I've solved it by finding the value of Z first. So is there any other ways that can solve it without having to find the value of Z? Given the $z+ \frac{ 1  }{ z  }  =i$
Calculate the $z  ^ { n  }  + \frac{ 1  }{ z  ^ { n  }    }$

Let $f(n)=z^n+1/z^n.$
We have $f(n+1)=f(n)f(1)-f(n-1)=if(n)-f(n-1).$
The set of roots of $x^2-ix+1=0$ is $\{ip,-i/p\}$ where $p=(1+\sqrt 5)/2.$
From the theory of linear homogeneous recursions, there are unique $A,B$ such that $f(n) =A(ip)^n +B(-i/p)^n$ for all $n$. We can find $A,B$ from the pair of equations $2=f(0)=A+B$ and $i=f(1)=A(ip)+B(-i/p).$ After the calculations we see that $A=B=1.$
If $n$ is even then $f(n)=L(n)(-1)^{n/2},$ and  if $n$ is odd then $f(n)=i\cdot L(n)(-1)^{(n-1)/2}$ where $L(n)$ is the $n$th Lucas number: $L(0)=2$ and $L(1)=1$ and $L(n+1)=L(n)+L(n-1)$ and also  $L(n)=p^n+(-1/p)^n.$
