AS June 2025 Q5
5. In an Argand diagram, the curve \(C\) with equation
\[\arg\left(\frac{z + 4}{z - 2\mathrm{i}}\right) = \frac{\pi}{4}\]represents an arc of a circle.
Given that \(z = x + \mathrm{i}y\), where \(x\) and \(y\) are real numbers,
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{z + 4}{z - 2\mathrm{i}} = \dfrac{x + \mathrm{i}y + 4}{x + \mathrm{i}y - 2\mathrm{i}} = \dfrac{(x + \mathrm{i}y + 4)(x - \mathrm{i}(y - 2))}{(x + \mathrm{i}y - 2\mathrm{i})(x - \mathrm{i}(y - 2))}\) | M1 | 3.1a |
| \(= \dfrac{x^2 + y^2 + 4x - 2y}{x^2 + (y - 2)^2} + \dfrac{2x - 4y + 8}{x^2 + (y - 2)^2}\mathrm{i}\) | M1 | 1.1b |
| \(\arg\left(\dfrac{z + 4}{z - 2\mathrm{i}}\right) = \dfrac{\pi}{4} \Rightarrow x^2 + y^2 + 4x - 2y = 2x - 4y + 8\) | M1 | 2.2a |
| \(\Rightarrow x^2 + y^2 + 2x + 2y - 8 = 0\) | A1 | 2.1 |
| (4) |
Notes
M1: Starts the problem by introducing \(z = x + \mathrm{i}y\) into \(\dfrac{z + 4}{z - 2\mathrm{i}}\) and multiplies numerator and denominator by the complex conjugate of the denominator
M1: Collects terms and establishes the real and imaginary parts
M1: Deduces from \(\arg\left(\dfrac{z + 4}{z - 2\mathrm{i}}\right) = \dfrac{\pi}{4}\) that the real and imaginary parts are equal
A1: Correct equation
Alternative
| Scheme | Marks | AO |
|---|---|---|
| Angle subtended at centre \(= \dfrac{\pi}{2} \Rightarrow r^2 + r^2 = (-4)^2 + 2^2 \Rightarrow r = \ldots\left(\sqrt{10}\right)\) | M1 | 3.1a |
| Midpoint \((-2, 1)\), gradient \(m = \dfrac{2}{4} \Rightarrow \perp\) bisector is \(y - 1 = -2(x + 2)\) | M1 | 1.1b |
| e.g. \(10 = (x - (-4))^2 + (-2x - 3)^2 \Rightarrow x = \ldots(-3, -1) \Rightarrow y = \ldots(3, -1)\) | M1 | 2.2a |
| \(\Rightarrow x^2 + y^2 + 2x + 2y - 8 = 0\) | A1 | 2.1 |
| (4) |
M1: Realises the angle subtended at the centre gives a right triangle and proceeded to find the radius (or its square).
M1: Attempts the perpendicular bisector of the two known points on the arc.
M1: Complete method to find the centre using their radius and bisector.
A1: Correct equation with justification as to which points were chosen for the centre.
| Scheme | Marks | AO |
|---|---|---|
| \(\Rightarrow x^2 + y^2 + 2x + 2y - 8 = 0 \Rightarrow r = \ldots,\ \text{centre} = \ldots\) | M1 | 1.1b |
| \(|z|_{\min} = \sqrt{10} - \sqrt{1^2 + 1^2}\) | M1 | 3.1a |
| \(\sqrt{10} - \sqrt{2}\) | A1 | 1.1b |
| (3) | ||
| (7 marks) |
Notes
M1: Uses their equation from part (a) to establish the centre and radius of the circle
M1: Adopts the correct strategy for the minimum value using their centre and radius
A1: Correct exact value.