Question

If $\omega$ is the nth root of unity and $z_1$ and $z_2$ any two complex numbers

then prove that $\sum _{k=0}^{n-1}|z_1+{\omega }^k{z}_2{|}^2=n\{|z_1{|}^2+|z_2{|}^2\}(n\in N)$.

Answer 1

$1,\omega ,{\omega }^2,{\omega }^3,.....,{\omega }^{n-1}$ are the n, nth roots of unity then
$\sum _{k=0}^{n-1}{\omega }^k=0$ and $\sum _{k=0}^{n-1}(\overline{\omega }{)}^k=0$
L.H.S. $\sum _{k=0}^{n-1}|z_1+{\omega }^k{z}_2{|}^2=\sum _{k=0}^{n-1}(z_1+{\omega }^k{z}_2)(\overline{z_1}+(\overline{\omega }{)}^k\overline{z_2})$
$=\sum _{k=0}^{n-1}{z}_1\overline{z_1}+z_1\overline{z_2}(\overline{\omega }{)}^k+\overline{z_1}{z}_2{\omega }^k+z_2\overline{z_2}({\omega }^k\left)\right(\overline{\omega }{)}^k$
$\sum _{k=0}^{n-1}|z_1{|}^2+\sum _{k=0}^{n-1}{z}_1\overline{z_2}(\overline{\omega }{)}^k+\sum _{k=0}^{n-1}\overline{z_1}{z}_2{\omega }^k+\sum _{k=0}^{n-1}|z_1{|}^2$
$=n|z_1{|}^2+0+0+n|z_2{|}^2$
$=n\{|z_1{|}^2+|z_2{|}^2\}$
$=R.H.S$