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.

Problem no 1.160 on my previous question.

Let $ABCD...PQ$ represent a regular polygon of $n$ sides inscribed in a circle of unit radius. Prove that the product of the lengths of the diagonals $AC, AD, ... , AP$ is $\frac14 n \csc^2 \left( \frac{\pi}{n} \right )$.

How do I proceed? I am thinking $\displaystyle \prod_{k=1}^{n-2} \left | 1 - e^{i\frac{2\pi k }{n}} \right |$, am I in right direction? And how do I simplify it?

share|improve this question
    
what is the source of the problems? –  lab bhattacharjee Feb 21 '13 at 15:57
    
@labbhattacharjee it's Schaums Series Complex Variables ... want it?? –  hasExams Feb 21 '13 at 15:57
    
Well, $k$ should go from $2$ to $n-2$ in that expression. –  Thomas Andrews Feb 21 '13 at 16:01
    
1  
I think the answer should be $\frac{1}{4}n\csc^2 \pi/n$. For example, when $n=4$, there is one diagonal of length $2$, and $\csc \pi/4 = \sqrt 2$ –  Thomas Andrews Feb 21 '13 at 16:13

1 Answer 1

up vote 2 down vote accepted

Define $p(z)=\prod_{k=1}^{n-1} \left(z-e^{2\pi ik/n}\right)=1+z+z^2+... + z^{n-1}$.

Then note that the value you are looking for is:

$$\left|\frac{p(1)} {(1-e^{2\pi i/n})(1-e^{-2\pi i/n})}\right|$$

But $p(1)=n$. And the denominator is $2-2\cos(2\pi/n)$. But $1-\cos 2x = 2\sin^2 x$, so we get that the value you are looking for is $$\frac{n}{4\sin^2 \frac{\pi}{n}}$$

share|improve this answer
    
why am i looking for that form?? –  hasExams Feb 21 '13 at 16:23
    
Form was a typo, meant "for." But you are looking for that value because $|p(1)|$ is the product of all the lengths from $AB\cdot AC\cdot AD\dots AQ$ and the denominator is dividing by the lengths that are not diagonals, $AB\cdot AQ$. –  Thomas Andrews Feb 21 '13 at 16:25
    
Oh!! I see .. also I understand the comment you made. I was reading improperly. –  hasExams Feb 21 '13 at 16:28

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.