FP2 June 2005 Q9
9. A complex number \(z\) is represented by the point \(P\) in the Argand diagram. Given that \[|z - 3\mathrm{i}| = 3,\]
(a) sketch the locus of \(P\). (2)
(b) Find the complex number \(z\) which satisfies both \(|z - 3\mathrm{i}| = 3\) and \(\arg(z - 3\mathrm{i}) = \tfrac{3}{4}\pi\). (4)
The transformation \(T\) from the \(z\)-plane to the \(w\)-plane is given by \[w = \frac{2\mathrm{i}}{z}.\]
(c) Show that \(T\) maps \(|z - 3\mathrm{i}| = 3\) to a line in the \(w\)-plane, and give the cartesian equation of this line. (5)
| Scheme | Marks |
|---|---|
| Circle | M1 |
| Correct circle. (centre \((0, 3)\), radius 3) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| Drawing correct half-line passing as shown | B1 |
| Find either \(x\) or \(y\) coord of \(A\). | M1A1 |
| \(z = -\dfrac{3\sqrt{2}}{2} + \left(3 + \dfrac{3\sqrt{2}}{2}\right)\mathrm{i}\) | A1 |
| (4) |
Notes
[Algebraic approach, i.e. using \(y = 3 - x\) and equation of circle will only gain M1A1, unless the second solution is ruled out, when B1 can be given by implication, and final A1, if correct]
| Scheme | Marks |
|---|---|
| \(|z - 3\mathrm{i}| = 3 \to \left|\dfrac{2\mathrm{i}}{\omega} - 3\mathrm{i}\right| = 3\) | M1 |
| \(\Rightarrow \dfrac{|2\mathrm{i} - 3\mathrm{i}\omega|}{|\omega|} = 3\) | A1 |
| \(\Rightarrow |\omega - \tfrac{2}{3}| = |\omega|\) | M1A1 |
| Line with equation \(u = \tfrac{1}{3}\ \ (x = \tfrac{1}{3})\) | A1 |
| (5) | |
| (11 marks) |
Notes
Some alternatives: (i)
| Scheme | Marks |
|---|---|
| \(\omega = \dfrac{2\mathrm{i}}{x + \mathrm{i}y} = \dfrac{2\mathrm{i}(x - \mathrm{i}y)}{x^2 + y^2} \Rightarrow u = \dfrac{2y}{x^2 + y^2},\ v = \dfrac{2x}{x^2 + y^2}\) | M1A1 |
| As \(x^2 + y^2 - 6y = 0,\ u = \dfrac{1}{3}\) | M1, A1A1 |
(ii)
| Scheme | Marks |
|---|---|
| \(\omega = \dfrac{2\mathrm{i}}{3\cos\theta + 3\mathrm{i}(1 + \sin\theta)} = \dfrac{2\mathrm{i}\{\cos\theta - \mathrm{i}(1 + \sin\theta)\}}{3\{\cos^2\theta + (1 + \sin\theta)^2\}}\) | M1A1 |
| \(= \dfrac{2}{3}\,\dfrac{(1 + \sin\theta) + \mathrm{i}\cos\theta}{2 + 2\sin\theta},\ = \dfrac{1}{3} + \mathrm{i}\,\dfrac{\cos\theta}{1 + \sin\theta},\) | M1A1 |
| So locus is line \(u = \dfrac{1}{3}\) | A1 |