If `z=x+i y` is a complex number with `x ,y in Qa n d|z|=1,` then show that `|z^(2n)-1|` is a rational number for every `n in Ndot`

Explanation

`|z|=1`
`impliesz=e^(itheta)=x+iy`
`impliesx=costheta,y=sintheta`
Now `costheta" and "sinthetainQ`. Also,
`|z^(2n)-1|^(2)=(z^(2n-1)(bar(z)^(2n)-1)`
`=(zbar(z))^(2n)-z^(2n)-bar(z)^(2n)+1`
`=2-(z^(2n)+bar(z)^(2n))`
`=2-2cos2n theta=4sin^(2)ntheta`
`implies|z^(2n)-1|=2|sinntheta|`
Now, `sinntheta=""^(n)C_(1)cos^(n-1)thetasintheta-""^(n)C_(3)cos^(n-3)thetasin^(3)theta+...`
`=" Rational number "(because sintheta, costheta" are rationals")`
`implies|z^(2n)-1|=" Rational number "`