Mathematics - Complex Number Question with Solution | TestHub

$\begin{array}{rl}& \alpha =\omega \\ & \therefore {\left({\omega }^k+\frac{1}{{\omega }^k}\right)}^2={\omega }^{2k}+\frac{1}{{\omega }^{2k}}+2\\ & ={\omega }^{2k}+{\omega }^k+2\because {\omega }^{3k}=1\end{array}$
$\begin{array}{rl}& \therefore \sum _{\mathrm{k}=1}^{\mathrm{n}}\left({\omega }^{2\mathrm{k}}+{\omega }^{\mathrm{k}}+2\right)=20\\ & \Rightarrow \left({\omega }^2+{\omega }^4+{\omega }^6+\dots +{\omega }^{2\mathrm{n}}\right)+(\omega +{\omega }^2+{\omega }^3+\dots +\\ & {\omega }^{\mathrm{n}})+2\mathrm{n}=20\end{array}$
Now if $n=3m,m\in I$
Then $0+0+2\mathrm{n}=20\Rightarrow \mathrm{n}=10$ (not satisfy)
if $n=3m+1$, then
$\begin{array}{rl}& {\omega }^2+\omega +2\mathrm{n}=20\\ & -1+2\mathrm{n}=20\Rightarrow \mathrm{n}=\frac{21}{2}(\text{ not possible })\end{array}$
if $n=3m+2$,
$\begin{array}{rl}& \left({\omega }^8+{\omega }^{10}\right)+\left({\omega }^4+{\omega }^5\right)+2\mathrm{n}=20\\ & \Rightarrow \left({\omega }^2+\omega \right)+\left(\omega +{\omega }^2\right)+2\mathrm{n}=20\\ & 2\mathrm{n}=22\\ & \mathrm{n}=11\text{ satisfy }\mathrm{n}=3\mathrm{m}+2\\ & \therefore \mathrm{n}=11\end{array}$