Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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
1  
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

4 Answers 4

up vote 14 down vote accepted

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.

share|improve this answer

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 $1$.

share|improve this answer

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

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

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.