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Show that if $z_1,z_2,z_3$ are complex numbers, $z_1+z_2+z_3=0$ and $z_1^2+z_2^2+z_3^2=0$ then: $|z_1|=|z_2|=|z_3|$

$$z_3=-z_1-z_2\
z_3^2=z_1^2+2z_1z_2+z_2^2$$
Since $z_3^2=-z_1^2-z_2^2$ you get
$$z_1^2+z_1z_2+z_2^2=0$$
Multiplying by $z_1-z_2$ you get
$$z_1^3=z_2^3$$

Applying absolute values you get
$$|z_1|^3=|z_2|^3$$
and hence
$$|z_1|=|z_2|$$

The equality $|z_1|=|z_3|$ can be obtained same way.

$$z_1 z_2 + z_2 z_3 + z_3 z_1 = big((z_1 + z_2 + z_3)^2 - (z_1^2 + z_2^2 + z_3^2)big)/2 = 0.$$
So $(z-z_1)(z-z_2)(z-z_3)=z^3-z_1 z_2 z_3$ for any $z$, and $z_1^3=z_2^3=z_3^3(=z_1 z_2 z_3)$.

Let $f=(t-z_1)(t-z_2)(t-z_3)$.

Then $f$ is a cubic polynomial in $t$, with roots $z_1,z_2,z_3$.

In expanded form, $f$ can be expressed as
$$f=t^3-at^2+bt-c$$
where
beginalign*
a&=z_1+z_2+z_3\[4pt]
b&=z_1z_2+z_2z_3+z_3z_1\[4pt]
c&=z_1z_2z_3\[4pt]
endalign*
From $z_1+z_2+z_3=0$, we get $a=0$, and since we also have $z_1^2+z_2^2+z_3^2=0$, the identity
$$(z_1+z_2+z_3)^2=z_1^2+z_2^2+z_3^2+2(z_1z_2+z_2z_3+z_3z_1)$$
yields $b=0$.

Thus, $f=t^3-c$, hence, since
$$f(z_1)=f(z_2)=f(z_3)=0$$
we get
$$z_1^3=z_2^3=z_3^3=c$$
so $|z_1|^3=|z_2|^3=|z_3|^3$, which yields $|z_1|=|z_2|=|z_3|$.