Show that complex numbers satisfying $z_1 + z_2 + z_3 = z_1^2 + z_2^2 + z_3^2 = 0$ have equal module $|z_1|=|z_2|=|z_3|$

Question

I have 3 complex number $z_1,\> z_2, \>z_3$ from C and i know that: $$z_1 + z_2 + z_3 = 0 , \>\>\>\>\> z_1^2 + z_2^2 + z_3^2 = 0$$ I have to show that $$|z_1|=|z_2|=|z_3|$$

Answer 1

From the given,

$$ z_1z_2 + z_2z_3 +z_3z_1 =\frac12 \left[(z_1 + z_2 + z_3)^2 - (z_1^2 + z_2^2 + z_3^2) \right]= 0 $$

Substitute lhs with $z_3=-(z_1 + z_2)$,

$$z_1z_2 - (z_1 +z_2)^2 = 0 \implies z_1^2 + z_1z_2 + z_2^2 =0$$

which yields $ {z_1} = \frac12(-1\pm\sqrt3 i){z_2}=e^{i \pi \pm i\frac\pi3}{z_2}\implies \left|{z_1}\right| = |{z_2}|$ and likewise $|{z_1}|= |{z_3}|$. Thus,

$$|z_1|=|z_2|=|z_3|$$

Answer 2

In

If $z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1$ , PT $z_1,z_2,z_3$ represent the vertices of an equilateral triangle.

put $a,b,c$ to $z_1,z_2,z_3$ respectively

See also

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