If z-1 2 z+1 is an imaginary number and if it represents a circle then its radius is

Step-by-step solution

Given z = x + iy where x, y R , the condition that z-1 2z+1 is purely imaginary implies its real part vanishes. Substituting z = x + iy yields: (x-1) + iy (2x+1) + 2iy Rationalizing the denominator by multiplying numerator and denominator by the complex conjugate (2x+1) - 2iy : [(x-1) + iy][(2x+1) - 2iy] (2x+1)^2 + 4y^2 Expanding the numerator: [(x-1)(2x+1) + 2y^2] + i[3y] = (2x^2 + 2y^2 - x - 1) + i(3y) Consequently: z-1 2z+1 = 2x^2 + 2y^2 - x - 1 (2x+1)^2 + 4y^2 + i 3y (2x+1)^2 + 4y^2 Setting the real part to zer