FP2 June 2007 Q12
12. The transformation \(T\) from the \(z\)-plane, where \(z = x + \mathrm{i}y\), to the \(w\)-plane, where \(w = u + \mathrm{i}v\), is given by \[w = \frac{z + \mathrm{i}}{z}, \qquad z \neq 0.\]
(a) The transformation \(T\) maps the points on the line with equation \(y = x\) in the \(z\)-plane, other than \((0, 0)\), to points on a line \(l\) in the \(w\)-plane. Find a cartesian equation of \(l\). (5)
(b) Show that the image, under \(T\), of the line with equation \(x + y + 1 = 0\) in the \(z\)-plane is a circle \(C\) in the \(w\)-plane, where \(C\) has cartesian equation \[u^2 + v^2 - u + v = 0.\] (7)
(c) On the same Argand diagram, sketch \(l\) and \(C\). (3)
| Scheme | Marks |
|---|---|
| Let \(z = \lambda + \lambda\mathrm{i}\); \(w = \dfrac{\lambda + (\lambda + 1)\mathrm{i}}{\lambda(1 + \mathrm{i})}\) | M1 |
| \(= \dfrac{\lambda + (\lambda + 1)\mathrm{i}}{\lambda(1 + \mathrm{i})} \times \dfrac{1 - \mathrm{i}}{1 - \mathrm{i}}\) | M1 |
| \(u + \mathrm{i}v = \dfrac{(2\lambda + 1) + \mathrm{i}}{2\lambda}\) | A1 |
| \(u = 1 + \dfrac{1}{2\lambda},\ v = \dfrac{1}{2\lambda}\) | M1 |
| Eliminating \(\lambda\) gives a line with equation \(v = u - 1\) or equivalent | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| Let \(z = \lambda - (\lambda + 1)\mathrm{i}\): \(w = \dfrac{\lambda - \lambda\mathrm{i}}{\lambda - (\lambda + 1)\mathrm{i}}\) | M1 |
| \(= \dfrac{\lambda - \lambda\mathrm{i}}{\lambda - (\lambda + 1)\mathrm{i}} \times \dfrac{\lambda + (\lambda + 1)\mathrm{i}}{\lambda + (\lambda + 1)\mathrm{i}}\) | M1 |
| \(u + \mathrm{i}v = \dfrac{\lambda(2\lambda + 1) + \lambda\mathrm{i}}{2\lambda^2 + 2\lambda + 1}\) | A1 |
| \(u = \dfrac{\lambda(2\lambda + 1)}{2\lambda^2 + 2\lambda + 1},\ v = \dfrac{\lambda}{2\lambda^2 + 2\lambda + 1}\) | M1 |
| \(\dfrac{u}{v} = 2\lambda + 1\) \(v = \dfrac{2\lambda}{4\lambda^2 + 4\lambda + 2} = \dfrac{(2\lambda + 1) - 1}{(2\lambda + 1)^2 + 1} = \dfrac{\frac{u}{v} - 1}{\left(\frac{u}{v}\right)^2 + 1}\) | M1 |
| Reducing to the circle with equation \(u^2 + v^2 - u + v = 0\ *\) cso | M1A1 |
| (7) |
Notes
Alternative 1
| Scheme | Marks |
|---|---|
| Let \(z = \lambda - (\lambda + 1)\mathrm{i}\): \(w = \dfrac{\lambda - \lambda\mathrm{i}}{\lambda - (\lambda + 1)\mathrm{i}}\) | M1 |
| \(= \dfrac{\lambda - \lambda\mathrm{i}}{\lambda - (\lambda + 1)\mathrm{i}} \times \dfrac{\lambda + (\lambda + 1)\mathrm{i}}{\lambda + (\lambda + 1)\mathrm{i}}\) | M1 |
| \(u + \mathrm{i}v = \dfrac{\lambda(2\lambda + 1) + \lambda\mathrm{i}}{2\lambda^2 + 2\lambda + 1}\) | A1 |
| \(u = \dfrac{\lambda(2\lambda + 1)}{2\lambda^2 + 2\lambda + 1},\ v = \dfrac{\lambda}{2\lambda^2 + 2\lambda + 1}\) | M1 |
| \(u^2 + v^2 - u + v = \left(\dfrac{\lambda(2\lambda + 1)}{2\lambda^2 + 2\lambda + 1}\right)^2 + \left(\dfrac{\lambda}{2\lambda^2 + 2\lambda + 1}\right)^2 - \dfrac{\lambda(2\lambda + 1)}{2\lambda^2 + 2\lambda + 1} + \dfrac{\lambda}{2\lambda^2 + 2\lambda + 1}\) \(= \dfrac{(4\lambda^4 + 4\lambda^3 + \lambda^2) + \lambda^2 - 2\lambda^2(2\lambda^2 + 2\lambda + 1)}{(2\lambda^2 + 2\lambda + 1)^2}\) | M1 |
| \(= 0*\) | M1A1 |
Alternative 2
| Scheme | Marks |
|---|---|
| Let \(z = \lambda - (\lambda + 1)\mathrm{i}\): \(u + \mathrm{i}v = \dfrac{\lambda - \lambda\mathrm{i}}{\lambda - (\lambda + 1)\mathrm{i}}\) | M1 |
| \((u + \mathrm{i}v)(\lambda - (\lambda + 1)\mathrm{i}) = \lambda - \lambda\mathrm{i}\) | M1 |
| \(u\lambda + v(\lambda + 1) + [v\lambda - u(\lambda + 1)]\mathrm{i} = \lambda - \lambda\mathrm{i}\) | A1 |
| Equating real & imaginary parts \(u\lambda + v(\lambda + 1) = \lambda\ \text{(i)} \qquad v\lambda - \lambda u - u = -\lambda\ \text{(ii)}\) | M1 |
| From (i) \(\lambda = \dfrac{v}{1 - u - v}\) From (ii) \(\lambda = \dfrac{u}{1 - u + v}\) \(\dfrac{v}{1 - u - v} = \dfrac{u}{1 - u + v}\) | M1 |
| Reducing to the circle with equation \(u^2 + v^2 - u + v = 0\ *\) | M1A1 |
(corrected from the printed mark scheme: in Alternative 2 the right-hand side of the third line is printed as \(\lambda - \lambda 1\))
| Scheme | Marks |
|---|---|
| ft their line | B1ft |
| Circle through origin, centre in correct quadrant | B1 |
| Intersection correctly placed | B1 |
| (3) | |
| (15 marks) |