Showing that for complex numbers $z_1,z_2,z_3$, $|z_1z_2+z_2z_3+z_3z_1|=|z_1+z_2+z_3|$ given that $|z_i|=1$. 
Let $z_1,z_2$ and $z_3$ be complex numbers such that $|z_i|=1$ for $i=1,2,3$. Show that $|z_1z_2+z_2z_3+z_3z_1|=|z_1+z_2+z_3|$

This is an exercise from Jonathan S. Golan's book on Linear Algebra. To check that it is true for particular values of $z_i$, it is quite obvious that it holds when $z_i$ are the cube roots of unity.
I have checked it by computing both sides with $z_i=e^{i\theta_i}$, but the computation is extremely long and tedious. I was wondering if there is a more elegant way of proving this.
Apologies if this is a duplicate.
 A: Note that $|z_1z_2z_3|=1$. Using the property that $|xy|=|x|\cdot |y|$, the LHS simplifies to
$$\left|\sum_\text{cyc} z_1z_2\right|$$
$$=\left|\sum_\text{cyc} \frac{z_1z_2z_3}{z_1}\right|$$
$$=|z_1z_2z_3|\cdot\left|\sum_\text{cyc} \frac{1}{z_1}\right|$$
$$=1\cdot\left|\sum_\text{cyc} \frac{1}{z_1}\right|$$
Note that since $|z_1|=1$, we have $\frac{1}{z_1}=\overline{z_1}$, where $\overline{x}$ is the complex conjugate of $x$. Hence, the LHS is equivalent to
$$\left|\sum_\text{cyc} \overline{z_1}\right|$$
$$=\left|\overline{\sum_\text{cyc} z_1}\right|$$
Using the fact that $|x|=|\overline{x}|$, this is equivalent to
$$\left|\sum_\text{cyc} z_1\right|$$
Hence, the LHS is equivalent to the RHS.
A: We can rotate $z_1, z_2, z_3$ in a way so $z_1=1$. So let's suppose that $z_1=1$. The equation simplifies to:
$$|z_2+z_2z_3+z_3|=|1+z_2+z_3|$$
Conjugation doesn't affect the absolute value, so we can rewrite the equation to:
$$|\overline{z_2}+\overline{z_2z_3}+\overline{z_3}|=|1+z_2+z_3|$$
Conjugation of a complex unit is equal to its inverse.
$$\left|\frac1{z_2}+\frac1{z_2z_3}+\frac1{z_3}\right|=|1+z_2+z_3|$$
$$\left|\frac1{z_2z_3}\right|\left|z_3+1+z_2\right|=|1+z_2+z_3|$$
And the absolute value of $\frac1{z_2z_3}$ is 1, so the equation holds.
A: We have that $|z_1|=|z_2|=|z_3|=1$. So $|z_1 z_2 z_3|=|z_1|| z_2|| z_3|=1$
So
$$|z_1 z_2+z_2 z_3+z_3 z_1|=|z_1 z_2 z_3||\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}|=
|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}|$$
$$=\left|\frac{\overline{z_1}}{|z_1|^2}+\frac{\overline{z_2}}{|z_2|^2}+\frac{\overline{z_3}}{|z_3|^2}\right|=|\overline{z_1}+\overline{z_2}+\overline{z_3}|$$
$$=|\overline{z_1+z_2+z_3}|=|z_1+z_2+z_3|.$$
A: The way you proposed to derive the identity can be made a bit less tedious by not trying to meet all the multiplications "head-on".  If we write $ \ z_{i} \ = \ e^{ \ i \ · \ \theta_{i}} \ \ , $ the sum $ \ z_1 + z_2 + z_3 \ $ has some (generally non-unit) modulus $ \ | \ z_1 + z_2 + z_3 \ | \ \ . $  On the Argand diagram, conjugation is equivalent to reflection about the " $ \ x-$axis", so certainly $ \ | \ z_1 + z_2 + z_3 \ | \  = \ | \ \overline{z_1} + \overline{z_2} + \overline{z_3} \ | \ = \ | \ e^{ \ -i \ · \ \theta_1} + e^{ \ -i \ · \ \theta_2} + e^{ \ -i \ · \ \theta_3} \ | $ $ = \left| \ \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \ \right| \ \ . $
But since $$ \ | \ z_1 \ · \ z_2 \ · \ z_3 \ | \ \  = \  \ | \ e^{ \ i \ · \ (\theta_1 + \theta_2 + \theta_3)}  \ | \ \ = \ \ 1 \ \ , $$
we may write
$$  | \ z_1 + z_2 + z_3 \ | \ \  = \ \ | \ z_1 \ · \ z_2 \ · \ z_3 \ | \ · \ \left| \ \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \ \right| $$
$$  = \  \  \left| \ (z_1  ·  z_2 \ ·  z_3) \ · \ \left(\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right) \ \right|  $$ $$ = \ \  | \ [ \ e^{ \ i \ · \ (\theta_1 + \theta_2 + \theta_3)} \ ] \ · \ (e^{ \ -i \ · \ \theta_1} \ + \ e^{ \ -i \ · \ \theta_2} \ + \ e^{ \ -i \ · \ \theta_3}) \  | $$
$$ = \ \  | \  e^{ \ i \ · \ (\theta_2 + \theta_3)} \ +  \ e^{ \ i \ · \ (\theta_1 + \theta_3)} \ +  \ e^{ \ i \ · \ (\theta_1 + \theta_2)}  \  | \ \ 
  = \ \  | \ z_2z_3 \   +    \ z_1z_3 \ + \ z_1z_2 \ |  \ \ .  $$
(Except for the choice of using the "polar form", this approach is not significantly different from the other algebraic calculations from the other respondents.)
