On complex numbers and absolute values

Exercise 1.31 of Analysis by Apostol states:

Given three complex numbers $z_1,z_2,z_3$ such that $|z_1| = |z_2| = |z_3| = 1$ and $z_1 + z_2 +z_3 = 0$. Show that these numbers are vertices of an equilateral triangle inscribed in the unit circle with centre at the origin.

Solution: It is clear that the three numbers are vertices of a triangle inscribed in the unit circle with centre in the origin. It remains to show that $|z_1-z_2| = |z_2 -z_3| = |z_3 - z_1|$

Note that $|2z_1 + z_3| = |2z_3 + z_1|$ by $z_1 + z_2 +z_3 = 0$

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I have a problem noticing that $|2z_1 + z_3| = |2z_3 + z_1|$ by $z_1 + z_2 +z_3 = 0$ can anybody help me out?

By direct computation, $$|2z_1+z_3|^2=(2z_1+z_3)(2\bar{z}_1+\bar{z}_3)=4|z_1|^2+2z_3\bar{z}_1+2z_1\bar{z}_3+|z_3|^2\\ =5+2z_3\bar{z}_1+2z_1\bar{z}_3.$$ Similarly, $$|z_1+2z_3|^2=(z_1+2z_3)(\bar{z}_1+2\bar{z}_3)=|z_1|^2+2z_3\bar{z}_1+2z_1\bar{z}_3+4|z_3|^2\\ =5+2z_3\bar{z}_1+2z_1\bar{z}_3.$$ So the equality you seek follows from $|z_1|=|z_3|$ rather than $z_1+z_2+z_3=0$.