Question

If $z_1,z_2$ are two complex numbers and ${\omega }^k,k=0,1,...,n-1$ are the nth roots of unity, then $\sum _{k=0}^{n-1}{|z_1+z_2{\omega }^k|}^2$

• A. $<n({\left|z_1\right|}^2+{\left|z_2\right|}^2)$
• B. $=n({\left|z_1\right|}^2+{\left|z_2\right|}^2)$
• C. $>n({\left|z_1\right|}^2+{\left|z_2\right|}^2)$
• D. can't say

Answer 1

Answer: (B)

$=n({\left|z_1\right|}^2+{\left|z_2\right|}^2)$

We have, ${|z_1+z_2{\omega }^k|}^2=(z_1+z_2{\omega }^k)\left(\overline{z_1+z_2{\omega }^k}\right)$

$=(z_1+z_2{\omega }^k)(\overline{z_1}+\overline{z_2}{\omega }^{-k})[{\omega }^k=e^{i\left(2\pi k/n\right)}\Rightarrow {\omega }^{\overline{k}}=e^{-i\left(2\pi k/n\right)}={\omega }^{-k}]$

$={\left|z_1\right|}^2+{\left|z_2\right|}^2+\overline{z_1}{z}_2{\omega }^k+z_1\overline{z_2}{\omega }^{-k}$

Therefore, we have

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

$=n({\left|z_1\right|}^2+{\left|z_2\right|}^2)[\sum _{k=0}^{n-1}{\omega }^k=\sum _{k=0}^{n-1}{\omega }^{-k}]$