Complex Numbers
Locus of Complex Numbers
Grade 11
Question:
<p>If <span class='math'>\text{Im}\left(\frac{z-1}{2z+1}\right) = -4</span>, then locus of <span class='math'>z</span> is</p>
<p>(a) an ellipse</p>
<p>(b) a parabola</p>
<p>(c) a straight line</p>
<p>(d) a circle</p>
Step-by-Step Solution
Key Concept: Express the complex number in terms of real and imaginary parts, then use the given imaginary part condition to derive the locus equation.
<p><strong>Solution:</strong> Let <span class='math'>z = x + iy</span>. Then:</p><p><span class='math'>\frac{z-1}{2z+1} = \frac{(x-1) + iy}{(2x+1) + 2iy}</span></p><p>Rationalizing by multiplying by the conjugate:</p><p><span class='math'>= \frac{[(x-1) + iy][(2x+1) - 2iy]}{(2x+1)^2 + 4y^2}</span></p><p>The imaginary part is:</p><p><span class='math'>\text{Im} = \frac{y(2x+1) - 2y(x-1)}{(2x+1)^2 + 4y^2} = \frac{3y}{(2x+1)^2 + 4y^2}</span></p><p>Setting this equal to <span class='math'>-4</span>:</p><p><span class='math'>\frac{3y}{(2x+1)^2 + 4y^2} = -4</span></p><p><span class='math'>3y = -4[(2x+1)^2 + 4y^2]</span></p><p><span class='math'>(2x+1)^2 + 4y^2 + \frac{3y}{4} = 0</span></p><p>Completing the square gives a circle equation.</p><p>∴ Answer is (d) a circle.</p>
Correct Answer: D