FP2 June 2014 Q6
6. 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 \[w = \frac{4(1 - \mathrm{i})z - 8\mathrm{i}}{2(-1 + \mathrm{i})z - \mathrm{i}}, \quad z \neq \frac{1}{4} - \frac{1}{4}\mathrm{i}\]
The transformation \(T\) maps the points on the line \(l\) with equation \(y = x\) in the \(z\)-plane to a circle \(C\) in the \(w\)-plane.
(a) Show that \[w = \frac{ax^2 + bx\mathrm{i} + c}{16x^2 + 1}\] where \(a\), \(b\) and \(c\) are real constants to be found. (6)
(b) Hence show that the circle \(C\) has equation \[(u - 3)^2 + v^2 = k^2\] where \(k\) is a constant to be found. (4)
| Scheme | Marks |
|---|---|
| \(w = \dfrac{4(1 - \mathrm{i})z - 8\mathrm{i}}{2(\mathrm{i} - 1)z - \mathrm{i}}\) | |
| Method 1: Substituting \(z = x + x\mathrm{i}\) at the start | |
| \(w = \dfrac{4(1 - \mathrm{i})(x + x\mathrm{i}) - 8\mathrm{i}}{2(\mathrm{i} - 1)(x + x\mathrm{i}) - \mathrm{i}}\) Substitutes for \(z\) | M1 |
| \(w = \dfrac{4(x + x\mathrm{i} - x\mathrm{i} + x) - 8\mathrm{i}}{2(x\mathrm{i} - x - x - x\mathrm{i}) - \mathrm{i}}\) M1: Attempt to expand numerator and denominator A1: Correct expression | M1A1 |
| \(\dfrac{8\mathrm{i} - 8x}{4x + \mathrm{i}} \cdot \dfrac{4x - \mathrm{i}}{4x - \mathrm{i}}\) M1: Multiplies numerator and denominator by the conjugate of their denom. No expansion needed A1: Uses correct conjugate (not ft) | M1A1 |
| \(= \dfrac{-32x^2 + 40x\mathrm{i} + 8}{16x^2 + 1}\) cso Award only if final answer is correct and follows correct working | B1 |
| NB: The B mark appears first on e-PEN but will be awarded last | |
| (6) |
Method 2: if they proceed without \(y = x\) (substitution may happen anywhere in the working)
| Scheme | Marks |
|---|---|
| \(w = \dfrac{(1 - \mathrm{i})z - 8\mathrm{i}}{2(-1 + \mathrm{i}) - \mathrm{i}} = \dfrac{4(1 - \mathrm{i})(x + \mathrm{i}y) - 8\mathrm{i}}{2(-1 + \mathrm{i})(x + \mathrm{i}y) - \mathrm{i}}\) Substitutes for \(z\) | M1 |
| \(= \dfrac{4(1 - \mathrm{i})x + 4(1 - \mathrm{i})\mathrm{i}y - 8\mathrm{i}}{2(-1 + \mathrm{i})x + 2(-1 + \mathrm{i})\mathrm{i}y - \mathrm{i}}\) Attempt to expand numerator and denominator | M1 |
| \(= \dfrac{4x + 4y + (4y - 4x - 8)\mathrm{i}}{-2x - 2y + (2x - 2y - 1)\mathrm{i}}\) Correct expression | A1 |
| \(= \dfrac{4x + 4y + (4y - 4x - 8)\mathrm{i}}{-2x - 2y + (2x - 2y - 1)\mathrm{i}} \times \dfrac{-2x - 2y - (2x - 2y - 1)\mathrm{i}}{-2x - 2y - (2x - 2y - 1)\mathrm{i}}\) M1: Multiplies numerator and denominator by the conjugate of their denom. No expansion needed. A1: Uses correct conjugate. (not ft) | M1A1 |
| \(= \dfrac{-16x^2 - 16y^2 + 12y - 12x + 8 + (20x + 20y)\mathrm{i}}{8x^2 + 8y^2 - 4x + 4y + 1}\) | |
| \(= \dfrac{-32x^2 + 40x\mathrm{i} + 8}{16x^2 + 1}\) cso Correct answer using \(y = x\) Award only if final answer is correct and follows correct working | B1 |
| NB: The B mark appears first on e-PEN but will be awarded last | (6) |
| Scheme | Marks |
|---|---|
| NB: The order of awarding the marks here has changed from the original mark scheme, but they must still be entered on e-PEN by their descriptors (M or A) | |
| \(u = \dfrac{-32x^2 + 8}{16x^2 + 1},\quad v = \dfrac{40x}{16x^2 + 1}\) Identifies \(u\) and \(v\) (Real and imaginary parts) May be implied by their working and may be in terms of \(x\) and \(y\). | M1 1st M mark on e-PEN |
| \(\left(\dfrac{8 - 32x^2}{16x^2 + 1} - 3\right)^2 + \left(\dfrac{40x}{16x^2 + 1}\right)^2\) Substitutes for their \(u\) and \(v\) in the given equation. May be in terms of \(x\) and \(y\). May have \(a\), \(b\), \(c\) instead of their values (which may be chosen by the candidate if unable to do (a)) | dM1 2nd M mark on e-PEN |
| \(= \left(\dfrac{8 - 32x^2 - 48x^2 - 3}{16x^2 + 1}\right)^2 + \left(\dfrac{40x}{16x^2 + 1}\right)^2\) \(= \dfrac{\left(5 - 80x^2\right)^2}{\left(16x^2 + 1\right)^2} + \dfrac{1600x^2}{\left(16x^2 + 1\right)^2}\) | |
| \(= \dfrac{6400x^4 + 800x^2 + 25}{\left(16x^2 + 1\right)^2}\) Combines to form a single correct fraction | A1 1st A mark on e-PEN |
| \(= \dfrac{25\left(16x^2 + 1\right)^2}{\left(16x^2 + 1\right)^2} = 25\) \(k = 5\) or \(k^2 = 25\) may (but need not) be seen explicitly | A1 2nd A mark on e-PEN |
| (4) | |
| (10 marks) |