A2 June 2025 Q9
9. The transformation \(T\) from the \(z\)-plane to the \(w\)-plane, where \(z = x + \mathrm{i}y\) and \(w = u + \mathrm{i}v\) is given by\[w = \frac{z}{z + 2\mathrm{i}} \qquad z \neq -2\mathrm{i}\]
The circle with equation \(|z| = 4\) is mapped by \(T\) onto the circle \(C\).
Determine
| Scheme | Marks | AO |
|---|---|---|
| \(w(z + 2\mathrm{i}) = z \Rightarrow wz + 2w\mathrm{i} = z \Rightarrow 2w\mathrm{i} = z - wz\) \(\Rightarrow 2w\mathrm{i} = z(1 - w) \Rightarrow z = \tfrac{2w\mathrm{i}}{(1 - w)}\) | M1 A1 | 3.1a 1.1b |
| \(|z| = 4 \Rightarrow \left|\dfrac{2w\mathrm{i}}{(1 - w)}\right| = 4 \Rightarrow |2w\mathrm{i}| = 4|1 - w|\) | dM1 | 1.1b |
| \(w = u + v\mathrm{i}\) \(|2u\mathrm{i} - 2v| = 4|(1 - u) - v\mathrm{i}| \Rightarrow (2u)^2 + (2v)^2 = 16\left[(1 - u)^2 + v^2\right]\) oe | ddM1 | 2.1 |
| \(\left\{4u^2 + 4v^2 = 16 - 32u + 16u^2 + 16v^2\right\}\) \(12u^2 - 32u + 12v^2 + 16 = 0\) oe | A1 | 1.1b |
| \(u^2 - \dfrac{8}{3}u + v^2 + \dfrac{4}{3} = 0 \Rightarrow \left(u - \dfrac{4}{3}\right)^2 - \dfrac{16}{9} + v^2 + \dfrac{4}{3} = 0\) | M1 | 1.1b |
| Centre \(\left(\dfrac{4}{3}, 0\right)\) or radius \(= \dfrac{2}{3}\) | A1 | 1.1b |
| Centre \(\left(\dfrac{4}{3}, 0\right)\) and radius \(= \dfrac{2}{3}\) | A1 | 2.2a |
| (8) | ||
| (8 marks) |
Notes
M1: A complete method of rearranging to make \(z\) the subject. Condone slips rearranging
A1: Correct expression for \(z\)
NB if the only error is incorrect sign, this will correct when taking modulus – allow recovering of later marks, losing just this first A mark.
dM1: Dependent on the previous method mark. Sets \(|\text{their } z| = 4\) and forms a linear equation
ddM1: Dependent on previous method mark. Uses \(w = u + v\mathrm{i}\) and Pythagoras to form an equation in \(u\) and \(v\) only. Condone \((2u)^2 + (2v)^2 = 4\left[(1 - u)^2 + v^2\right]\) or slips in signs on \(1 - (u + v\mathrm{i})\). They may factor out the 2 first, which is fine.
A1: Correct equation
M1: Having reached a circle equation, completes the square on \(u\) (and \(v\) if appropriate).
