Correct answer is (a) |z|=1
For explanation: Let us consider the signal to be x(n)
Z{x(n)}=\(\sum_{n=-∞}^∞ x(n)z^{-n} and X(ω)=\sum_{n=-∞}^∞ x(n)e^{-jωn}\)
Now, represent the ‘z’ in the polar form
=> z=r.e^jω
=>Z{x(n)}=\(\sum_{n=-∞}^∞ x(n)r^{-n} e^{-jωn}\)
Now Z{x(n)}= X(ω) only when r=1=>|z|=1.