# Proving that $\left(1+\frac{z_{1}}{z_{2}}\right)\left(1+\frac{z_{2}}{z_{3}}\right)…\left(1+\frac{z_{n}}{z_{1}}\right)\in\mathbb R$

I need to prove that $\left(1+\frac{z_{1}}{z_{2}}\right)\left(1+\frac{z_{2}}{z_{3}}\right)...\left(1+\frac{z_{n}}{z_{1}}\right)\in\mathbb R$ where $|z_{1}|=|z_{2}|=...=|z_{n}|=1$.

This can be done relatively easily by induction, but I'm looking for more elegant solution.

Any ideas?

Thanks!

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Any solution must use induction at some point since you need to prove the statement for all $n \in \mathbb{N}$. – user17762 Dec 24 '12 at 6:40
I didn't try it, but generally showing $z$ is real can be done by showing that $\bar{z}=z$. – Quinn Culver Dec 24 '12 at 6:48

Let $z_k = e^{i \phi_k}$. Hence, we have $$f=\left(1+ \dfrac{z_1}{z_2} \right)\left(1+ \dfrac{z_2}{z_3} \right)\cdots\left(1+ \dfrac{z_n}{z_1} \right) = \prod_{k=1}^n \left(1+e^{ia_k} \right)$$ where $\displaystyle \sum_{k=1}^n a_k = 0$. Now $$\bar{f} = \prod_{k=1}^n \left(1+e^{-ia_k} \right) = \dfrac{\displaystyle \prod_{k=1}^n \left(1+e^{ia_k} \right)}{\displaystyle \prod_{k=1}^n e^{ia_k}} = \dfrac{\displaystyle \prod_{k=1}^n \left(1+e^{ia_k} \right)}{\exp\left({i \displaystyle \sum_{k=1}^n a_k} \right)} = \dfrac{f}{e^0} = f$$ Hence, $f$ is real.

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Its very simple: $$1+{z_k\over z_{k+1}}=z_k\ \bar z_{k+1}\Bigl(1+{\bar z_k\over \bar z_{k+1}}\Bigr)\ ,$$ and the cyclic product $\prod_k z_k\bar z_{k+1}$ is $1$.

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Let $z_j=\cos 2t_j+i\sin 2t_j$

So, $$1+\frac{z_j}{z_k}=1+\frac{\cos 2t_j+i\sin 2t_j}{\cos 2t_k+i\sin 2t_k}$$ $$=1+\cos2(t_i-t_j)+i\sin2(t_i-t_j)=2\cos(t_i-t_j)\{\cos(t_i-t_j)+i\sin(t_i-t_j)\}=2\cos(t_i-t_j)e^{i(t_i-t_j)}$$

Putting $i,j=1,2;2,3;\cdots;n,1,$ and taking the product

$$\left(1+\frac{z_{1}}{z_{2}}\right)\left(1+\frac{z_{2}}{z_{3}}\right)...\left(1+\frac{z_{n}}{z_{1}}\right)=2\cos(t_n-t_1)\prod_{1\le i\le n-1}2\cos(t_i-t_{i+1})$$

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 should be 2\cos(t_i-t_j)e^(i(t_i-t_j))? – Apprentice Queue Dec 24 '12 at 8:48 @ApprenticeQueue, thanks, rectified. – lab bhattacharjee Dec 24 '12 at 8:58

Here's a geometric way of looking at it. Multiplication by a complex number can be considered as a rotation by its argument, with a scaling by its modulus. Since each $z_k$ is on the unit circle, multiplication by $z_k+z_{k+1}\;\;(\text{modulo}\;n;\; k=1\;$,...,$\;n)$ imparts a rotation by the average of the arguments of $z_k$ and $z_{k+1}$, while division by $z_{k+1}$ rotates backwards by its argument. The cyclic combination of these rotations cancels out to a null rotation, leaving only a scaling effect.

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