Question

If $\frac{1}{{\alpha }_k+i}$ are 8 vertices of a regular octogon where ${\alpha }_kϵR,1,2,3,....8$ (where $i=\left(\sqrt{-1}\right)$ then area of the regular octagon is:-

• A. $1$
• B. $\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{\sqrt{2}}$

$z=\frac{1}{{\alpha }_k+i}\text{Put}{\alpha }_k=\cot \theta$
$z=\frac{\sin \theta }{\cos \theta +i\sin \theta }=x+iy$
$2x=\sin 2\theta ,2y+1=\cos 2\theta$
$x^2+{(y+\frac{1}{2})}^2=\frac{1}{4}$ is circumcircle of octagon
$\Rightarrow$ Area of octogon $=8\times \frac{1}{2}{R}^2\sin \left(\frac{2\pi }{8}\right)=\frac{1}{\sqrt{2}}$