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Construction of Incircle

If z= ω,ω2,...

Question

If $z = ω, ω^{2}$ ,where is a nonreal.complex cube root of unity, are two vertices of an equilateral triangle in the Argand plane then the third vertex may be represented by

A

z = 2

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B

z = 0

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C

z = 1

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D

z = - 1

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Solution

The correct option is B $z = 1$
Let the third vertex be $z_{3}$
Since $z_{3}$ $w$ and $w^{2}$ forms an equilateral triangle, hence
$z_{3}^{2} + w^{2} + w^{4} = z_{3} w + w^{2} z_{3} + w^{3}$
Hence
$z_{3}^{2} + w^{2} + w^{3} . w = z_{3} (w + w^{2}) + 1$
$z_{3}^{2} + w^{2} + w = z_{3} (- 1) + 1$
$z_{3}^{2} - 1 = - z_{3} + 1$
$z_{3}^{2} + z_{3} - 2 = 0$
$(z_{3} + 2) (z_{3} - 1) = 0$
Hence
$z_{3} = - 2$ and $z_{3} = 1$ . ...(i)
Since they form an equilateral triangle,
$| z_{3} - w | = | w - w^{2} | = | z_{3} - w^{2} |$ ...(ii)
From i and ii, we get $z_{3} = 1$ .
Hence the third vertex is 1.

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