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$z_1, z_2, ... z_n$ 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.

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

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

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