$G = \left[ \begin{array}{c|c} A & B \\ \hline C & D \\ \end{array} \right] = C(zI-A)^{-1}B+D$
Suppose:
1. $A,B,C,D$ are all real matrix.
2. $z = e^{j\theta}$, i.e. $r=1$ for simplicity.
3. $G^*$ denotes transpose conjugate of $G$.
$G^* = B^T(\frac{1}{z}I-A^T)^{-1}C^T+D^T=B^Tz^2(zI-z^2A^T)^{-1}C^T+D^T$
$G^* = \left[
\begin{array}{c|c}
z^2A & zC^T \\
\hline
zB^T & D^T \\
\end{array}
\right]
$
It seems like that. But my advisor said that there should be no frequency term $z$ in the state space representation.
How do I represent it?
You can use the fact that $z$ is an operator that shifts functions in time. So your system becomes
$$\begin{align*} x[k+1] &= A^T x[k+2] + C^T u[k+1] \\ x[k+2] &= A^{-T} x[k+1] - A^{-T} C^T u[k+1] \\ \\ y[k] &= B^T x[k+1] + D^T u[k] \\ y[k+1] &= B^T x[k+2] + D^T u[k+1] \\ &= B^T A^{-T} x[k+1] + (D^T - B^T A^{-T} C^T) u[k+1] \end{align*}$$
$$G^* = \left[ \begin{array}{c|c} A^{-T} & - A^{-T} C^T \\ \hline B^T A^{-T} & D^T - B^T A^{-T} C^T \\ \end{array} \right]$$