FP2 June 2017 Q8
8. The transformation \(T\) from the \(z\)-plane to the \(w\)-plane is given by \[w = \frac{z + 3\mathrm{i}}{1 + \mathrm{i}z}, \quad z \neq \mathrm{i}\]
The transformation \(T\) maps the circle \(|z| = 1\) in the \(z\)-plane onto the line \(l\) in the \(w\)-plane.
(a) Find a cartesian equation of the line \(l\). (5)
The circle \(|z - a - b\mathrm{i}| = c\) in the \(z\)-plane is mapped by \(T\) onto the circle \(|w| = 5\) in the \(w\)-plane.
(b) Find the exact values of the real constants \(a\), \(b\) and \(c\). (6)
| Scheme | Marks |
|---|---|
| \(w = \dfrac{z + 3\mathrm{i}}{1 + \mathrm{i}z}\) | |
| \(z = \dfrac{w - 3\mathrm{i}}{1 - \mathrm{i}w}\) oe M1: Attempt to make \(z\) the subject A1: Correct equation | M1A1 |
| \(|z| = 1 \Rightarrow \left|\dfrac{w - 3\mathrm{i}}{1 - \mathrm{i}w}\right| = 1 \Rightarrow |w - 3\mathrm{i}| = |1 - w\mathrm{i}|\) \(\therefore\ |u + \mathrm{i}v - 3\mathrm{i}| = |(u + \mathrm{i}v)\mathrm{i} - 1|\) Uses \(|z| = 1\) and introduce “\(u + \mathrm{i}v\)” (or \(x + \mathrm{i}y\)) for \(w\) | M1 |
| \(u^2 + (v - 3)^2 = u^2 + (v + 1)^2\) Correct use of Pythagoras on either side. | M1 |
| \(v = 1\) oe \(v = 1\) or \(y = 1\) | A1 |
| (5) |
Notes
Alternative 1 for (a)
| Scheme | Marks |
|---|---|
| eg \(w(1) = \dfrac{1 + 3\mathrm{i}}{1 + \mathrm{i}} = 2 + \mathrm{i}\) M1: Maps one point on the circle using the given transformation A1:Correct mapping | M1A1 |
| eg \(w(-\mathrm{i}) = \dfrac{2\mathrm{i}}{2} = \mathrm{i}\) Maps a second point on the circle | M1 |
| \(v = 1\) oe M1: Forms Cartesian equation using their 2 points A1: \(v = 1\) or \(y = 1\) | M1A1 |
Alternative 2 for (a)
| Scheme | Marks |
|---|---|
| \(z = \dfrac{w - 3\mathrm{i}}{1 - \mathrm{i}w}\) oe M1: Attempt to make \(z\) the subject A1: Correct equation | M1A1 |
| \(|z| = 1 \Rightarrow \left|\dfrac{w - 3\mathrm{i}}{1 - \mathrm{i}w}\right| = 1 \Rightarrow |w - 3\mathrm{i}| = |1 - w\mathrm{i}|\) \(|w - 3\mathrm{i}| = |w + \mathrm{i}| = |w - (-\mathrm{i})|\) Uses \(|z| = 1\) and changes to form \(|w - \ldots| = |w - \ldots|\) or draws a diagram | M1 |
| Perpendicular bisector of points \((0, 3)\) and \((0, -1)\) Uses a correct geometrical approach | M1 |
| \(v = 1\) oe \(v = 1\) or \(y = 1\) | A1 |
Alternative 3 for (a)
| Scheme | Marks |
|---|---|
| Let \(z = x + \mathrm{i}y,\ |z| = 1 \Rightarrow x^2 + y^2 = 1\) | |
| \(w = \dfrac{z + 3\mathrm{i}}{1 + \mathrm{i}z} = \dfrac{x + \mathrm{i}y + 3\mathrm{i}}{1 + \mathrm{i}(x + \mathrm{i}y)} = \dfrac{x + \mathrm{i}(y + 3)}{(1 - y) + \mathrm{i}x}\) | |
| \(w = \dfrac{x + \mathrm{i}(y + 3)}{(1 - y) + \mathrm{i}x}\times\dfrac{(1 - y) - \mathrm{i}x}{(1 - y) - \mathrm{i}x}\) Substitute \(z = x + \mathrm{i}y\) and multiply numerator and denominator by complex conjugate of their denominator | M1 |
| \(w = \dfrac{x(1 - y) - \mathrm{i}x^2 + \mathrm{i}(y + 3)(1 - y) - \mathrm{i}^2x(y + 3)}{(1 - y)^2 - \mathrm{i}x(1 - y) + \mathrm{i}x(1 - y) - \mathrm{i}^2x^2}\) | |
| \(w = \dfrac{\left[x(1 - y) + x(y + 3)\right] + \mathrm{i}\left[-x^2 + (y + 3)(1 - y)\right]}{(1 - y)^2 + x^2}\) M1: Multiply out and collect real and imaginary parts in numerator. Denominator must be real. A1: all correct | M1 A1 |
| \(w = \dfrac{\left[x - xy + xy + 3x\right] + \mathrm{i}\left[-x^2 + y - y^2 + 3 - 3y\right]}{1 - 2y + y^2 + x^2}\) | |
| \(w = \dfrac{[4x] + \mathrm{i}[-1 + 3 - 2y]}{2 - 2y}\) Applies \(x^2 + y^2 = 1\) | M1 |
| \(w = \dfrac{4x + \mathrm{i}[2 - 2y]}{2 - 2y} = \dfrac{4x}{2 - 2y} + \mathrm{i}\) | |
| \(y = 1\) \(y = 1\) or \(v = 1\) | A1 |
| Scheme | Marks |
|---|---|
| \(|w| = 5 \Rightarrow \left|\dfrac{z + 3\mathrm{i}}{1 + \mathrm{i}z}\right| = 5 \Rightarrow |z + 3\mathrm{i}| = 5|1 + \mathrm{i}z|\) \(\therefore\ |x + \mathrm{i}y + 3\mathrm{i}| = 5|(x + \mathrm{i}y)\mathrm{i} + 1|\) Uses \(|w| = 5\) and introduce “\(x + \mathrm{i}y\)” | M1 |
| \(x^2 + (y + 3)^2 = 25\left(x^2 + (1 - y)^2\right)\) M1: Correct use of Pythagoras Allow 25 or 5 A1: Correct equation | M1A1 |
| \(x^2 + y^2 - \dfrac{7}{3}y + \dfrac{2}{3} = 0\) | |
| \(x^2 + \left(y - \dfrac{7}{6}\right)^2 = \dfrac{25}{36}\) Attempt circle form or attempt \(r^2\) from the line above. | M1 |
| \(a = 0,\ b = \dfrac{7}{6},\ c = \dfrac{5}{6}\) A1: 2 correct A1: All correct | A1, A1 |
| (6) | |
| (11 marks) |
Notes
Or, for the last 3 marks:
| Scheme | Marks |
|---|---|
| \(\left|z - 0 - \dfrac{7}{6}\mathrm{i}\right| = \dfrac{5}{6}\) | M1A1A1 |
| If 0 not shown score M1A1A0 | |
| No need to list \(a\), \(b\), \(c\) separately if answer in this form. |