If Re ( (z - 1)/(2z + i) ) = 1, where z = x + iy, then the point (x, y) lies on a :

Solution

Maths sir · 4 steps

Rationalise with the conjugate of the denominator, set the real part to 1, and see what curve falls out.

1. Rationalise. With z = x + iy, the fraction is ((x-1) + iy)/(2x + i(2y+1)). Multiply the top and bottom by 2x - i(2y+1) so the denominator becomes real.

   Re((z-1)/(2z+i)) = (2x(x-1) + y(2y+1))/(4x² + (2y+1)²)

2. Set it to 1. 2x² - 2x + 2y² + y = 4x² + 4y² + 4y + 1. Collecting terms gives 2x² + 2y² + 2x + 3y + 1 = 0. The x² and y² terms survive with equal coefficients, so this is a circle, not a line.
3. Centre and radius. Divide by 2: x² + y² + x + 3/2y + 1/2 = 0. The centre is (-1/2, -3/4) and r² = 1/4 + 9/16 - 1/2 = 5/16, so r = √5/4.
4. So. The diameter is 2r = √5/2, which is option (c).

The idea to keep. For a condition on Re or Im of a quotient, multiply by the conjugate of the denominator. If x² and y² remain with equal coefficients, it is a circle: divide by that coefficient before reading the centre.

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