Find the perimeter of the triangle whose vertices are the roots of the equation $(z + αβ)^{3} = α^{3}$ , where α,ß are given complex number

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$3 \sqrt{3} | β |$

$\sqrt{3} | α |$

$3 | α |$

$3 \sqrt{3} | α |$

answer is D.

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Detailed Solution

$(z + αβ)^{3} = α^{3} \Rightarrow z + αβ = α, αω, {αω}^{2} z_{1} = α - α β, z_{2} = α ω - α β, z_{3} = α ω^{2} - α β . These are the vertices denoted by A, B, C respectively. Length of AB = |z_{1} - z_{2}| = | a | | 1 - ω | = | a | |1 - \frac{- 1 - i \sqrt{3}}{2}| = | α | |\frac{3 + i \sqrt{3}}{2}| = \sqrt{3} | α | Length of BC = |z_{2} - z_{3}| = | α | |1 - ω^{2}| = | α | |1 - \frac{- 1 + i \sqrt{3}}{2}| = | α | |\frac{3 - i \sqrt{3}}{2}| = \sqrt{3} | α | Length of CA = |z_{3} - z_{1}| = | α | | ω | | 1 - ω | = | α | |\frac{- 1 - i \sqrt{3}}{2}| |1 - \frac{- 1 - i \sqrt{3}}{2}| = \begin{aligned} | α | |\frac{3 + i \sqrt{3}}{2}| \\ = \sqrt{3} | α | \end{aligned}$ Roots are vertices of equilateral triangle lying on circle with radius |α| Length of side