Question

If the fourth roots of unity are$z_1,z_2,z_3$ and $z_4$ , then ${z_1}^2+{z_2}^2+{z_3}^2+{z_4}^2$ is equal to

• A. $0$
• B. $2$
• C. $3$
• D. $4$
• E. None of these

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Given that $z_1,z_2,z_3$ and $z_4$ are fourth roots of unity.

$\begin{array}{crcc}\therefore & z^4& =& 1\\ \Rightarrow & z^4-1& =& 0\\ \Rightarrow & \left(z^2+1\right)\left(z^2-1\right)& =& 0\end{array}$

Case 1: $\begin{array}{ccc}{z}^2+1& =& 0\end{array}$

$\begin{array}{crcl}& z^2& =& -1\\ & z& =& \pm \sqrt{-1}\\ & z& =& +i,-i\end{array}$

Case 2: $\begin{array}{ccc}{z}^2-1& =& 0\end{array}$

$\begin{array}{crcl}& z^2& =& 1\\ & z& =& \pm \sqrt{1}\\ & z& =& +1,-1\end{array}$

$\therefore$$z_1,z_2,z_3$ and $z_4$ are $+1,-1,+i,-i$ respectively,

$\begin{array}{cccl}\therefore & {z_1}^2+{z_2}^2+{z_3}^2+{z_4}^2& =& {\left(+1\right)}^2+{\left(-1\right)}^2+{\left(+i\right)}^2+{\left(-i\right)}^2\\ & & =& 1+1-1-1\\ \Rightarrow & {z_1}^2+{z_2}^2+{z_3}^2+{z_4}^2& =& 0\end{array}$

Hence, the correct answer is option (A) i.e. $0$