Mathematics - Complex Number Question with Solution | TestHub

$C+is=e^{i\theta }+e^{i(2\theta )}+e^{i(4\theta )}$, where $\theta =\frac{2\pi }{7}$

i.e $C+iS=\alpha +{\alpha }^2+{\alpha }^4$, where $\alpha =e^{i\theta }$....(i)

so, $C-iS=\bar{\alpha }+\left({\bar{\alpha }}^2\right)+\left({\bar{\alpha }}^4\right)={\alpha }^6+{\alpha }^5+{\alpha }^3$....(ii) $(\sin ce\bar{\alpha }=\frac{\alpha \bar{\alpha }}{\alpha }=\frac{1}{\alpha }=\frac{{\alpha }^7}{\alpha }={\alpha }^6$ etc $)$

From Eqs (i) and (ii) , we get $2C=\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5+{\alpha }^6=\frac{\alpha \left({\alpha }^6-1\right)}{\alpha -1}$ $\Rightarrow 2C=\frac{1-\alpha }{\alpha -1}\left(\therefore {\alpha }^7=1\right)$

$\Rightarrow C=-\frac{1}{2}$

Again $(C+iS)(C-iS)=\left(\alpha +{\alpha }^2+{\alpha }^4\right)\left({\alpha }^6+{\alpha }^5+{\alpha }^3\right)$ $\Rightarrow C^2+S^2=1+{\alpha }^6+{\alpha }^4+\alpha +1+{\alpha }^5+{\alpha }^3+{\alpha }^2+1\left(a^7=1\right)$ $\Rightarrow {\left(-\frac{1}{2}\right)}^2+S^2=2+\left(1+\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5+{\alpha }^6\right)$ $\Rightarrow S^2=2+0-\frac{1}{4}=\frac{7}{4}$

$\Rightarrow S=\frac{\sqrt{7}}{2}$