# Prove that $\frac{(z_1 + z_2)(z_2 + z_3)…(z_{n-1} + z_n)(z_n + z_1)}{z_1 \cdot z_2 \cdot … \cdot z_n}$ is real

$$z_1, z_2, ... z_n$$ are complex numbers such that $$|z_1| = |z_2| = ... = |z_n|$$. How to prove that $$\frac{(z_1 + z_2)(z_2 + z_3)...(z_{n-1} + z_n)(z_n + z_1)}{z_1 \cdot z_2 \cdot ... \cdot z_n}$$ is real? I've tried writing $$z_1, z_2, ..., z_n$$ in polar form, but couldn't figure out too much from abundance of sines/cosines.

• If $\,|z_1|=|z_2|>0\,$ then $\,2\arg(z_1+z_2) = \arg(z_1)+\arg(z_2).$ – Somos Oct 15 '18 at 20:02

## 2 Answers

Note that $$w$$ is real if and only if $$w=\bar w$$, and that when $$|w|=1$$, $$\bar w = 1/w$$. It then suffices to show your expression is invariant when you replace every $$z_i$$ by its inverse. But this can be seen by noting your expression is

$$w = \left( 1 + \frac{z_2}{z_1}\right)\left(1+\frac{z_3}{z_2}\right)\cdots \left(1+\frac{z_{n}}{z_{n-1}}\right)\left(1+\frac{z_1}{z_n}\right)$$

which is built to be invariant under such map.

Hint: We may assume $$|z_k|=1$$. Then compute the conjugate of the expression using that $$z_k \bar z_k =1$$.