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Given the $z+ \frac{ 1 }{ z } =i$

Calculate the $z ^ { n } + \frac{ 1 }{ z ^ { n } }$

Ray Akabe
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    You could write it as a recurrence then solve it. Technically that doesn't involve finding $,z,$ explicitly, though the actual calculations are quite similar nevertheless. – dxiv Feb 10 '22 at 02:39
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    See sections 4.1 & 4.2 of https://en.wikipedia.org/wiki/Linear_recurrence_with_constant_coefficients to elucidate my A. – DanielWainfleet Feb 10 '22 at 20:19

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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 $L(n)=p^n+(-1/p)^n.$

DanielWainfleet
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