Asked by Kaur
If (costheta + i sintheta) (cos2theta+ isin2theta) (cos3theta + i sin3theta).............
(cos2ntheta + isin2ntheta) = 1, then value of theta is
(cos2ntheta + isin2ntheta) = 1, then value of theta is
Answers
Answered by
oobleck
cisθ + i sinθ = e^(iθ)
so you have
e^iθ + e^2iθ + ... + e^2niθ
e^iθ + (e^iθ)^2 + ... + (e^iθ)^2n
so if x = e^iθ you now have
x(1+x+x^2+...+x^2n)
= x (x^2n +1)/(x-1)
so
e^iθ (e^2niθ - 1) = e^iθ - 1
e^iθ (e^2niθ - 2) = -1 = e^iπ
see what you can do with that when you equate to 1+0i
so you have
e^iθ + e^2iθ + ... + e^2niθ
e^iθ + (e^iθ)^2 + ... + (e^iθ)^2n
so if x = e^iθ you now have
x(1+x+x^2+...+x^2n)
= x (x^2n +1)/(x-1)
so
e^iθ (e^2niθ - 1) = e^iθ - 1
e^iθ (e^2niθ - 2) = -1 = e^iπ
see what you can do with that when you equate to 1+0i
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