Question

If $\omega$ and ${\omega }^2$ (where $\omega$ is a non-real cube root of unity) are two vertices of an equilateral triangle in the Argand plane, then the third vertex $z$ may be represented by

• A. $z=1$
• B. $z=0$
• C. $z=-2$
• D. $z=-1$

Answer 1

Answer: (A), (C)

$z=-2$
$z=1$

Since $w$ is the cube root of unity.
$\therefore w^3=1\&1+w+w^2=0$,
where, $w=\frac{-1+i\sqrt{3}}{2}\&w^2=\frac{-1-i\sqrt{3}}{2}$.
$w\&w^2$ are the two vertices of equilateral triangle.
Let $z=x+iy$ be third vertex.
$\therefore |z-w|=|z-w^2|=|w-w^2|=|\frac{-1+i\sqrt{3}}{2}-\frac{-1-i\sqrt{3}}{2}|=\sqrt{3}$
$\Rightarrow |x+\frac{1}{2}+i(y-\frac{\sqrt{3}}{2})|=|x+\frac{1}{2}+i(y+\frac{\sqrt{3}}{2})|\Rightarrow {(x+\frac{1}{2})}^2+{(y-\frac{\sqrt{3}}{2})}^2={(x+\frac{1}{2})}^2+{(y+\frac{\sqrt{3}}{2})}^2\Rightarrow y=0$
And, $|z-w|=\sqrt{3}\Rightarrow |x+\frac{1}{2}+i(y-\frac{\sqrt{3}}{2})|=\sqrt{3}\Rightarrow {(x+\frac{1}{2})}^2+{(y-\frac{\sqrt{3}}{2})}^2=3\Rightarrow {(x+\frac{1}{2})}^2=\frac{9}{4}\Rightarrow x=1,-2$
$\therefore z=1,-2$
Hence, option A and C are correct.