\:\:\:\:\:\:\:\mathcal{Z}\:\:\:\:\:\:\:\:} \frac{1}{2} \sum_{n=0}^\infty\,\Bigg[ \bigg(\alpha\,e^{i\Omega_0n}z^{-1}\bigg)^n+\bigg(\alpha\,e^{-i\Omega_0n}z^{-1}\bigg)^n\Bigg]$$
Buradan seriyi kolayca hesaplayabiliriz.
$$\alpha^n\cos(\Omega_0n)u[n]\xrightarrow{\:\:\:\:\:\:\:\:\mathcal{Z}\:\:\:\:\:\:\:\:}\frac{1}{2}\Bigg[ \frac{1}{1-\alpha\,e^{i\Omega_0}\,z^{-1}}+\frac{1}{1-\alpha\,e^{-i\Omega_0}\,z^{-1}}\Bigg] $$
$$\large\color{#A00000}{\boxed{\alpha^n\cos(\Omega_0n)u[n]\xrightarrow{\:\:\:\:\:\:\:\:\mathcal{Z}\:\:\:\:\:\:\:\:}\frac{1-\alpha\cos(\Omega_0)\,z^{-1}}{1-2\alpha\,\cos(\Omega_0)\,z^{-1}+\alpha^2\,z^{-2}}\:\:\:,\:\:\:|z|>|\alpha|}}$$