    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

$0$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

$2$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

$3$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

$4$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
E

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## 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.$\begin{array}{crcc}\therefore & {z}^{4}& =& 1\\ ⇒& {z}^{4}-1& =& 0\\ ⇒& \left({z}^{2}+1\right)\left({z}^{2}-1\right)& =& 0\end{array}$Case 1: $\begin{array}{ccc}{z}^{2}+1& =& 0\end{array}$ $\begin{array}{crcl}& {z}^{2}& =& -1\\ & z& =& ±\sqrt{-1}\\ & z& =& +i,-i\end{array}$Case 2: $\begin{array}{ccc}{z}^{2}-1& =& 0\end{array}$ $\begin{array}{crcl}& {z}^{2}& =& 1\\ & z& =& ±\sqrt{1}\\ & z& =& +1,-1\end{array}$$\therefore$${z}_{1},{z}_{2},{z}_{3}$ and ${z}_{4}$ are $+1,-1,+i,-i$ respectively, $\begin{array}{cccl}\therefore & {{z}_{1}}^{2}+{{z}_{2}}^{2}+{{z}_{3}}^{2}+{{z}_{4}}^{2}& =& {\left(+1\right)}^{2}+{\left(-1\right)}^{2}+{\left(+i\right)}^{2}+{\left(-i\right)}^{2}\\ & & =& 1+1-1-1\\ ⇒& {{z}_{1}}^{2}+{{z}_{2}}^{2}+{{z}_{3}}^{2}+{{z}_{4}}^{2}& =& 0\end{array}$Hence, the correct answer is option (A) i.e. $0$  Suggest Corrections  3      Similar questions
Join BYJU'S Learning Program
Select...  Related Videos   MATHEMATICS
Watch in App  Explore more
Join BYJU'S Learning Program
Select...