Question

If the point \( P \) represents the complex number \( z = x + iy \) in the Argand plane and if \[ \frac{z\bar{z} + 1}{z - 1} \] is a purely imaginary number, then the locus of \( P \) is:

• A. \( x^2 + y^2 + x - y = 0 \) and \( (x, y) \neq (1,0) \)
• B. \( x^2 + y^2 - x + y = 0 \) and \( (x, y) \neq (1,0) \)
• C. \( x^2 + y^2 - x + y = 0 \) and \( (x, y) = (1,0) \)
• D. \( x^2 + y^2 + x + y = 0 \)

Hint:

To find the locus of a complex number satisfying a given condition, express the equation in terms of \( x \) and \( y \), simplify, and ensure any singularities are excluded from the domain.

Answer 1

The Correct Option is B

We are given the complex number \( z = x + iy \) and its conjugate \( \bar{z} = x - iy \). The given expression is: \[ \frac{z\bar{z} + 1}{z - 1}. \] Step 1: Expand the given expression Since \( z\bar{z} = x^2 + y^2 \), we rewrite: \[ \frac{x^2 + y^2 + 1}{(x + iy) - 1} = \frac{x^2 + y^2 + 1}{x - 1 + iy}. \] Let this expression be purely imaginary, say \( i k \), meaning the real part must be zero: \[ \frac{x^2 + y^2 + 1}{x - 1 + iy} = i k. \] Step 2: Rationalizing Multiplying numerator and denominator by the conjugate of the denominator: \[ \frac{(x^2 + y^2 + 1)(x - 1 - iy)}{(x - 1)^2 + y^2} = i k. \] Expanding the numerator: \[ (x^2 + y^2 + 1)(x - 1) - i (x^2 + y^2 + 1) y. \] For the expression to be purely imaginary, the real part must be zero: \[ (x^2 + y^2 + 1)(x - 1) = 0. \] Since \( x^2 + y^2 + 1 \neq 0 \) for real \( x, y \), we get: \[ x^2 + y^2 - x + y = 0. \] Step 3: Excluding the singularity at \( (1,0) \) The denominator of the original fraction must not be zero: \[ (x - 1) + iy \neq 0 \Rightarrow (x, y) \neq (1,0). \] Thus, the required locus is: \[ x^2 + y^2 - x + y = 0, \quad (x, y) \neq (1,0). \] Thus, the correct answer is: \[ \boxed{x^2 + y^2 - x + y = 0, \quad (x, y) \neq (1,0).} \]