Question

If $1,\omega ,{\omega }^2,\cdots \cdots {\omega }^{n-1}$ are the $n^{\text{th}}$ roots of unity and $z_1$ and $z_2$ are any two complex numbers, then $\sum _{k=0}^{n-1}|z_1+{\omega }^k{z}_2{|}^2$ is equal to

• A. $n[|z_1{|}^2+|z_2{|}^2]$
• B. $(n-1)[|z_1{|}^2+|z_2{|}^2]$
• C. $(n+1)[|z_1{|}^2+|z_2{|}^2]$
• D. $(n+1)[|z_1{|}^2-|z_2{|}^2]$

Answer 1

The correct option is A $n[|z_1{|}^2+|z_2{|}^2]$

Since $1,\omega ,{\omega }^2,\cdots \cdots {\omega }^{n-1}$ are the $n^{\text{th}}$ roots of unity

$\therefore \sum _{k=0}^{n-1}{\omega }^k=0\text{and}\sum _{k=0}^{n-1}(\overline{\omega }{)}^k=0$

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

$=\sum _{k=0}^{n-1}[z_1\overline{z_1}+z_1\overline{z_2}(\overline{\omega }{)}^k+\overline{z_1}{z}_2(\omega {)}^k+z_2\overline{z_2}(\omega {)}^k\left(\overline{\omega }{)}^k\right]$
$=\sum _{k=0}^{n-1}|z_1{|}^2+\sum _{k=0}^{n-1}{z}_1\overline{z_2}(\overline{\omega }{)}^k+\sum _{k=0}^{n-1}\overline{z_1}{z}_2(\omega {)}^k+\sum _{k=0}^{n-1}|z_2{|}^2$

$=n|z_1{|}^2+0+0+n|z_2{|}^2=n(|z_1{|}^2+|z_2{|}^2)$