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Question on complex numbers.. Q. Let Po,P1,P2,P3,P4,.........,P(n-1) are vertices of a regular polygon in a unit circle,then prove that- (a)PoP1*PoP2*PoP3*..............PoP(n-1)=n (b)sin pi/n * sin 2pi/n * sin 3pi/n*........*sin[(n-1)pi/n]=2/2^(n-1).




  • Question on complex numbers..
    Q. Let Po,P1,P2,P3,P4,.........,P(n-1) are vertices of a regular polygon in a unit circle,then prove that-
    (a)PoP1*PoP2*PoP3*..............PoP(n-1)=n
    (b)sin pi/n * sin 2pi/n * sin 3pi/n*........*sin[(n-1)pi/n]=2/2^(n-1).




Grade:upto college level

1 Answers

Amit Saxena
35 Points
10 years ago

 we can see these pt. P0,P1,P2,P3...Pn-1  as the roots of unity as lie on a unit circle
and vertices of a REGULAR polygon in the argand plane the distance of all these pts. will be same  and also they are seperated by equal angles 2pie/n
so it has n roots e^(ik2pie /n)   k = 0 .........n-1
(z-z1)....(z-zn-1) = z^n-1 +z^n-2 .........+1

now take z = 1,which is P0 , we getPoP1*PoP2*PoP3*..............PoP(n-1)=n
where  (z-z1) = P1P0 and so on

origin is centre of the regular polygon. and line joining origin and P0 is the real axis

b) (1-z1)(1-z2)....(1-zn-1) = n
  mod (1-zk) = mod {1-cosk(2pie/n) - sin(2pie/n)k}
mod(1-zk) = 2 sin (piek /18)

multiplying all these values we get from k = 0 to n-1


sin pi/n * sin 2pi/n * sin3pi/n*........*sin[(n-1)pi/n]=2/2^(n-1)

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