Let z = cos theta + i sin theta Then the value of sigma(m=1 to 15) Im(z2m-1) at theta - Maths - Complex Numbers and Quadratic Equations

Question

Let z = cos theta + i sin theta Then the value of sigma(m=1 to
15) Im(z2m-1) at theta = 2o is
(A) 1/ sin 2o
(B) 1/ 3sin 2o
(C) 1/ 2sin 2o
(D) 1/ 4sin 2o

Ajanta Trivedi · Accepted Answer

it is given that z=cosθ+i
sinθ⇒z2m-1=(cosθ+isinθ)2m-1
by demoiver's law:
z2m-1=cos(2m-1)θ+isin(2m-1)θim(z2m-1)=sin(2m-1)θ
∑m=115im(z2m-1)=∑m=115[sin(2m-1)θ]=sinθ+sin3θ+sin5θ+
+sin(2*15-1)θ=1-cos2*15*θ2sinθ    [since sinx+sin3x+
+sin(2k-1)x=1-cos(2kx)2sinx=1-cos30*θ2sinθ=1-cos30*22sin2  [since θ=2=1-cos602sin2=1-1/22sin2=12*12sin2=14sin2
thus option (d) is correct

hope this helps you

Let z = cos theta + i sin theta. Then the value of sigma(m=1 to 15) Im(z2m-1) at theta = 2o is(A) 1/ sin 2o (B) 1/ 3sin 2o (C) 1/ 2sin 2o (D) 1/ 4sin 2o

Let z = cos theta + i sin theta. Then the value of sigma_{(m=1 to 15)} Im(z^2m-1) at theta = 2^o is
(A) 1/ sin 2^o
(B) 1/ 3sin 2^o
(C) 1/ 2sin 2^o
(D) 1/ 4sin 2^o