Question

If $z=\omega ,{\omega }^2$,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. $z=2$
• B. $z=0$
• C. $z=1$
• D. $z=-1$

Answer 1

Answer: (C)

$z=1$

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.