Moduli and conjugates


Let z_{1}, z_{2}, z_{3} be complex numbers such that


Prove that |z_{1}|=|z_{2}|=|z_{3}|.


Method I:

Substituting z_{1}+z_{2}=-z_{3} in z_{1}z_{2}+z_{3}(z_{1}+z_{2})=0 gives z_{1}z_{2}=z_{3}^{2}, so |z_{1}|.|z_{2}|=|z_{3}|^{2}.

Likewise, |z_{2}|.|z_{3}|=|z_{1}|^{2} and |z_{3}|.|z_{1}|=|z_{2}|^{2}. Then,

|z_{1}|^{2}+|z_{2}|^{2}+|z_{3}|^{2}=|z_{1}|.|z_{2}|+|z_{2}|.|z_{3}|+|z_{3}|.|z_{1}|, that is,

(|z_{1}|-|z_{2}|)^{2}+(|z_{2}|-|z_{3}|)^{2}+(|z_{3}|-|z_{1}|)^{2}=0 yielding |z_{1}|=|z_{2}|=|z_{3}|.

Method II:

Using the relations between the roots and the coefficients, it follows that z_{1}, z_{2}, z_{3} are the roots of polynomial z^{3}-p, where p=z_{1}z_{2}z_{3}. Hence, z_{1}^{3}-p=z_{2}^{3}-p=z_{3}^{3}-p=0 implying z_{1}^{3}=z_{2}^{3}=z_{3}^{3}, and the conclusion follows.

More complex fun later,

Nalin Pithwa

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