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.
Writing $z_k = e^{i\theta_k}$, $k = 1,2,\ldots, n$, we write the above expression as
\begin{align}&(1 + e^{i(\theta_2 - \theta_1)})(1 + e^{i(\theta_3 - \theta_2)})\cdots (1 + e^{i(\theta_n - \theta_{n-1})})(1 + e^{i(\theta_1 - \theta_n)})\\ &= e^{-i(\theta_2 - \theta_1)/2}e^{-i(\theta_3 - \theta_2)/2}\cdots e^{-i(\theta_n - \theta_{n-1})/2}(e^{-i(\theta_2- \theta_1)/2} + e^{i(\theta_2 - \theta_1)/2})\cdots (e^{-i(\theta_1 - \theta_n)/2} + e^{i(\theta_1 - \theta_n)/2})\\ &= e^{-i[(\theta_2 - \theta_1) + (\theta_3 - \theta_2) + \cdots + (\theta_1 - \theta_n)]/2}\cdot 2^n\cos((\theta_2 - \theta_1)/2)\cos((\theta_3 - \theta_2)/2)\cdots \cos((\theta_1 - \theta_n)/2)\\ &= 2^n\cos((\theta_2 - \theta_1)/2)\cos((\theta_3 - \theta_2)/2)\cdots \cos((\theta_1 - \theta_n)/2). \end{align}
The last expression is a real number.
Use induction:
for $n=2$, we have $$\big( 1 + \frac{z_2}{z_1} \big) \cdot \big( 1 + \frac{z_1}{z_2} \big)= 2 + \frac{z_2}{z_1} + \frac{z_1}{z_2} \in\mathbb{R} $$ since $ \dfrac{z_2}{z_1} = \exp\big( i \theta \big)$ for some $\theta\in\mathbb{R}$ while $\dfrac{z_1}{z_2} = \exp\big( - i\theta \big)$ and $ \exp\big( i\theta \big) + \exp\big( - i\theta \big) = 2 \cos\theta$ is real.
Assume the result holds true for $k = n \geq 2$, that is, $$\big( 1 + \frac{z_2}{z_1} \big) \cdot \big( 1 + \frac{z_3}{z_2} \big)\ldots \big( 1 + \frac{z_n}{z_{n-1}} \big) \cdot \big( 1 + \frac{z_1}{z_n} \big)\in\mathbb{R} , $$ then for $k = n+1$, we have \begin{align} & \qquad \big( 1 + \frac{z_2}{z_1} \big) \cdot \big( 1 + \frac{z_3}{z_2} \big)\ldots \big( 1 + \frac{z_n}{z_{n-1}} \big) \cdot \big( 1 + \frac{z_{n+1}}{z_n} \big) \big( 1 + \frac{z_1}{z_{n+1}} \big) \\ & = \left[ \big( 1 + \frac{z_2}{z_1} \big) \cdot \big( 1 + \frac{z_3}{z_2} \big)\ldots \big( 1 + \frac{z_n}{z_{n-1}} \big) \cdot \big( 1 + \frac{z_1}{z_n} \big) \right]\times \frac{ \big( 1 + \dfrac{z_{n+1}}{z_n} \big) \big( 1 + \dfrac{z_1}{z_{n+1}} \big)}{ \big( 1 + \dfrac{z_1}{z_n} \big)} \end{align} now it suffices to show $ \frac{ \big( 1 + \dfrac{z_{n+1}}{z_n} \big) \big( 1 + \dfrac{z_1}{z_{n+1}} \big)}{ \big( 1 + \dfrac{z_1}{z_n} \big)}$ is a real number.
Clearly, $ \frac{ \big( 1 + \dfrac{z_{n+1}}{z_n} \big) \big( 1 + \dfrac{z_1}{z_{n+1}} \big)}{ \big( 1 + \dfrac{z_1}{z_n} \big)} = \frac{ \big( 1 + \dfrac{z_{n+1}}{z_n} \big) \big( 1 + \dfrac{z_1}{z_{n+1}} \big) \big( 1 + \dfrac{z_n}{z_1} \big)}{ \big( 1 + \dfrac{z_1}{z_n} \big) \big( 1 + \dfrac{z_n}{z_1} \big)}$ is real, since the denominator is real by step-1 and the nominator is real by some easy calculation. Q.E.D
