Question

If \( z = x+iy \) and the point \( P \) denotes \( z \) in the Argand plane, then the locus of \( P \) satisfying the condition \( \text{Im}\left(\frac{z-3i}{z+2}\right) = 1, z \neq -2 \) is:

• A. \( 7x - 2y + 10 = 0 \)
• B. \( x^2 + xy + y^2 + 7x - 2y + 10 = 0 \)
• C. \( x^2 - xy + y^2 + 7x - 2y + 10 = 0 \)
• D. \( x^2 + y^2 + 7x - 2y + 10 = 0 \)

Hint:

For \( \text{Im}\left(\frac{z-z_1}{z-z_2}\right) = k \), if \( k \neq 0 \), the locus is always a circle. If \( k=0 \), the locus is a straight line. Here, \( k=1 \), so look for the equation representing a circle.

Answer 1

The Correct Option is D

Concept: To find the imaginary part of a quotient \( \frac{z_1}{z_2} \), multiply the numerator and denominator by the conjugate of the denominator \( \overline{z}_2 \).
• \( \frac{x+iy}{a+ib} = \frac{(x+iy)(a-ib)}{a^2+b^2} \)
• \( \text{Im}\left(\frac{z_1}{z_2}\right) = \frac{\text{Im}(z_1\overline{z}_2)}{|z_2|^2} \)

Step 1: Expressing the fraction in terms of \( x \) and \( y \).
\[ \frac{z-3i}{z+2} = \frac{x + i(y-3)}{(x+2) + iy} \] Multiply numerator and denominator by \( (x+2) - iy \): \[ \frac{[x + i(y-3)][(x+2) - iy]}{(x+2)^2 + y^2} \]

Step 2: Extracting the imaginary part.
The imaginary part of the numerator is: \[ x(-y) + (y-3)(x+2) = -xy + xy + 2y - 3x - 6 = 2y - 3x - 6. \] Setting the imaginary part of the whole expression to 1: \[ \frac{-3x + 2y - 6}{(x+2)^2 + y^2} = 1 \]

Step 3: Simplifying to find the locus.
\[ -3x + 2y - 6 = x^2 + 4x + 4 + y^2 \] Rearranging terms to one side: \[ x^2 + y^2 + 4x + 3x - 2y + 4 + 6 = 0 \] \[ x^2 + y^2 + 7x - 2y + 10 = 0 \]