Moduli and conjugates

Problem:

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

$z_{1}+z_{2}+z_{3}=z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}=0$

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

Proof:

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

This entry was written by Nalin Pithwa, posted on May 22, 2016 at 1:21 pm, filed under Complex Numbers, IITJEE Mains. Bookmark the permalink. Follow any comments here with the RSS feed for this post. Post a comment or leave a trackback: Trackback URL.

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