Mathematics - Complex Number Question with Solution | TestHub

$$\begin{array}{rlrlrl}& Z=\frac{-1}{2}i(1+i\sqrt{3})=i{\omega }^2& Z^{101}=i\omega & {\left(Z^{101}+i^{109}\right)}^{106}=-{\left(i{\omega }^2\right)}^{106}=-{\omega }^2& \therefore -{\omega }^2={\left(i{\omega }^2\right)}^n=i^n{\omega }^{2n}& {\omega }^{2n-2}{i}^n=-1\end{array}$$

This is possible only when $N=4r+2$ and $2n-2$ is multiple of 3 i.e., $2(4r+2)-2$ is multiply of 3 i.e., $3r+2$ is multiple of $3\Rightarrow r=2$ $\therefore n=10\therefore \frac{2}{5}k=4$