Question

Show that $\sum _{r=0}^{n-1}{|z_1+a^r{z}_2|}^2=n({\left|z_1\right|}^2+{\left|z_2\right|}^2)$, where $\alpha ;r=0,1,2,...,(n-1)$, are the $n$th roots of unity and $z_1,z_2$ are any two complex numbers.

Answer 1

$\sum _{r=0}^{n-1}{|z_1+a^r{z}_2|}^2=\sum _{r=0}^{n-1}(z_1+a^r{z}_2)(\overline{z_1}+a^r\overline{z_2})$ Since, $|z{|}^2=z\overline{z}$

$=\sum _{r=0}^{n-1}{\left|z_1\right|}^2+\sum _{r=0}^{n-1}{\left|z_2\right|}^2+(z_1\overline{z_2}+\overline{z_1}{z}_2)\sum _{r=0}^{n-1}{a}^r=n({\left|z_1\right|}^2+{\left|z_2\right|}^2)$

Since, $a$ is the nth root of unity

Therefore, ${\sum }_{r=0}^{n-1}{a}^r=0$

Hence, $\sum _{r=0}^{n-1}{|z_1+a^r{z}_2|}^2=n({\left|z_1\right|}^2+{\left|z_2\right|}^2)$