Let $\frac{1}{a_{1} - 2 i}, \frac{1}{a_{2} - 2 i}, \frac{1}{a_{3} - 2 i}, \frac{1}{a_{4} - 2 i}, \frac{1}{a_{5} - 2 i}, \frac{1}{a_{6} - 2 i}, \frac{1}{a_{7} - 2 i}, \frac{1}{a_{8} - 2 i}$ are vertices of regular octagon. If the area of octagon is A, then the value of $8 \sqrt{2} A$ is (where $a_{j} \in R$ for $j = 1, 2, 3, 4, 5, 6, 7, 8$ and $i = \sqrt{- 1}$ ) _____

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$\frac{1}{4 \sqrt{2}}$

$\frac{1}{4 \sqrt{3}}$

$\frac{1}{5 \sqrt{2}}$

$\frac{1}{1 \sqrt{2}}$

answer is C.

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Detailed Solution

$x + i y = \frac{a_{r} + 2 i}{{(a_{r})}^{2} + 4}$

$\Rightarrow x = \frac{a_{r}}{a_{r}^{2} + 4}, y = \frac{2}{a_{r}^{2} + 4}$

$x^{2} + y^{2} = \frac{1}{2} y$

$x^{2} + y^{2} - \frac{1}{2} y = 0$

So all points lie on circle $x^{2} + y^{2} - \frac{y}{2} = 0$

and $r = \frac{1}{4}$

$∴$ Area of regular octagon = $8 (\frac{1}{2} \times r^{2} \times \sin (\frac{2 π}{8}))$ = $8 \times \frac{1}{2} \times \frac{1}{16} \times \frac{1}{\sqrt{2}} = \frac{1}{4 \sqrt{2}}$

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