A1: Correct centre or radius of \(C\) from correct work
A1: Centre and radius both correct from correct work. Accept centre as coordinates or as \(\dfrac{4}{3} + 0\mathrm{i}\)
Alt I
| Scheme | Marks | AO |
|---|---|---|
| \(w(z + 2\mathrm{i}) = z \Rightarrow wz + 2w\mathrm{i} = z \Rightarrow 2w\mathrm{i} = z - wz\) \(\Rightarrow 2w\mathrm{i} = z(1 - w) \Rightarrow z = \tfrac{2w\mathrm{i}}{(1 - w)}\) | M1 A1 | 3.1a 1.1b |
| \(z = \dfrac{2u\mathrm{i} - 2v}{1 - u - v\mathrm{i}} \times \dfrac{1 - u + v\mathrm{i}}{1 - u + v\mathrm{i}} = \ldots\) | dM1 | 1.1b |
| \(|z| = 4 \Rightarrow \left(\tfrac{2v}{(u - 1)^2 + v^2}\right)^2 + \left(\tfrac{2u^2 + 2v^2 - 2u}{(u - 1)^2 + v^2}\right)^2 = 16\) \(\Rightarrow 12v^4 + 24u^2v^2 - 56uv^2 + 28v^2 + 12u^4 - 56u^3 + 92u^2 - 64u + 16 = 0\) \(\Rightarrow 4\left(v^2 + u^2 - 2u + 1\right)\left(3v^2 + 3u^2 - 8u + 4\right) = 0\) | ddM1 | 2.1 |
| \(\left\{4u^2 + 4v^2 = 16 - 32u + 16u^2 + 16v^2\right\}\) \(12u^2 - 32u + 12v^2 + 16 = 0\) oe | A1 | 1.1b |
| \(u^2 - \dfrac{8}{3}u + v^2 + \dfrac{4}{3} = 0 \Rightarrow \left(u - \dfrac{4}{3}\right)^2 - \dfrac{16}{9} + v^2 + \dfrac{4}{3} = 0\) | M1 | 1.1b |
| Centre \(\left(\dfrac{4}{3}, 0\right)\) or radius \(= \dfrac{2}{3}\) | A1 | 1.1b |
| Centre \(\left(\dfrac{4}{3}, 0\right)\) and radius \(= \dfrac{2}{3}\) | A1 | 1.1b |
| (8) |
M1: A complete method of rearranging to make \(z\) the subject. Condone slips rearranging
A1: Correct expression for \(z\)
dM1: Dependent on the previous method mark. Applies \(w = u + v\mathrm{i}\) and rationalises the denominator with correct conjugate used.
ddM1: Dependent on previous method mark. Extracts real and imaginary components, sets \(|\text{their } z| = 4\) and applies \(\text{Re}^2 + \text{Im}^2 = 16\) (condone “=4”) and proceeds to simplify to a quadratic expression in \(u\) and \(v\). This mark is not likely to be successfully earned but is possible. The most likely approach is shown but identifying and cancelling a factor \((u - 1)^2 + v^2\) is also possible.
A1: Correct equation
M1: Having reached a circle equation, completes the square on \(u\) (and \(v\) if appropriate).
A1: Correct centre or radius of \(C\) from correct work
A1: Centre and radius both correct from correct work. Accept centre as coordinates or as \(\dfrac{4}{3} + 0\mathrm{i}\)
Alt II
| Scheme | Marks | AO |
|---|---|---|
| \(w = \tfrac{x + \mathrm{i}y}{x + \mathrm{i}y + 2\mathrm{i}} \times \tfrac{x - (y + 2)\mathrm{i}}{x - (y + 2)\mathrm{i}} = \ldots\) \(= \tfrac{x^2 - x(y + 2)\mathrm{i} + \mathrm{i}xy + y(y + 2)}{x^2 + (y + 2)^2}\) | M1 A1 | 3.1a 1.1b |
| \(|z| = 4 \Rightarrow x^2 + y^2 = 16 \Rightarrow w = \dfrac{x^2 + y^2 + 2y - \mathrm{i}(xy + 2x - xy)}{x^2 + y^2 + 4y + 4} = \dfrac{16 + 2y - 2x\mathrm{i}}{4y + 20}\) | dM1 | 1.1b |
