We have $A=\left[\begin{array}{cc}1+i& 1\\ -i& 0\end{array}\right]$

$A^2=A·A=\left[\begin{array}{cc}1+i& 1\\ -i& 0\end{array}\right]\left[\begin{array}{cc}1+i& 1\\ -i& 0\end{array}\right]=\left[\begin{array}{cc}i& 1+i\\ -i+1& -i\end{array}\right]$

$A^3=A^2·A=\left[\begin{array}{cc}i& 1+i\\ 1-i& -i\end{array}\right]\left[\begin{array}{cc}1+i& 1\\ -i& 0\end{array}\right]=\left[\begin{array}{cc}0& i\\ 1& 1-i\end{array}\right]$

$A^4=A^3·A=\left[\begin{array}{cc}0& i\\ 1& 1-i\end{array}\right]\left[\begin{array}{cc}1+i& 1\\ -i& 0\end{array}\right]\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]=I$

$\therefore A^4=I$

So, $A^5=A^4·A=I·A=A$ 

$A^6=A^4·A^2=I·A^2=A^2$ and so on

$\because A^1=A^5=A^9=.....=A^{97}=A$

Hence, possible values of $n$, such that $A^n=A$

$=\left\{1,5,9,\dots ,97\right\}$

Clearly, above sequence is in A.P. where

$a=1, d=4 \& t_n=97$  $\Rightarrow a+\left(n-1\right)d=97$

$\Rightarrow 1+\left(n-1\right)4=97$  $\Rightarrow n=25$

$\therefore$ The number of elements in the given set $=25$.