given: $$ \left | z_{1}\right |=\left | z_{2}\right |=...=\left | z_{n}\right |=1 $$
How do I prove:
$$ (1+\frac{z_{2}}{z_{1}})(1+\frac{z_{3}}{z_{2}})*...*(1+\frac{z_{n}}{z_{n-1}})(1+\frac{z_{1}}{z_{n}}) \in \mathbb{R} $$
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given: $$ \left | z_{1}\right |=\left | z_{2}\right |=...=\left | z_{n}\right |=1 $$
How do I prove:
$$ (1+\frac{z_{2}}{z_{1}})(1+\frac{z_{3}}{z_{2}})*...*(1+\frac{z_{n}}{z_{n-1}})(1+\frac{z_{1}}{z_{n}}) \in \mathbb{R} $$
?
You need to consider of the angle from the origin of each complex number $1 + \frac{z_{k+1}}{z_k}$. That's $\frac{1}{2}\arg \frac{z_{k+1}}{z_k} = \frac{1}{2}(\arg z_{k+1} - \arg z_k)$. Add these up and see they cancel out to give $0$.