Question

Let $z=\frac{\sqrt{3}-i}{2}$ then $(z^{107}+i^{107}{)}^{25}$ is equal to

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

Answer 1

The correct option is A z

Given $z=\frac{\sqrt{3}-i}{2}=\frac{i-i^2\sqrt{3}}{2}$

$\Rightarrow z=i\left(\frac{-1-i\sqrt{3}}{2}\right)=i{\omega }^2=\left(\frac{i}{\omega }\right)$

$\therefore z^{107}=i^{107}{\omega }^{214}=-i\omega$ and

$\therefore i^{107}=-i$

$\therefore z^{107}+i^{107}=-i\omega -i=-i(1+\omega )=i{\omega }^2$

$\therefore (z^{107}+i^{107}{)}^{25}=(i{\omega }^2{)}^{25}=i^{25}{\omega }^{50}=i{\omega }^2=z$