Question

If $z=\frac{\sqrt{3}+i}{2}$, then the value of $z^{69}$ is

• A. 1
• B. -1
• C. 0
• D. -i

Answer 1

Given, $z=\frac{\sqrt{3}}{2}+\frac{i}{2}$
$iz=\frac{-1}{2}+i\frac{\sqrt{3}}{2}=\omega$
And $z^{69}=z^{4(17)}z=(iz{)}^{4(17)}z=(\omega {)}^{68}z\left(\because i^{4n}=1\right)$
$=\frac{{\omega }^{69}}{i}=\frac{{\left({\omega }^3\right)}^{23}}{i}=-i$