FP2 June 2015 Q5
5. A transformation \(T\) from the \(z\)-plane to the \(w\)-plane is given by \[w = \frac{z}{z + 3\mathrm{i}}, \quad z \neq -3\mathrm{i}\]
The circle with equation \(|z| = 2\) is mapped by \(T\) onto the curve \(C\).
The region \(|z| \leqslant 2\) in the \(z\)-plane is mapped by \(T\) onto the region \(R\) in the \(w\)-plane.
| Scheme | Marks |
|---|---|
| \(w = \dfrac{z}{z + 3\mathrm{i}}\) | |
| \(w(z + 3\mathrm{i}) = z \qquad z = \dfrac{3\mathrm{i}w}{1 - w}\) or \(\dfrac{-3\mathrm{i}w}{w - 1}\) | M1A1 |
| \(|z| = 2 \qquad \left|\dfrac{3\mathrm{i}w}{1 - w}\right| = 2\) | dM1 |
| \(|3\mathrm{i}w| = 2|1 - w|\) | |
| \(w = u + \mathrm{i}v \quad 9\left(u^2 + v^2\right) = 4\left((1 - u)^2 + v^2\right)\) | ddM1A1 |
| \(9u^2 + 9v^2 = 4\left(1 - 2u + u^2 + v^2\right)\) | |
| (i) \(5u^2 + 5v^2 + 8u - 4 = 0\) \(\left(u + \dfrac{4}{5}\right)^2 + v^2 = \dfrac{36}{25}\) | dddM1 |
| (ii) So a circle, Centre \(\left(-\dfrac{4}{5}, 0\right)\) Radius \(\dfrac{6}{5}\) (oe fractions or decimals) | A1A1 |
| (8) |
Notes
M1: re-arrange to \(z = \ldots\)
A1: correct result
dM1: dep (on first M1) using \(|z| = 2\) with their previous result
ddM1: dep ( on both previous M marks) use \(w = u + \mathrm{i}v\) (or eg \(w = x + \mathrm{i}y\)) and find the moduli. Moduli to contain no is and must be +. Allow 9 or 3 and 4 or 2
A1: for a correct equation in \(u\) and \(v\) or any other pair of variables
dddM1: dep (on all previous M marks) re-arrange to the form of the equation of a circle (same coeffs for the squared terms
A1A1: deduce circle and give correct centre and radius. Completion of square may not be shown. Deduct 1 for each error or omission. (Enter A1A0 on e-PEN)
Special Case: If \(z = \dfrac{3\mathrm{i}w}{w - 1}\) obtained, give M1A0 but all other marks can be awarded.
ALTERNATIVE for 5(a):
| Scheme | Marks |
|---|---|
| Let \(z = x + \mathrm{i}y\) | |
| \(w = \dfrac{x + \mathrm{i}y}{x + \mathrm{i}(y + 3)} = \dfrac{(x + \mathrm{i}y)\left(x - \mathrm{i}(y + 3)\right)}{\left(x + \mathrm{i}(y + 3)\right)\left(x - \mathrm{i}(y + 3)\right)}\) | M1 |
| \(= \dfrac{x^2 + y^2 + 3y - 3\mathrm{i}x}{x^2 + y^2 + 6y + 9}\) | A1 |
| \(\dfrac{3y + 4 - 3\mathrm{i}x}{6y + 13}\) as \(|z| = 2 \Rightarrow x^2 + y^2 = 4\) | dM1 |
| \(w = u + \mathrm{i}v\) so \(u = \dfrac{3y + 4}{6y + 13} \qquad v = \dfrac{-3x}{6y + 13}\) | ddM1 A1 |
| Using \(u = \dfrac{\frac{1}{2}(6y + 13)}{6y + 13} - \dfrac{\frac{5}{2}}{6y + 13}\) | |
| \(u^2 + v^2 = \dfrac{9y^2 + 24y + 16 + 9x^2}{(6y + 13)^2} = \dfrac{24y + 52}{(6y + 13)^2} = \dfrac{4}{6y + 13}\) \(= \dfrac{8}{5}\left(\dfrac{1}{2} - u\right)\) | dddM1 |
| \(\therefore\ 5u^2 + 5v^2 + 8u = 4\) | |
| Then as main scheme: Circle, centre, radius | A1A1 |
| (8) |
M1 Rationalise the denominator - must use conjugate of the denominator
A1 Expand brackets and obtain correct numerator and denominator
dM1 Use \(x^2 + y^2 = 4\) in their expression to remove the squares
ddM1 Equating real and imaginary parts
A1 Correct expressions for \(u\) and \(v\) in terms of \(x\) and \(y\)
dddM1 Uses \(u^2 + v^2 = \ldots\) to eliminate \(x\) and \(y\) and obtain an equation of the circle
A1A1 As main scheme
| Scheme | Marks |
|---|---|
| Circle drawn on an Argand diagram in correct position ft their centre and radius | B1ft |
| Region inside correct circle shaded no ft | B1 |
| (2) | |
| (10 marks) |
Notes
Mark diagram only - ignore any working shown.
B1ft: No numbers needed but circle must be in the correct region (or on the correct axis) for their centre and the centre and radius must be consistent (ie check how the circle crosses the axes) B0 if the equation in (a) is not an equation of a circle.
B1: Region inside the correct circle shaded. (no ft here)