Edexcel FP2 2014 June — Question 6 11 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2014
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeComplex transformations (Möbius)
DifficultyChallenging +1.2 This is a standard Möbius transformation question from FP2 requiring algebraic manipulation to find image curves. Part (a) involves finding a preimage line (routine substitution and simplification), while part (b) requires showing a line maps to a circle and finding its equation. These are well-practiced techniques in Further Maths, though requiring more steps than typical A-level questions.
Spec4.02k Argand diagrams: geometric interpretation4.02m Geometrical effects: multiplication and division
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 $$w = \frac { z } { i z + 1 } , \quad z \neq i$$ The transformation \(T\) maps the line \(l\) in the \(z\)-plane onto the line with equation \(v = - 1\) in the \(w\)-plane.
  1. Find a cartesian equation of \(l\) in terms of \(x\) and \(y\). The transformation \(T\) maps the line with equation \(y = \frac { 1 } { 2 }\) in the \(z\)-plane onto the curve \(C\) in the \(w\)-plane.
    1. Show that \(C\) is a circle with centre the origin.
    2. Write down a cartesian equation of \(C\) in terms of \(u\) and \(v\).
Question 6:
Part (a)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(w = (u - \mathrm{i}) = \dfrac{x + \mathrm{i}y}{\mathrm{i}x - y + 1}\)M1 Substitute \(z = x + \mathrm{i}y\)
\(w = (u-\mathrm{i}) = \dfrac{x+\mathrm{i}y}{\mathrm{i}x - y+1} \times \dfrac{(1-y-\mathrm{i}x)}{(1-y-\mathrm{i}x)}\)M1 Multiply numerator and denominator by conjugate of denominator
\(w = (u-\mathrm{i}) = \dfrac{x - xy + xy + \mathrm{i}(y - y^2 - x^2)}{(1-y)^2 + x^2}\)A1 Correct equation with real denominator on rhs
\(-1 = \dfrac{y - y^2 - x^2}{(1-y^2)+x^2}\)M1 Use \(w = u - \mathrm{i}\) and equate imaginary part to \(-1\)
\(-1 + 2y - y^2 - x^2 = y - y^2 - x^2\)
\(y = 1\)A1 (5)
Alternative 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(z = \dfrac{w}{1 - w\mathrm{i}}\)M1 Re-arrange to \(z = \ldots\)
\(x + \mathrm{i}y = \dfrac{u - \mathrm{i}}{1-(u-\mathrm{i})\mathrm{i}}\)M1 Replace \(w\) with \(u - \mathrm{i}\), realise denominator
\(= \dfrac{u-\mathrm{i}}{-u\mathrm{i}}\)A1 Correct rhs (may still have \(v\))
\(= \mathrm{i} + \dfrac{1}{u}\)M1 Separate real and imaginary parts
\(y = 1\)A1 (5)
Alternative 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\w + 2\mathrm{i}\ = \
\(\left\\dfrac{z}{\mathrm{i}z+1} + 2\mathrm{i}\right\ = \left\
\(\left\\dfrac{z - 2z + 2\mathrm{i}}{\mathrm{i}z+1}\right\ = \left\
\(\z - 2\mathrm{i}\ = \
\(\Rightarrow y = 1\)A1 (5)
Part (b)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(u + \mathrm{i}v = \dfrac{x + \frac{1}{2}\mathrm{i}}{\mathrm{i}\left(x+\frac{1}{2}\mathrm{i}\right)+1} = \dfrac{x+\frac{1}{2}\mathrm{i}}{\mathrm{i}x + \frac{1}{2}}\)M1 Substitute \(z = x + \frac{1}{2}\mathrm{i}\) or work with \(x + \mathrm{i}y\)
\(u + \mathrm{i}v = \dfrac{x+\frac{1}{2}\mathrm{i}}{\mathrm{i}x+\frac{1}{2}} \times \dfrac{\frac{1}{2}-x\mathrm{i}}{\frac{1}{2}-x\mathrm{i}}\)M1 Multiply numerator and denominator by conjugate of denominator
\(u + \mathrm{i}v = \dfrac{x + \mathrm{i}\left(\frac{1}{4} - x^2\right)}{\frac{1}{4}+x^2}\)A1 Correct equation with real denominator
\(u = \dfrac{x}{\frac{1}{4}+x^2}, \quad v = \dfrac{\frac{1}{4}-x^2}{\frac{1}{4}+x^2}\)M1 Use their \(u\), \(v\) and find \(u^2 + v^2\)
\(u^2 + v^2 = \dfrac{x^2 + \left(\frac{1}{4}-x^2\right)^2}{\left(\frac{1}{4}+x^2\right)^2} = \dfrac{\frac{1}{16}-\frac{1}{2}x^2+x^2+x^4}{\left(\frac{1}{4}+x^2\right)^2} = 1\)M1 Simplify and cancel including use of \(y = \frac{1}{2}\)
\(u^2 + v^2 = 1\), Centre \(O\)M1, A1 (6)
Alternative 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(w = u + \mathrm{i}v = \dfrac{x+\frac{1}{2}\mathrm{i}}{\mathrm{i}x+\frac{1}{2}}\)M1
\(\w\ = \dfrac{\sqrt{x^2+\frac{1}{4}}}{\sqrt{x^2+\frac{1}{4}}} = 1\)
\(u^2 + v^2 = 1\)M1 Deduce circle centre \(O\)
Centre \(O\)A1
Alternative 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\z - \mathrm{i}\ = \
\(\left\\dfrac{w}{1-\mathrm{i}w} - \dfrac{\mathrm{i}(1-\mathrm{i}w)}{1-\mathrm{i}w}\right\ = \left\
\(\w - \mathrm{i} + \mathrm{i}^2 w\ = \
\(\Rightarrow \w\ = 1\)
\(u^2 + v^2 = 1\)M1
Centre \(O\)A1 (6)
Alternative 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(z = \dfrac{w}{1-\mathrm{i}w} = \dfrac{u+\mathrm{i}v}{1-(u+\mathrm{i}v)\mathrm{i}}\)M1
Realise the denominatorM1
Correct resultA1
Set imaginary part \(= \frac{1}{2}\) and simplifyM1
\(u^2 + v^2 = 1\)M1
Centre \(O\)A1 (6) 
# Question 6:

