1. ## complex number question

i can do first part of this question. i need help with the last part

find roots of $z^3 - 1 = 0$
show if one complex root is $w$ show the other is $w^2$
show $Re(\frac{1}{1+w^r}) = 1/2$ for r=1,2,3

i need help for next part

if $z_1^2 +z_2^2+z_3^2- z_1 z_2-z_2 z_3-z_3 z_1 = 0$

show
$z_1 = -wz_2 -w^2 z_3$ or $z_1 = -wz_3 -w^2 z_2$
deduce three complex numbers are in a equaliteral triangle

2. Originally Posted by stud_02
if $z_1^2 +z_2^2+z_3^2- z_1 z_2-z_2 z_3-z_3 z_1 = 0$

show
$z_1 = -wz_2 -w^2 z_3$ or $z_1 = -wz_3 -w^2 z_2$
One way (not very elegant) is to write the equation as a quadratic in $z_1$ and solve it:

$z_1^2 - (z_2+z_3)z_1 + (z_2^2+z_3^2 - z_2z_3) = 0$,

\begin{aligned}z_1 &= \tfrac12\bigl(z_2+z_3 \pm\sqrt{6z_2z_3-3z_2^2-3z_3^2}\bigr)\\[0.5ex] &= \tfrac12\bigl(z_2+z_3 \pm\sqrt3i(z_2-z_3)\bigr) \\[0.5ex] &= \tfrac12(1\pm\sqrt3i)z_2 +\tfrac12(1\mp\sqrt3i)z_3.\end{aligned}

That gives the required result, since the complex cube roots of 1 are $\tfrac12(-1 \pm\sqrt3i)$.