Question

The point P denotes the complex number \(z = x + iy\) in the Argand plane. If \( \frac{2z - i}{z - 2} \) is purely real number, then the equation of the locus of P is.

• A. \(2x^2 + 2y^2 - 4x - y = 0\)
• B. \(x + 4y - 2 = 0 \& (x,y) \neq (2,0) \)
• C. \(x - 4y - 2 = 0 \& (x,y) \neq (2,0) \)
• D. \(x^2 + y^2 - 4x - 2y = 0\)

Hint:

To ensure a complex function is real, the imaginary part must be zero. This can lead directly to the locus in the complex plane.

Answer 1

The Correct Option is B

We are given \( z = x + iy \) and \( \frac{2z - i}{z - 2} \) is purely real. For a fraction to be real, its imaginary part must be zero. 1. Compute \( \frac{2z - i}{z - 2} \): \[ \frac{2(x + iy) - i}{(x + iy) - 2} = \frac{2x + i(2y - 1)}{(x - 2) + iy} \] 2. Multiply numerator and denominator by the conjugate of the denominator: \[ \frac{(2x + i(2y - 1))((x - 2) - iy)}{(x - 2)^2 + y^2} \] 3. The imaginary part of the numerator must be zero: \[ -x - 4y + 2 = 0 \implies x + 4y = 2 \] 4. Exclude \( (x, y) = (2, 0) \) to avoid division by zero. Thus, the locus is \( x + 4y - 2 = 0 \) with \( (x, y) \neq (2, 0) \), which matches option (2). \(\boxed{B}\)