Question

If $\frac{1}{{\alpha }_k+i}({\alpha }_k\in R)$ are $8$ vertices of a regular octagon for $k=1,2,3,\dots ,8$, where $i=\sqrt{-1}$, then the area of the octagon is

• A. $\sqrt{2}$
• B. $2\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{2\sqrt{2}}$

Answer 1

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

$z=\frac{1}{{\alpha }_k+i}$
Assuming ${\alpha }_k=\cot \theta ,\theta \in (0,\pi )$
$\Rightarrow z=\frac{\sin \theta }{\cos \theta +i\sin \theta }\Rightarrow z=\frac{\sin \theta (\cos \theta -i\sin \theta )}{1}\Rightarrow 2z=\sin 2\theta -i(1-\cos 2\theta )$

Let $z=x+iy$
Then $2x=\sin 2\theta ,2y+1=\cos 2\theta$
$x^2+{(y+\frac{1}{2})}^2=\frac{1}{4}$ is the circumcircle of the octagon and circumradius, $R=\frac{1}{2}$


Area of the octagon is,
$A=8\times \frac{1}{2}{R}^2\sin \left(\frac{2\pi }{8}\right)=\frac{1}{\sqrt{2}}$