Complex number modulus identity (on unit circle)

Question

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?

Answer 1

$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 .