FP2 June 2014 (R) Q6
6. The transformation \(T\) maps points from the \(z\)-plane, where \(z = x + \mathrm{i}y\), to the \(w\)-plane, where \(w = u + \mathrm{i}v\).
The transformation \(T\) is given by \[w = \frac{z}{\mathrm{i}z + 1}, \quad z \neq \mathrm{i}\]
The transformation \(T\) maps the line \(l\) in the \(z\)-plane onto the line with equation \(v = -1\) in the \(w\)-plane.
The transformation \(T\) maps the line with equation \(y = \dfrac{1}{2}\) in the \(z\)-plane onto the curve \(C\) in the \(w\)-plane.
| Scheme | Marks |
|---|---|
| \(w = (u - \mathrm{i} =)\,\dfrac{x + \mathrm{i}y}{\mathrm{i}x - y + 1}\) | M1 |
| \(w = (u - \mathrm{i} =)\,\dfrac{x + \mathrm{i}y}{\mathrm{i}x - y + 1}\times\dfrac{(1 - y - \mathrm{i}x)}{(1 - y - \mathrm{i}x)}\) | M1 |
| \(w = (u - \mathrm{i} =)\,\dfrac{x - xy + xy + \mathrm{i}\left(y - y^2 - x^2\right)}{(1 - y)^2 + x^2}\) | A1 |
| \(-1 = \dfrac{y - y^2 - x^2}{(1 - y)^2 + x^2}\) | M1 |
| \(-1 + 2y - y^2 - x^2 = y - y^2 - x^2\) | |
| \(y = 1\) | A1 |
| (5) |
Notes
M1 substitute \(z = x + \mathrm{i}y\)
M1 multiply the numerator and the denominator by conjugate of the denominator
A1 correct equation with real denominator on rhs
M1 use \(w = u - \mathrm{i}\) and equate imaginary part in (their) equation to -1
A1 deducing \(y = 1\)
Alternative 1
| Scheme | Marks |
|---|---|
| \(z = \dfrac{w}{1 - w\mathrm{i}}\) | M1 |
| \(x + \mathrm{i}y = \dfrac{u - \mathrm{i}}{1 - (u - \mathrm{i})\mathrm{i}}\) | M1 |
| \(= \dfrac{u - \mathrm{i}}{-u\mathrm{i}}\) | A1 |
| \(= \mathrm{i} + \dfrac{1}{u}\) | M1 |
| \(y = 1\) | A1 |
| (5) |
M1 re-arrange to \(z = \ldots\)
M1 replace \(w\) with \(u - \mathrm{i}\) or replace with \(u + \mathrm{i}v\) and realise the denominator
A1 correct rhs May still have \(v\)
M1 separate real and imaginary parts in (their) above equation. This may be implied by a correct answer.
A1 deducing \(y = 1\)
Alternative 2
| Scheme | Marks |
|---|---|
| \(|w + 2\mathrm{i}| = |w|\) | M1 |
| \(\left|\dfrac{z}{\mathrm{i}z + 1} + 2\mathrm{i}\right| = \left|\dfrac{z}{\mathrm{i}z + 1}\right|\) | M1 |
| \(\left|\dfrac{z - 2z + 2\mathrm{i}}{\mathrm{i}z + 1}\right| = \left|\dfrac{z}{\mathrm{i}z + 1}\right|\) | A1 |
| \(|z - 2\mathrm{i}| = |z|\) | M1 |
| \(\Rightarrow y = 1\) | A1 |
| (5) |
Alternative 2: M1 M1 A1 M1 A1
Alternative 3: M1 M1 A1 M1 A1
(Corrected from the printed mark scheme: the denominator in the line \(-1 = \ldots\) is printed as \(\left(1 - y^2\right) + x^2\); it is \((1 - y)^2 + x^2\), as in the line above.)
