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Trigo + Complex

Let $$A_1,A_2,...A_n$$ be a regular polygon of $$n$$ sides whose centre is origin $$O$$.Llet the complex numbers representing the vertices $$A_1,A_2.....A_n$$be$$z_1,z_2....z_n$$ respectively.The radius is of length unity.Then, $\prod_{n=2}^{n}|A_1A_n|=\mathbb{?}$

2 years, 2 months ago

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Let the circumradius be $$R$$. Now we consider all $$z$$ that satisfy the circle and write a certain expansion:

$z^n-R^n=(z-R)(z^{n-1} + Rz^{n-2}+Rz^{n-3} + \dots + zR^{n-2}+R^{n-1})$

Then we have $$z_1=R$$ and $$,z_2,z_3\dots z_n$$ as roots of $$(z^{n-1} + Rz^{n-2}+Rz^{n-3} + \dots + zR^{n-2}+R^{n-1})$$ , So we can write :

$(z-z_2)(z-z_3) \dots (z-z_n)=z^{n-1} + Rz^{n-2}+Rz^{n-3} + \dots + zR^{n-2}+R^{n-1})$

Since $$z_1=R$$ is one of the $$z$$'s , we substitute $$z=R$$ in RHS . Thus the equation becomes:

$|(z-z_1)||(z-z_2)||(z-z_3)| \dots |(z-z_n)|=nR^{n-1} \dots (1)$

To calculate $$R$$ we use cosine rule and get $$R=\dfrac{1}{2\sin\left(\dfrac{\pi}{n}\right)}$$. So finally

$\displaystyle\prod_{i=2}^n |A_1A_n|=\dfrac{n}{2\sin^{n-1}\left(\dfrac{\pi}{n}\right)}$

Edit: The problem is edited and the circumradius is unity and not the side. Hence substituting $$R=1$$ in $$(1)$$ we get:

$\displaystyle\prod_{i=2}^n |A_1A_n|=n\times 1^{n-1} = \boxed{n}$

- 2 years, 2 months ago

Answer is $$n$$.

- 2 years, 2 months ago

Are you sure that the side equals unity? Or the circumradius?

- 2 years, 2 months ago

Oh!Sorry!R=1!

- 2 years, 2 months ago

Yes , now the answer is $$n$$.

- 2 years, 2 months ago

I asked this question a while ago and Alan Yan wrote a nice solution

- 2 years, 2 months ago

Thanks a lot sir!

- 2 years, 2 months ago