We know that,

DFT of h(n) is given by:

$H(k)=∑_{n=0}^{N-1}h(n) e^{\frac{-j2πkn}{N}}$ …………………(1)

From the equation of IDFT we have,

$h(n)=\frac{1}{N} ∑_{k=0}^{N-1}H(k) e^{\frac{j2πkn}{N}}$…………………(2)

By the definition of Z transform we have,

              $H(z)=∑_{n=0}^{N-1}h(n) z^{-n}$…………………(3)

Put the value of h(n) from equation (2) to equation (3), we get

$H(z)=∑_{n=0}^{N-1}\frac{1}{N} ∑_{k=0}^{N-1}H(k) e^{\frac{j2πkn}{N}} z^{-n}$

$H(z)=\frac{1}{N} ∑_{k=0}^{N-1}H(k) ∑_{n=0}^{N-1}e^{\frac{j2πkn}{N}} z^{-n} $

Now by sum of finite GP series we have;

We know that, $e^{j2πk}=1$

$H(z)=\frac{(1-z^{-N})}{N} ∑_{k=0}^{N-1}[\frac{H(k)}{(1-e^{\frac{j2πk}{N}} z^{-1} )}$]

$H(z)=\frac{(1-z^{-N})}{N} [\frac{H(0)}{(1-z^{-1} )}+\frac{H(1)}{(1-e^{\frac{j2π}{N}} z^{-1} )}+\frac{H(2)}{(1-e^{\frac{j4π}{N}} z^{-1} )}+⋯………………+\frac{H(N-1)}{(1-e^{\frac{(j2π(N-1))}{N}}} z^{-1} )$]

Frequency Sampling Realization;