| Scheme | Marks |
|---|---|
| \(u + \mathrm{i}v = \dfrac{x + \frac{1}{2}\mathrm{i}}{\mathrm{i}\left(x + \frac{1}{2}\mathrm{i}\right) + 1} = \dfrac{x + \frac{1}{2}\mathrm{i}}{\mathrm{i}x + \frac{1}{2}}\) | M1 |
| \(u + \mathrm{i}v = \dfrac{x + \frac{1}{2}\mathrm{i}}{\mathrm{i}x + \frac{1}{2}}\times\dfrac{\frac{1}{2} - x\mathrm{i}}{\frac{1}{2} - x\mathrm{i}}\) | M1 |
| \(u + \mathrm{i}v = \dfrac{x + \mathrm{i}\left(\frac{1}{4} - x^2\right)}{\frac{1}{4} + x^2}\) | A1 |
| \(u = \dfrac{x}{\frac{1}{4} + x^2} \qquad v = \dfrac{\frac{1}{4} - x^2}{\frac{1}{4} + x^2}\) | M1 |
| \(u^2 + v^2 = \dfrac{x^2 + \left(\frac{1}{4} - x^2\right)^2}{\left(\frac{1}{4} + x^2\right)^2} = \dfrac{\frac{1}{16} - \frac{1}{2}x^2 + x^2 + x^4}{\left(\frac{1}{4} + x^2\right)^2} = 1\) | M1 |
| \(u^2 + v^2 = 1\), Centre \(O\) * | M1,A1 |
| (6) | |
| (11 marks) |
Notes
M1 substitute \(z = x + \mathrm{i}1/2\) or work with \(x + \mathrm{i}y\)
M1 multiply the numerator and the denominator by conjugate of the denominator
A1 correct equation with real denominator on rhs
M1 use their \(u\), \(v\) and find \(u^2 + v^2\)
M1 simplify and cancel including use of \(y = \dfrac{1}{2}\)
A1cso \(u^2 + v^2 = 1\) Centre \(O\)
Alternative 1
| Scheme | Marks |
|---|---|
| \(w = u + \mathrm{i}v = \dfrac{x + \frac{1}{2}\mathrm{i}}{\mathrm{i}x + \frac{1}{2}},\ (= r\mathrm{e}^{\mathrm{i}\theta})\) or \(\dfrac{x + \mathrm{i}y}{\mathrm{i}x + y}\) | M1 |
| \(|w| = \dfrac{\sqrt{x^2 + \frac{1}{4}}}{\sqrt{x^2 + \frac{1}{4}}},\ = 1\) | M1M1,A1 |
| \(u^2 + v^2 = 1\ \therefore\) | M1 |
| Centre \(O\) * | A1 |
M1 as above
M1 find \(|w|\)
M1 substitute \(y = \dfrac{1}{2}\)
A1 \(|w| = 1\)
M1 deduce circle centre \(O\)
A1cso \(u^2 + v^2 = 1\)
Alternative 2
| Scheme | Marks |
|---|---|
| \(|z - \mathrm{i}| = |z|\) | M1 |
| \(\left|\dfrac{w}{1 - \mathrm{i}w} - \dfrac{\mathrm{i}(1 - \mathrm{i}w)}{1 - \mathrm{i}w}\right| = \left|\dfrac{w}{1 - \mathrm{i}w}\right|\) | M1 |
| \(\left|w - \mathrm{i} + \mathrm{i}^2w\right| = |w|\) | A1 |
| \(\Rightarrow |w| = 1\) | M1 |
| \(u^2 + v^2 = 1\) | M1 |
| Centre \(O\) | A1 |
| (6) |
Alternative 3
| Scheme | Marks |
|---|---|
| \(w = \dfrac{z}{\mathrm{i}z + 1}\) | |
| \(z = \dfrac{w}{1 - \mathrm{i}w} = \dfrac{u + \mathrm{i}v}{1 - (u + \mathrm{i}v)\mathrm{i}}\) | M1 |
| Realise the denominator | M1 |
| Correct result | A1 |
| Set imaginary part \(= \dfrac{1}{2}\) and simplify expression | M1 |
| \(u^2 + v^2 = 1\) | M1 |
| Centre \(O\) | A1 |
| (6) |