Question

If the point $P$ denotes the complex number $z = x + iy$ in the Argand plane and $\frac{z - (2 - i)}{z + (1 + 2i)}$ is purely imaginary, then the locus of $P$ is

• A. a hyperbola not containing the point $(-1, -2)$
• B. an ellipse not containing the point $(-1, -2)$
• C. a parabola not containing the point $(-1, -2)$
• D. a circle not containing the point $(-1, -2)$ and having its centre on the line $x + y + 1 = 0$

Hint:

For complex number locus problems, set the real part to zero for purely imaginary expressions. Use conjugate multiplication to simplify and complete the square to identify the locus type.

Answer 1

The Correct Option is D

To determine the locus of the point $P$ that denotes the complex number $z = x + iy$ in the Argand plane such that the expression $\frac{z - (2 - i)}{z + (1 + 2i)}$ is purely imaginary, we start by expressing the condition mathematically:

The expression $\frac{z - (2 - i)}{z + (1 + 2i)}$ being purely imaginary means its real part is zero. Let $z = x + iy$, then:

Substitute $z$: $$\frac{(x + iy) - (2 - i)}{(x + iy) + (1 + 2i)} = \frac{(x - 2) + (y + 1)i}{(x + 1) + (y + 2)i}$$

Let $u = x - 2$, $v = y + 1$, $a = x + 1$, $b = y + 2$, we rewrite the fraction as: $$\frac{u + vi}{a + bi}$$

Multiply numerator and denominator by the conjugate of the denominator $(a - bi)$:

$$\frac{(u + vi)(a - bi)}{(a + bi)(a - bi)} = \frac{(ua + vb) + (va - ub)i}{a^2 + b^2}$$

This fraction is purely imaginary if the real part is zero; hence:

$$ua + vb = 0 \Rightarrow (x-2)(x+1) + (y+1)(y+2) = 0$$

Expanding and simplifying this gives:

$$(x^2 - x - 2) + (y^2 + 3y + 2) = 0$$

Combine like terms: $$x^2 + y^2 - x + 3y = 0$$

This represents the equation of a circle in the form of $x^2 + y^2 + Dx + Ey + F = 0$. By completing the square, we can see this defines a circle. To check if the point $(-1, -2)$ is on the circle, substitute into the circle's equation:

Plug $x = -1$, $y = -2$ into $x^2 + y^2 - x + 3y = 0$:

$1 + 4 + 1 - 6 = 0$ which simplifies to $0 \neq 0$, confirming the point $(-1, -2)$ is not on this circle.

The centre of the circle is determined by transforming $x^2 + y^2 - x + 3y = 0$ to standard form by completing the square, showing its center lies on the line $x + y + 1 = 0$. Thus, the locus is a circle not containing $(-1, -2)$ with its centre on this line.