Let A = { θ ∈ [0, 2π] : 1 + 10Re( (2 cos θ + i sin θ)/(cos θ - 3i sin θ) ) = 0 } . Then Σ_(θ ∈ A) θ² is equal to

Solution

Maths sir · 4 steps

Rationalise the fraction first; the real part then turns the condition into tan² θ = 1.

1. Rationalise. Write c = cos θ and s = sin θ. Multiply the top and bottom by the conjugate of the denominator, c + 3is. The denominator becomes c² + 9s², which is real and never zero.

   (2c + is)/(c - 3is) = ((2c² - 3s²) + 7ics)/(c² + 9s²)

2. Use the condition. The real part is (2c² - 3s²)/(c² + 9s²). Then 1 + 10 × Re = 0 gives c² + 9s² + 20c² - 30s² = 0, that is 21c² = 21s², so tan² θ = 1.
3. List the angles. In [0, 2π], tan θ = ± 1 gives θ = π/4, 3π/4, 5π/4, 7π/4. Each one keeps cos θ ≠ 0, so all four are valid.
4. So. Σ θ² = π²/16(1 + 9 + 25 + 49) = 84π²/16 = 21/4π², which is option (a).

The idea to keep. To find the real part of a fraction, multiply by the conjugate of the denominator. The bottom becomes real, and the real part can be read straight off the top.

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