First slide

Geometry of complex numbers

Question

The roots of the cubic equation ${(z + a b)}^{3} = a^{3}$ such that $a \neq 0$ represent vertices of a triangle then length of side of triangle is..

Easy

A

$\frac{1}{\sqrt{3}} | a - b |$
B

$\sqrt{3} | a |$
C

$\sqrt{3} | b |$
D

$| a |$

Solution

given ${(z + a b)}^{3} = a^{3}$
$\Rightarrow z + a b = a {(1)}^{1 / 3}$
$\Rightarrow z + a b = a, z + a b = a w, z + a b = a w^{2}$
$∴ Z_{1} = a - a b, Z_{2} = a w - a b$ $Z_{3} = a w^{2} - a b$
Where $w = \frac{- 1 + \sqrt{3} i}{2}, w^{2} = \frac{- 1 - \sqrt{3} i}{2}$
Now $Z_{1} - Z_{2} = a [1 - w]$
$= a [\frac{3}{2} - \frac{i \sqrt{3}}{2}]$
$\Rightarrow |Z_{1} - Z_{2}| = \sqrt{3} | a |$
similarly $| Z_{2} - Z_{3} | = | Z_{3} - Z_{1} | = \sqrt{3} a$

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