For any three complex numbers $z_1, z_2, z_3$ on the unit circle, $|z_1 + z_2 + z_3| = |z_1 z_2 + z_1 z_3 + z_2 z_3|$.

I am able to prove this by putting each number in modulus-argument form and then expanding algebraically, but this is somewhat inelegant and tedious (supposing the arguments are $\alpha, \beta, \gamma$ respectively):

$$ LHS^2 =(\cos \alpha + \cos \beta + \cos \gamma)^2 + (\sin \alpha + \sin \beta + \sin \gamma)^2 = \cos^2\alpha + \cos^2\beta+ \cos^2\gamma + 2(\cos \alpha \cos \beta + \cos \alpha \cos \gamma+ \cos \gamma \cos \beta) +\sin^2\alpha + \sin^2\beta+ \sin^2\gamma + 2(\sin \alpha \sin \beta + \sin \alpha \sin \gamma+ \sin \gamma \sin \beta) =3+2(\cos (\alpha -\beta) + \cos (\alpha -\gamma) + \cos (\beta -\gamma)) $$ And similarly for RHS.

Question: What other methods are there to prove this identity?

  • 1
    $\begingroup$ on the right hand side take out a factor of $z_1z_2z_3$ then note that $\frac{1}{z_1} = z_1$ conjugate etc. $\endgroup$ – Paul Oct 20 '14 at 9:56
  • $\begingroup$ @Paul That definitely deserves to be posted as an answer in my opinion, would you prefer to repost as an answer and for me to retract my post, since you posted first? $\endgroup$ – Sherlock Holmes Oct 20 '14 at 10:17
  • $\begingroup$ @Sherlock - not at all, don't worry about it:) $\endgroup$ – Paul Oct 20 '14 at 10:23

$RHS$$=|z_1z_2+z_2z_3+z_3z_1|=|z_1z_2z_3||\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}| \\ =|\bar{z_1}+\bar{z_2}+\bar{z_3}|=|\overline{z_1+z_2+z_3}|=|z_1+z_2+z_3|=LHS \\ \square$

Edit: Kudos to Paul, who posted the general principle as a comment first .


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