Trigo + Complex

Let A1,A2,...AnA_1,A_2,...A_n be a regular polygon of nn sides whose centre is origin OO.Llet the complex numbers representing the vertices A1,A2.....AnA_1,A_2.....A_nbez1,z2....znz_1,z_2....z_n respectively.The radius is of length unity.Then, n=2nA1An=?\prod_{n=2}^{n}|A_1A_n|=\mathbb{?}

Note by Adarsh Kumar
4 years ago

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Let the circumradius be RR. Now we consider all zz that satisfy the circle and write a certain expansion:

znRn=(zR)(zn1+Rzn2+Rzn3++zRn2+Rn1)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 z1=Rz_1=R and ,z2,z3zn,z_2,z_3\dots z_n as roots of (zn1+Rzn2+Rzn3++zRn2+Rn1)(z^{n-1} + Rz^{n-2}+Rz^{n-3} + \dots + zR^{n-2}+R^{n-1}) , So we can write :

(zz2)(zz3)(zzn)=zn1+Rzn2+Rzn3++zRn2+Rn1)(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 z1=Rz_1=R is one of the zz's , we substitute z=Rz=R in RHS . Thus the equation becomes:

(zz1)(zz2)(zz3)(zzn)=nRn1(1)|(z-z_1)||(z-z_2)||(z-z_3)| \dots |(z-z_n)|=nR^{n-1} \dots (1)

To calculate RR we use cosine rule and get R=12sin(πn)R=\dfrac{1}{2\sin\left(\dfrac{\pi}{n}\right)}. So finally

i=2nA1An=n2sinn1(πn)\displaystyle\prod_{i=2}^n |A_1A_n|=\dfrac{n}{2\sin^{n-1}\left(\dfrac{\pi}{n}\right)}

I may be wrong... Adarsh please check

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

i=2nA1An=n×1n1=n \displaystyle\prod_{i=2}^n |A_1A_n|=n\times 1^{n-1} = \boxed{n}

Nihar Mahajan - 4 years ago

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Answer is nn.

Adarsh Kumar - 4 years ago

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

Nihar Mahajan - 4 years ago

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@Nihar Mahajan Oh!Sorry!R=1!

Adarsh Kumar - 4 years ago

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@Adarsh Kumar Yes , now the answer is nn.

Nihar Mahajan - 4 years ago

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

Otto Bretscher - 4 years ago

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Thank you for the link!

Nihar Mahajan - 4 years ago

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Thanks a lot sir!

Adarsh Kumar - 4 years ago

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