Let $z^{16} - 1 = 0$ have 16 roots, then :

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If $α$ is one of the root of equation, then $\overline{α}$ is also the root of the same equation

Let $α_{1}, α_{2}, α_{3} \dots \dots \dots α_{16}$ be the roots of given equation, then $α_{1} α_{2} + α_{2} α_{3} + \dots \dots \dots α_{15} α_{16} = 0$

Area bounded by the polygon represented by the roots of given equation is $8 \sin \frac{π}{8}$

$\frac{1}{α_{1}} + \frac{1}{α_{2}} + \dots \dots \dots + \frac{1}{α_{16}} = 0$

answer is A, B, C, D.

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Detailed Solution

$α_{1}, α_{2}, \dots \dots, α_{16} are, 16 th$ roots of unity. They divide a unit circle in a regular polygon of 16 sides

$∴ Area of Δ = \frac{1}{2} (base \times height) = \frac{1}{2} (2 \cos θ) (\sin θ)$
$∴ Area of polygon = 16 \times Area of Δ$
$= 16 (\frac{1}{2}) \sin (\frac{π}{8}) = 8 \sin (\frac{π}{8})$
We know that $α_{1} α_{2} + \dots . + α_{15} α_{16} = 0$
$\begin{array}{l} & α_{i} {\bar{α}}_{i} = 1 \forall i = 1, 2 \dots \dots, 16 \\ ∵ α_{1} + α_{2} + \dots \dots . . + α_{16} = 0 \\ {\bar{α}}_{1} + {\bar{α}}_{2} + \dots \dots + {\bar{α}}_{16} = 0 \\ \frac{1}{α_{1}} + \frac{1}{α_{2}} + \dots . . + \frac{1}{α_{16}} = 0 \end{array}$