Question

If $p=a+b\omega +c{\omega }^2,q=b+c\omega +a{\omega }^2$, and $r=c+a\omega +b{\omega }^2$, where $a,b,c\ne 0$ and $\omega$ is the complex cube root of unity, then

• A. If p, q, r lie on the circle $|z|=2$, the triangle formed by these points is equilateral
• B. $p^2+q^2+r^2=a^2+b^2+c^2$
• C. $p^2+q^2+r^2=2(pq+qr+rp)$
• D. None of these

Answer 1

Answer: (A), (C)

The correct options are
A If p, q, r lie on the circle $|z|=2$, the triangle formed by these points is equilateral
C $p^2+q^2+r^2=2(pq+qr+rp)$

$p+q+r=a+b\omega +c{\omega }^2$
$+b+c\omega +a{\omega }^2$
$\underset{–}{+c+a\omega +b{\omega }^2}$
$\therefore p+q+r=(a+b+c)(1+\omega +{\omega }^2)=0$ (1)
$p,q,r$ lie on the circle $|z|=2$, whose circumcenter is origin.
Also, $\frac{(p+q+r)}{3}=0$. Hence the centroid coincides with circumcenter. So, the triangle is equilateral. Now,
$(p+q+r{)}^2=0$
$\Rightarrow p^2+q^2+r^2=-2pqr[\frac{1}{p}+\frac{1}{q}+\frac{1}{r}]$
$=-2pqr[\frac{1}{a+b\omega +c{\omega }^2}+\frac{1}{b+c\omega +a{\omega }^2}+\frac{1}{c+a\omega +b{\omega }^2}]$
$=-2pqr[\frac{1}{{\omega }^2(a\omega +b{\omega }^2+c)}+\frac{1}{\omega (b{\omega }^2+c+a\omega )}+\frac{1}{c+a\omega +b{\omega }^2}]$
$=\frac{-2pqr}{a\omega +b{\omega }^2+c}[\frac{1}{{\omega }^2}+\frac{1}{\omega }+\frac{1}{1}]=0\left(2\right)$
Hence, $p^2+q^2+r^2=2(pq+qr+rp)$