Proving the a matrix is bounded for every power I want to show that the following matrix satisfies $\|A^n(z)\| \leq C$ for every $z \in \mathbb{C}$ with $Re(z) \leq 0$, for $n= 0,1,2,...$, and some constant $C$. How do I approach this kind of problems?
$$A(z) = \begin{bmatrix} \frac{z^2+6z+8}{2(2-z)^2} & -\frac{z^2+10z+24}{8(2-z)^2}  \\ \frac{2z(z+2)}{(2-z)^2} & -\frac{z(z+6)}{2(2-z)^2}   \end{bmatrix}$$

Given matrix $M=\{M_{ij}\}$ and $N=\{N_{ij}\}$ where $M_{ij}\in\Bbb C$ and $N_{ij}\in\Bbb R_{>0}$, write $M\prec N$ if $|M_{ij}|<N_{ij}$ for all $i,j$. It is straightforward to verify that if $M\prec N$ and $M'\prec N'$, then $MM'\prec NN'$.
It is enough to prove the following claim.
Claim: Let $z \in \mathbb{C}$ with $\Re(z) \leq 0$. Then $A^n(z) \prec \begin{bmatrix} \frac12& \frac18  \\ 2 & \frac12\end{bmatrix}$ for all $n=1,2,3,\cdots$.
Proof:
Since $\Re(z)<0$, we have $|z+2|<|2-z|$, $\ |z+4|<|2-z|$ and $|z+6|<|2-z|$. $$A(z) = \begin{bmatrix} \frac{(z+2)(z+4)}{2(2-z)^2} & -\frac{(z+4)(z+6)}{8(2-z)^2}  \\ \frac{2z(z+2)}{(2-z)^2} & -\frac{z(z+6)}{2(2-z)^2}   \end{bmatrix}\prec \begin{bmatrix} \frac12& \frac18  \\ 2 & \frac12\end{bmatrix},$$ which means the claim is true for $n=1$.
Suppose the claim is true for $n$, i.e., $A^n(z)\prec \begin{bmatrix} \frac12& \frac18  \\ 2 & \frac12\end{bmatrix}$. Then
$$A^{n+1}(z)=A^n(z)A(z)\prec \begin{bmatrix} \frac12& \frac18  \\ 2 & \frac12\end{bmatrix}\begin{bmatrix} \frac12& \frac18  \\ 2 & \frac12\end{bmatrix}=\begin{bmatrix} \frac12& \frac18  \\ 2 & \frac12\end{bmatrix}.$$ Hence, the claim is true for $n+1$, too. $\quad\checkmark$.