| \(u = \tfrac{8 + y}{2y + 10} = \tfrac{1}{2} + \tfrac{3}{2y + 10},\ v = \tfrac{-x}{2y + 10}\) \(\Rightarrow u^2 + v^2 = \tfrac{(8 + y)^2 + x^2}{(2y + 10)^2} = \tfrac{64 + 16y + y^2 + x^2}{4(y + 5)^2} = \tfrac{64 + 16y + 16}{4(y + 5)^2}\) \(= \tfrac{16y + 80}{4(y + 5)^2} = \tfrac{16(y + 5)}{4(y + 5)^2} = \tfrac{4}{y + 5} = 4 \times \tfrac{2}{3}\left(u - \tfrac{1}{2}\right)\) | ddM1 | 2.1 |
| \(\Rightarrow u^2 + v^2 = \dfrac{8}{3}\left(u - \dfrac{1}{2}\right)\) | A1 | 1.1b |
| \(u^2 - \dfrac{8}{3}u + v^2 + \dfrac{4}{3} = 0 \Rightarrow \left(u - \dfrac{4}{3}\right)^2 - \dfrac{16}{9} + v^2 + \dfrac{4}{3} = 0\) | M1 | 1.1b |
| Centre \(\left(\dfrac{4}{3}, 0\right)\) or radius \(= \dfrac{2}{3}\) | A1 | 1.1b |
| Centre \(\left(\dfrac{4}{3}, 0\right)\) and radius \(= \dfrac{2}{3}\) | A1 | 2.2a |
| (8) |
M1: Replaces \(z\) by \(x + \mathrm{i}y\) and rationalises the denominator.
A1: Correct expression.
dM1: Dependent on the previous method mark. Uses \(|z| = 4 \Rightarrow x^2 + y^2 = 16\) in the equation to simplify coefficients
ddM1: Dependent on previous method mark. Extracts the real and imaginary parts of the equation and attempts to find \(u^2 + v^2\) eliminates all \(x\)’s and \(y\)’s
A1: Correct equation.
M1: Having reached a circle equation, completes the square on \(u\) (and \(v\) if appropriate).
A1: Correct centre or radius of \(C\) from correct work
A1: Centre and radius both correct from correct work
Alt III
| Scheme | Marks | AO |
|---|---|---|
| \(|z| = 4 \Rightarrow \pm 4, \pm 4\mathrm{i}\) on circle so e.g \(\tfrac{\pm 4}{\pm 4 + 2\mathrm{i}}, \tfrac{\pm 4\mathrm{i}}{\pm 4\mathrm{i} + 2\mathrm{i}}\) on \(C\) \(\Rightarrow \tfrac{2}{3}, 2, \tfrac{4}{5} \pm \tfrac{2}{5}\mathrm{i}\) on \(C\) | M1 A1 | 3.1a 1.1b |
| Perpendicular bisector of points \(\dfrac{2}{3}, 2\) is \(x = \dfrac{4}{3}\) | dM1 | 1.1b |
| Perpendicular bisector of points \(\dfrac{4}{5} \pm \dfrac{2}{5}\mathrm{i}\) is \(y = 0\) | ddM1 | 2.1 |
| \(x = \dfrac{4}{3}\) and \(y = 0\) | A1 | 1.1b |
| Centre is intersection of these lines = … | M1 | 1.1b |
| Centre \(\left(\dfrac{4}{3}, 0\right)\) or radius \(= \dfrac{2}{3}\) | A1 | 1.1b |
| Centre \(\left(\dfrac{4}{3}, 0\right)\) and radius \(= \dfrac{2}{3}\) | A1 | 2.2a |
| (8) |
M1: Identifies at least 2 points on the original circle and finds the images on \(C\)
A1: Correct simplified two points on \(C\)
dM1: Dependent on the previous method mark. Finds the perpendicular bisector of two of their points found on \(C\)
ddM1: Dependent on previous method mark. Finds a third point on \(C\) (or two more points on \(C\)) and attempts the perpendicular bisector for another pair of points.
A1: Correct equations for the bisectors.
M1: Solves the equations of the bisectors to find the centre of the circle.
A1: Correct centre or radius of \(C\) from correct work
A1: Centre and radius both correct from correct work.
There may be variations on approach (e.g. using diametrically opposite points to work out centre and radius). If solutions are not fully correct by such approaches, use the review system for guidance.