The sum of all possible values of θ ∈[-π, 2 π], for which (1+i cos θ)/(1-2 i cos θ) is purely imaginary, is equal to :

Solution

Maths sir · 4 steps

Rationalise the fraction, set the real part to zero, then list every angle in the interval and add.

1. Rationalise. Write c = cos θ and multiply the top and bottom by the conjugate 1 + 2ic. The denominator becomes 1 + 4c², which is always positive.

   (1+ic)/(1-2ic) = (1+ic)(1+2ic)/(1+4c²) = ((1-2c²) + 3ic)/(1+4c²)

2. Purely imaginary. The real part must be zero, so 1 - 2c² = 0, which gives cos θ = ±1/√2. Then the imaginary part 3c is not zero, so the number really is purely imaginary.
3. List the angles. cos θ = ±1/√2 means θ = π/4 + kπ/2. In [-π, 2π] these are -3π/4, -π/4, π/4, 3π/4, 5π/4, 7π/4. The next ones, -5π/4 and 9π/4, lie outside.
4. So. The four angles ±π/4 and ±3π/4 cancel in pairs, leaving 5π/4 + 7π/4 = 3π. The answer is (b).

The idea to keep. Purely imaginary means real part zero (and imaginary part non-zero). Rationalise first, then read off the real part. When summing angles, pair the ± ones so they cancel.

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