Let A={θ ∈(0,2 π): (1+2 i sin θ)/(1-i sin θ) is purely imaginary }. Then the sum of the elements in A is :

Solution

Maths sir · 4 steps

Rationalise with the conjugate, set the real part to zero, and collect every angle in (0, 2π).

1. Rationalise. Write s = sin θ and multiply the top and bottom by 1 + is, the conjugate of the denominator. The denominator becomes 1 + s², which is real and positive.

   (1+2is)/(1-is) = (1+2is)(1+is)/(1+s²) = ((1-2s²) + 3is)/(1+s²)

2. Real part zero. Purely imaginary means the real part is 0, so 1 - 2s² = 0. That gives sin² θ = 1/2, so sin θ = ±1/√2. The imaginary part 3s/(1+s²) is then non-zero, so the number is not 0.
3. List the angles. sin θ = 1/√2 gives π/4, 3π/4, and sin θ = -1/√2 gives 5π/4, 7π/4.
4. So. The sum is (π + 3π + 5π + 7π)/4 = 16π/4 = 4π, which is option (d).

The idea to keep. For a quotient of complex numbers, multiply by the conjugate of the denominator first. Then purely imaginary means real part zero, and purely real means imaginary part zero.

Answer from the NTA final answer key. Solution by PrepWiser.