<p>If the imaginary part of \ \(\dfrac{2z+1}{iz+1}\) is \ \(-2\), then find the locus of the point \ \(z\) representing in the complex plane.</p>
Step-by-Step Solution
Key Concept: Separate the complex number into real and imaginary parts by multiplying by the conjugate of the denominator, then use the condition that imaginary part equals -2 to find the relationship between real and imaginary components of z.
<p><strong>Step 1:</strong> Let z = x + iy where x, y ∈ ℝ</p><p><strong>Step 2:</strong> Substitute into the expression:
$$\frac{2z+1}{iz+1} = \frac{2(x+iy)+1}{i(x+iy)+1} = \frac{2x+1+2iy}{1+ix-y}$$</p><p><strong>Step 3:</strong> Rationalize by multiplying by conjugate of denominator (1-ix+y):
$$\frac{(2x+1+2iy)(1-ix+y)}{(1-y+ix)(1-y-ix)}$$</p><p><strong>Step 4:</strong> Expand numerator:
$$(2x+1+2iy)(1-ix+y) = (2x+1)(1+y) + 2iy(1+y) - i(2x+1)x - 2y x$$
$$= (2x+1+2xy+y-2xy-x) + i(2y+2y^2-2x^2-x)$$
$$= (x+y+1) + i(2y+2y^2-2x^2-x)$$</p><p><strong>Step 5:</strong> Expand denominator:
$$(1-y)^2 + x^2 = 1-2y+y^2+x^2$$</p><p><strong>Step 6:</strong> The imaginary part is:
$$\text{Im} = \frac{2y+2y^2-2x^2-x}{1-2y+y^2+x^2} = -2$$</p><p><strong>Step 7:</strong> Therefore:
$$2y+2y^2-2x^2-x = -2(1-2y+y^2+x^2)$$
$$2y+2y^2-2x^2-x = -2+4y-2y^2-2x^2$$
$$2y+2y^2-x = -2+4y-2y^2$$
$$4y^2-2y-x+2 = 0$$
$$x = 4y^2-2y+2$$</p><p><strong>Step 8:</strong> This can be rewritten as:
$$x = 4(y^2-\frac{y}{2})+2 = 4(y-\frac{1}{4})^2-\frac{1}{4}+2 = 4(y-\frac{1}{4})^2+\frac{7}{4}$$</p><p>∴ The locus is a parabola: <strong>x = 4y² - 2y + 2</strong> or equivalently <strong>4(x - 7/4) = (2y - 1)²</strong></p>
Correct Answer: B