Question

If $z=\frac{\sqrt{3}+i}{2},$ then $z^{69}$ is equal to

• A. $1$
• B. 2
• C. $-2$
• D. $-1$

Answer 1

Answer: (C)

$z=\frac{\sqrt{3}+i}{2}$
$=i\left[\frac{\sqrt{3}}{2i}+\frac{1}{2}\right]$
$=i\left[\frac{1}{2}-\frac{i\sqrt{3}}{2}\right]$
$=-i\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)$
$=-i\omega$ $z^{69}={(-i\omega )}^{69}$
$={(-i)}^{69}.{\omega }^{69}$
$={(-1)}^{69}.i{({\omega }^3)}^{23}$
$={(-1)}^{23}=-i$