If the set {Re((z-z̄+z z̄)/(2-3 z+5 z̄)): z ∈ ℂ, Re(z)=3} is equal to
the interval (α, β], then 24(β-α) is equal to :

Solution

Maths sir · 4 steps

Put z = 3 + iy, rationalise, and the real part becomes a function of y² alone.

1. Substitute. Take z = 3 + iy with y real. Then z - z̄ = 2iy and zz̄ = 9 + y², so the numerator is 9 + y² + 2iy. The denominator is 2 - 3(3 + iy) + 5(3 - iy) = 8 - 8iy.
2. Take the real part. Multiply top and bottom by 1 + iy. The bottom becomes 8(1 + y²), and the real part of the top is (9 + y²) - 2y² = 9 - y².

   Re = (9 - y²)/8(1 + y²) = 1/8(10/(1 + y²) - 1)

3. Find the range. As y² runs over [0, ∞), 10/(1 + y²) falls from 10 towards 0. At y = 0 the value is 9/8, which is reached. As y² → ∞ it tends to -1/8 but never reaches it. So the set is (-1/8, 9/8].
4. So. β - α = 9/8 + 1/8 = 5/4, and 24 × 5/4 = 30. The answer is (d).

The idea to keep. When Re(z) is fixed, write z = a + iy and reduce everything to one real variable. Then find the range by splitting the fraction as a constant plus a simple decreasing term.

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