Equality for modulus of 3 complex numbers

Question

Let $z_1,z_2,z_3 \in \mathbb{C}$ $z_1,z_2,z_3 \neq0 $ such that $z_1+z_2+z_3=z^7_1+z^7_2+z^7_3=0$

Prove that $\mid z_1 \mid=\mid z_2 \mid =\mid z_3 \mid $

Any ideas? I was thinking that $z_1,z_2,z_3$ are affixes of the vertices of an equilateral triangle.

Answer 1

Let $z_1=a$, $z_2=b$ and $z_3=c$.

Thus, $c=-a-b$ and $$0=-a^7-b^7-c^7=(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2,$$ which gives $$a^2+ab+b^2=0,$$ which gives $$a^3-b^3=0,$$ which gives $$|a|^3=|b|^3,$$ which gives $$|a|=|b|.$$ Now, $$|c|^2=|-a-b|^2=|a^2+2ab+b^2|=|ab|=|a|^2,$$ which gives $$|a|=|b|=|c|.$$ $$(a+b)^7-a^7-b^7=7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6=$$ $$=7ab(a^5+b^5+3a^4b+3ab^4+5a^3b^2+5a^2b^3)=$$ $$=7ab(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4+3ab(a^2-ab+b^2)+5a^2b^2)=$$ $$=7ab(a+b)(a^4+2a^3b+3a^2b^2+2ab^3+b^4)=$$ $$=7ab(a+b)(a^2+ab+b^2)^2.$$

Answer 2

Eliminating $z_3$ gives $$z_1^7+z_2^7-(z_1+z_2)^7=0.$$ This is $$z_1^6z_2+3z_1^5z_2^2+5z_1^4z_2^3+5z_1^3z_2^4+3z_1^2z_2^5+z_1z_2^6=0$$ and this factors as $$z_1z_2(z_1+z_2)(z_1^2+z_1z_2+z_2^2)^2=0.$$ So we could have some $z_k=0$, but otherwise $z_1^2+z_1z_2+z_2^2=0$ etc.

Answer 3

Sorry I should comment but I don't have enough reputation.

Credits to Lord Shark the Unknown for reducing the problem to $z_1 z_2 (z_1 + z_2)(z_1^2 + z_1 z_2 + z_2^2)^2 = 0$, which is simplified to $z_1^2 + z_1 z_2 + z_2^2 = 0$.
Solving gives $\frac{z_1}{z_2} = -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i$ which is a 3rd root of unity.

You can complete the proof from here.