Prove that the complex expression is real Let $|z_1|=\dots=|z_n|=1$ on the complex plane.
Prove that:
$$
\left(1+\frac{z_2}{z_1}\right)
\left(1+\frac{z_3}{z_2}\right)
\dots
\left(1+\frac{z_n}{z_{n-1}}\right)
\left(1+\frac{z_1}{z_n}\right)
\in\mathbb{R}
$$
I have tried induction and writing every "subexpression" as $(1+e^{i(\theta_n-\theta_{n-1})})$.
Any ideas?

We can do this without appealing to trigonometry by noticing that a number is real if any only if it equals its conjugate and that the conjugate of a number on the unit circle equals its reciprocal. In particular, let
$$x=\left(1+\frac{z_2}{z_1}\right)
\left(1+\frac{z_3}{z_2}\right)
\dots
\left(1+\frac{z_n}{z_{n-1}}\right)
\left(1+\frac{z_1}{z_n}\right)$$
then, we take the conjugate of the above, noting that $\frac{z_2}{z_1}$ is the conjugate of $\frac{z_1}{z_2}$ and that conjugates preserve addition and multiplication:
$$\bar x = \left(1+\frac{z_1}{z_2}\right)
\left(1+\frac{z_2}{z_3}\right)
\dots
\left(1+\frac{z_{n-1}}{z_{n}}\right)
\left(1+\frac{z_n}{z_1}\right)$$
Then, we can clearly multiply each side by $1=\frac{z_2}{z_1}\cdot\frac{z_3}{z_2}\ldots \frac{z_{n}}{z_{n-1}}\cdot \frac{z_1}{z_n}$ and get, after rearranging:
$$\bar x = \left(\frac{z_2}{z_1}\cdot \left(1+\frac{z_1}{z_2}\right)\right)\cdot\left(\frac{z_3}{z_2}\cdot \left(1+\frac{z_2}{z_3}\right)\right)\ldots \left(\frac{z_n}{z_{n-1}}\cdot \left(1+\frac{z_{n-1}}{z_n}\right)\right)\cdot\left(\frac{z_1}{z_n}\cdot \left(1+\frac{z_n}{z_1}\right)\right)$$
and simplifying each term yields:
$$\bar x = \left(1+\frac{z_2}{z_1}\right)
\left(1+\frac{z_3}{z_2}\right)
\dots
\left(1+\frac{z_n}{z_{n-1}}\right)
\left(1+\frac{z_1}{z_n}\right)$$
Where the right hand side was the definition of $x$, so $\bar x = x$, meaning $x$ is real.
