The correct choice is (d) z^-(M-1).H(z^-1)=±H(z)
Easy explanation: We know that H(z)=\(\sum_{k=0}^{M-1} h(k)z^{-k}\) and h(n)=±h(M-1-n) n=0,1,2…M-1
When we incorporate the symmetric and anti-symmetric conditions of the second equation into the first equation and by substituting z^-1 for z, and multiply both sides of the resulting equation by z^-(M-1) we get z^-(M-1).H(z^-1)=±H(z)