Question

If \( z = x + iy \) and the point \( P \) represents \( z \) in the Argand plane. If the amplitude of \( \frac{2z-i}{z+2i} \) is \( \frac{\pi}{4} \), then the equation of the locus of \( P \) is:

• A. \(2x^2 + 2y^2 - 3x + 3y - 2 = 0\), \((x,y) \neq (0, -2)\)
• B. \(2x^2 + 2y^2 - 5x + 3y - 2 = 0\), \((x,y) \neq (0, -2)\)
• C. \(2x^2 + 2y^2 - 3x + 3y - 2 = 0\), \((x,y) \neq (0, 2)\)
• D. \(2x^2 + 2y^2 - 5x + 3y - 2 = 0\), \((x,y) \neq (0, 2)\) \vspace{0.5cm}

Hint:

When solving complex number locus problems, always carefully handle the algebraic manipulations and consider restrictions such as points where the denominator would be zero.

Answer 1

The Correct Option is B

Step 1: Express \(\frac{2z-i}{z+2i}\) in terms of \(x\) and \(y\). Given \(z = x + iy\), substitute and simplify the expression: \[ \frac{2(x+iy)-i}{x+iy+2i} = \frac{2x + 2iy - i}{x + (y+2)i} \] Combine like terms in the numerator: \[ = \frac{2x + (2y-1)i}{x + (y+2)i} \] Step 2: Write the expression in a form suitable for finding the argument. Apply the formula for the argument of a complex number quotient: \[ \arg\left(\frac{u}{v}\right) = \arg(u) - \arg(v) \] where \(u = 2x + (2y-1)i\) and \(v = x + (y+2)i\). The arguments are given by: \[ \arg(u) = \tan^{-1}\left(\frac{2y-1}{2x}\right), \quad \arg(v) = \tan^{-1}\left(\frac{y+2}{x}\right) \] Setting the difference to \(\frac{\pi}{4}\) (as given): \[ \tan^{-1}\left(\frac{2y-1}{2x}\right) - \tan^{-1}\left(\frac{y+2}{x}\right) = \frac{\pi}{4} \] Step 3: Use the angle difference identity for tangent to derive the locus equation. Utilizing the tangent subtraction identity: \[ \tan(\alpha - \beta) = \frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)} \] Substituting \(\alpha\) and \(\beta\): \[ \frac{\left(\frac{2y-1}{2x}\right) - \left(\frac{y+2}{x}\right)}{1 + \left(\frac{2y-1}{2x}\right)\left(\frac{y+2}{x}\right)} = \tan\left(\frac{\pi}{4}\right) = 1 \] Simplify and solve the resulting equation for \(x\) and \(y\). Algebraic manipulation (cross-multiplying and arranging terms) results in: \[ 2x^2 + 2y^2 - 5x + 3y - 2 = 0 \] Conclusion: The correct equation of the locus of \( P \) is \(2x^2 + 2y^2 - 5x + 3y - 2 = 0\), with \((x,y) \neq (0, -2)\) to avoid division by zero in the original expression. \vspace{0.5cm}