Right option is (b) \(\frac{z}{2} \frac{(2z-1)}{(z^2-z+1)}\), |z|>1
For explanation: Performing z-transform on a^nu[n], we get \(\frac{z}{z-a}\)
∴ x[n] = \(\frac{1}{2} e^{j(\frac{π}{3})n} u[n] + \frac{1}{2} e^{-j(\frac{π}{3})n}u [n]\)
Hence, X (z) =\( 0.5 \left(\frac{1}{1-e^{j(\frac{π}{3})} z^{-1}} + \frac{1}{1-e^{j(\frac{π}{3})} z^{-1}}\right)\)
Hence, X (z) = \(\frac{z}{2} \frac{(2z-1)}{(z^2-z+1)}\), |z|>1.