Question

Let $z^3=\overline{z}$, wherer $z$ is a complex number not equal to zero. Then $z$ is a solution of

• A. $z^2=1$
• B. $z^3=1$
• C. $z^4=1$
• D. $z^9=1$

Answer 1

The correct option is C $z^4=1$

Let $Z=r{e}^{i\theta }$ then $z^3=\overline{z}$
$\Rightarrow r^3{e}^{3i\theta }=r{e}^{-i\theta }$
or $r^2{e}^{4i\theta }=1$
$r^2{e}^{4i\theta }=1e^{2\pi i}$
On comparison $r=1$ & $\theta =\pi /2$
Hence, $Z=r{e}^{i\theta }=1.e^{\frac{i\pi }{2}}=i$
So, only option (c) satisfies it.