Mathematics - Complex Number Question with Solution | TestHub

${\alpha }^{11}=1\Rightarrow (\alpha -1)\left({\alpha }^{10}+{\alpha }^9+{\alpha }^8+\dots +{\alpha }^2+\alpha +1\right)=0$-- (1) Now $X={\alpha }^6$ is a root of $1+x+x^2+\dots +x^{10}=0$ $\Rightarrow 1+\left(\lambda +{\lambda }^2+{\lambda }^3+{\lambda }^4+{\lambda }^5\right)\cdot 2=0$ $\Rightarrow \mathrm{Re}\left(\lambda +{\lambda }^2+\dots +{\lambda }^5\right)=-\frac{1}{2}$ Now, $\beta ,{\beta }^2,{\beta }^3,\dots {\beta }^{10}$ are also roots equation (1) $(x-\beta )\left(x-{\beta }^2\right)\dots \left(x-{\beta }^{10}\right)=1+x+x^2+..+x^{10}$

Now putting $x=\mu$, we get $(\mu -\beta )\left(\mu -{\beta }^2\right)\dots \left(\mu -{\beta }^{10}\right)=0$