A2 October 2020 Q5
5. A transformation \(T\) from the \(z\)-plane to the \(w\)-plane is given by
\[w = \frac{1 - 3z}{z + 2\mathrm{i}} \qquad z \ne -2\mathrm{i}\]The circle with equation \(|z + \mathrm{i}| = 3\) is mapped by \(T\) onto the circle \(C\).
| Scheme | Marks | AO |
|---|---|---|
| \(w = \dfrac{1 - 3z}{z + 2\mathrm{i}} \Rightarrow w(z + 2\mathrm{i}) = 1 - 3z \Rightarrow z = \ldots\) or \(\ldots(z + \mathrm{i}) = \ldots\) | M1 | 2.1 |
| \(z = \dfrac{1 - 2\mathrm{i}w}{w + 3}\) or \((w + 3)(z + \mathrm{i}) = 1 - 2\mathrm{i}w + (w + 3)\mathrm{i}\ \ (= 1 + (3 - w)\mathrm{i})\) | A1 | 1.1b |
| \(\left|\dfrac{1 - 2\mathrm{i}w}{w + 3} + \mathrm{i}\right| = 3 \Rightarrow |1 - 2\mathrm{i}w + \mathrm{i}(w + 3)| = 3|w + 3|\) or \(3|w + 3| = |1 + (3 - w)\mathrm{i}|\) * | M1 A1* | 3.1a 2.1a |
| (4) |
Notes
M1: Attempts to make \(z\) the subject or to extract \(z + \mathrm{i}\) as a term.
A1: Correct expression for \(z\) or correct equation with \(z + \mathrm{i}\) as only term(s) in \(z\).
M1: Applies \(|z + \mathrm{i}| = 3\) to their equation and eliminates fractions.
A1*: Correctly completes to the given result with no errors.
| Scheme | Marks | AO |
|---|---|---|
| (i) \(w = u + \mathrm{i}v \Rightarrow |(3 - u)\mathrm{i} + (v + 1)| = 3|u + 3 + \mathrm{i}v|\) | M1 | 1.1b |
| \(\Rightarrow (3 - u)^2 + (v + 1)^2 = 9\left[(u + 3)^2 + v^2\right]\) | M1 A1 | 2.1 1.1b |
| (ii) \(\Rightarrow 8u^2 + 60u + 8v^2 - 2v + 71 = 0 \Rightarrow (u + \ldots)^2 + (v + \ldots)^2 = \ldots\) | M1 | 1.1b |
| \(\left[\left(u + \tfrac{15}{4}\right)^2 + \left(v - \tfrac{1}{8}\right)^2 = \tfrac{333}{64} \Rightarrow\right]\) Centre is \(\left(-\dfrac{15}{4}, \dfrac{1}{8}\right)\) | A1 | 2.2a |
| Radius is \(\dfrac{3\sqrt{37}}{8}\) | A1 | 2.2a |
| (6) | ||
| (10 marks) |
Notes
(i)
M1: Uses \(w = u + \mathrm{i}v\) or \(w = x + \mathrm{i}y\) (or other suitable notation) in the given equation
M1: Squares and applies Pythagoras to modulus to form an equation in just \(u\) and \(v\). Allow if the 3 is not squared, or slips on sign inside the brackets (e.g. \((v - 1)^2\) instead of \((v + 1)^2\)). However, there must be no i's involved in the equation, and must be sum of square terms on each side, for this mark to be awarded (cannot be recovered – this is for rigorous argument).
A1: Correct equation from correct work – see note on the previous M. (This is a Cartesian equation so satisfies the demand.)
(ii) Allow recovery for these three marks if the 2nd M was withheld dues to i's in the equation but which were later recovered.
M1: Expands their equation and gathers terms, then completes square or uses other valid method to find the centre and/or radius of the circle.
A1: Correct centre \(\left(-\dfrac{15}{4}, \dfrac{1}{8}\right)\) Accept as coordinates or complex number. Allow \(\left(-\dfrac{15}{4}, \dfrac{1}{8}\mathrm{i}\right)\)
A1: Correct radius \(\dfrac{3\sqrt{37}}{8}\)