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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?

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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
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
up vote 3 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 $$\frac{n}{4\sin^2 \frac{\pi}{n}}$$

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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

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