Question

If $z=x+iy$ is a complex number with $x,y\in Q$ and $\left|z\right|=1$. then show that $|z^{2n}-1|$ is a rational number for every $n\in N$.

Answer 1

$\left|z\right|=1⟹z=e^{i\theta }=x+iy$
$⟹x=cos\theta ,y=isin\theta$
Now $cos\theta$ and $sin\theta \in Q.$ Also,
${|z^{2n}-1|}^2=(z^{2n}-1)({\overline{z}}^{2n}-1)$
$=(z\overline{z}{)}^{2n}-z^{2n}-{\overline{z}}^{2n}+1$
$=2-(z^{2n}+{\overline{z}}^{2n})$
$=2-2co{s}^{2}n\theta =4si{n}^{2}n\theta$
$=|z^{2n}-1|=2\left|sinn\theta \right|$
Now,
$sinn\theta {=}^n{C}_{1}co{s}^{n-1}\theta sin\theta {-}^n{C}_{3}co{s}^{n-3}\theta si{n}^3\theta +...$
= Rational number ($\therefore sin\theta ,cos\theta$ are rationals)
$⟹|z^{2n}-1|=$ Rational number
Ans: 1