Right option is (b) e^j2π(k+α)/M
For explanation I would say: The system function of the second filter in the cascade of an FIR realization by frequency sampling method is given by
H2(z)=\(\sum_{k=0}^{M-1} \frac{H(k+α)}{1-e^{\frac{j2π(k+α)}{M}} z^{-1}}\)
We obtain the poles of the above system function by equating the denominator of the above equation to zero.
=>\(1-e^{\frac{j2π(k+α)}{M}} z^{-1}\)=0
=>z=pk=\(e^{\frac{j2π(k+α)}{M}}\), k=0,1….M-1