If the locus of z ∈ ℂ, such that Re ( (z - 1)/(2z + i) ) + Re ( (z̅ - 1)/(2z̅ - i) ) = 2 , is a circle of radius r and center (a, b), then 15ab/r² is equal to :

Solution

Maths sir · 4 steps

The second term is just the conjugate of the first, so the equation collapses to one real part equal to 1.

1. Spot the conjugate. Take the conjugate of (z-1)/(2z+i) and you get (z̅-1)/(2z̅-i), which is exactly the second fraction. A number and its conjugate have the same real part, so the equation becomes 2 Re((z-1)/(2z+i)) = 2, that is, Re((z-1)/(2z+i)) = 1.
2. Find the real part. Put z = x + iy and multiply the top and bottom by 2x - i(2y+1). The real part is (2x² - 2x + 2y² + y)/(4x² + (2y+1)²).

   2x² - 2x + 2y² + y = 4x² + 4y² + 4y + 1 ⇒ x² + y² + x + 3/2y + 1/2 = 0

3. Read the circle. The centre is (-1/2, -3/4), so ab = 3/8. The radius satisfies r² = 1/4 + 9/16 - 1/2 = 5/16. The point z = -i/2 is left out because it makes the denominator zero, but the locus is still this circle.
4. So. 15ab/r² = (15 × 3/8)/(5/16) = 45/8 × 16/5 = 18, which is option (d).

The idea to keep. Before expanding anything, check whether one term is the conjugate of another. Since Re(w) = Re(w̅), the two terms merge and half the algebra disappears.

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