How to prove that at least two of three complex numbers have the same absolute value?

Question

Let $a, b,c$ complex numbers such that $a + b + c =0$ and $a^n + b^n + c^n =0$ for some integer $ n \geq 2$. Then two of them have the same absolute values. I try a recursive method but doesn't work. Can you give me an idea? Thank you!

Answer 1

We start with the following Lemma.

For the roots $z_i$ of polynomials $$\begin{align} P_k(z)&=\left(z+\frac{1}{2}\right)^{2k}+\left(z-\frac{1}{2}\right)^{2k}+1\\ Q_k(z)&=\left(z+\frac{1}{2}\right)^{2k+1}-\left(z-\frac{1}{2}\right)^{2k+1}-1\end{align} $$ with integer $k>0$ at least one of the following statements is valid: $$ \left|z_i-\frac{1}{2}\right|=1,\;\left|z_i+\frac{1}{2}\right|=1, \text{ or } \Re(z_i)=0\tag{1}. $$

Observe that both polynomials are of degree $2k$ and have correspondingly $2k$ complex roots. The Lemma for $P_k(z)$ was recently proved elsewhere. The zeros of polynomial $Q_k(z)$, disregarding two trivial real ones $\pm\frac{1}{2}$ obviously satisfying the condition (1), have exactly the same structure (including the special character of the roots $\pm\frac{i\sqrt3}{2}$) as the zeros of $P_k(z)$, which can be proved in completely same way (is left as an exercise).

Consider now the equations: $$ a+b+c=0,\quad a^n+b^n+c^n=0. $$

Without lost of generality we can assume that neither of $a,b,c$ is $0$. Indeed, if one of them, say $c$ is zero, the claim trivially follows. Substituting $a$ from the first equation into the second one, one obtains: $$ (-b-c)^n+b^n+c^n=0,\;\text{ or }\; T_k(z):=\left(-z-\frac{1}{2}\right)^{n}+\left(z-\frac{1}{2}\right)^{n}+1=0, $$ where $z=\frac{b}{c}+\frac{1}{2}$ was introduced. Observe that in terms of $z$ the claim: $$ \left|\frac{a}{b}\right|=1,\text{ or }\left|\frac{b}{c}\right|=1,\text{ or }\left|\frac{c}{a}\right|=1$$ translates into: $$ \left|\frac{z+\frac{1}{2}}{z-\frac{1}{2}}\right|=1,\text{ or }\left|z-\frac{1}{2}\right|=1,\text{ or }\left|z+\frac{1}{2}\right|=1,$$ the first of the equalities being equivalent to $\Re(z)=0$.

Thus, the claim follows directly from the identities $T_{2k}(z)\equiv P_k(z)$ and $T_{2k+1}(z)\equiv-Q_k(z)$.