## Part (a)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $w = (u - \mathrm{i}) = \dfrac{x + \mathrm{i}y}{\mathrm{i}x - y + 1}$ | M1 | Substitute $z = x + \mathrm{i}y$ |
| $w = (u-\mathrm{i}) = \dfrac{x+\mathrm{i}y}{\mathrm{i}x - y+1} \times \dfrac{(1-y-\mathrm{i}x)}{(1-y-\mathrm{i}x)}$ | M1 | Multiply numerator and denominator by conjugate of denominator |
| $w = (u-\mathrm{i}) = \dfrac{x - xy + xy + \mathrm{i}(y - y^2 - x^2)}{(1-y)^2 + x^2}$ | A1 | Correct equation with real denominator on rhs |
| $-1 = \dfrac{y - y^2 - x^2}{(1-y^2)+x^2}$ | M1 | Use $w = u - \mathrm{i}$ and equate imaginary part to $-1$ |
| $-1 + 2y - y^2 - x^2 = y - y^2 - x^2$ | | |
| $y = 1$ | A1 | (5) |

**Alternative 1:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = \dfrac{w}{1 - w\mathrm{i}}$ | M1 | Re-arrange to $z = \ldots$ |
| $x + \mathrm{i}y = \dfrac{u - \mathrm{i}}{1-(u-\mathrm{i})\mathrm{i}}$ | M1 | Replace $w$ with $u - \mathrm{i}$, realise denominator |
| $= \dfrac{u-\mathrm{i}}{-u\mathrm{i}}$ | A1 | Correct rhs (may still have $v$) |
| $= \mathrm{i} + \dfrac{1}{u}$ | M1 | Separate real and imaginary parts |
| $y = 1$ | A1 | (5) |

**Alternative 2:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\|w + 2\mathrm{i}\| = \|w\|$ | 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) |

## Part (b)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $u + \mathrm{i}v = \dfrac{x + \frac{1}{2}\mathrm{i}}{\mathrm{i}\left(x+\frac{1}{2}\mathrm{i}\right)+1} = \dfrac{x+\frac{1}{2}\mathrm{i}}{\mathrm{i}x + \frac{1}{2}}$ | M1 | Substitute $z = x + \frac{1}{2}\mathrm{i}$ or work with $x + \mathrm{i}y$ |
| $u + \mathrm{i}v = \dfrac{x+\frac{1}{2}\mathrm{i}}{\mathrm{i}x+\frac{1}{2}} \times \dfrac{\frac{1}{2}-x\mathrm{i}}{\frac{1}{2}-x\mathrm{i}}$ | M1 | Multiply numerator and denominator by conjugate of denominator |
| $u + \mathrm{i}v = \dfrac{x + \mathrm{i}\left(\frac{1}{4} - x^2\right)}{\frac{1}{4}+x^2}$ | A1 | Correct equation with real denominator |
| $u = \dfrac{x}{\frac{1}{4}+x^2}, \quad v = \dfrac{\frac{1}{4}-x^2}{\frac{1}{4}+x^2}$ | M1 | Use their $u$, $v$ and find $u^2 + v^2$ |
| $u^2 + v^2 = \dfrac{x^2 + \left(\frac{1}{4}-x^2\right)^2}{\left(\frac{1}{4}+x^2\right)^2} = \dfrac{\frac{1}{16}-\frac{1}{2}x^2+x^2+x^4}{\left(\frac{1}{4}+x^2\right)^2} = 1$ | M1 | Simplify and cancel including use of $y = \frac{1}{2}$ |
| $u^2 + v^2 = 1$, Centre $O$ | M1, A1 | (6) |

**Alternative 1:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $w = u + \mathrm{i}v = \dfrac{x+\frac{1}{2}\mathrm{i}}{\mathrm{i}x+\frac{1}{2}}$ | M1 | |
| $\|w\| = \dfrac{\sqrt{x^2+\frac{1}{4}}}{\sqrt{x^2+\frac{1}{4}}} = 1$ | M1M1, A1 | Find $\|w\|$; substitute $y = \frac{1}{2}$ |
| $u^2 + v^2 = 1$ | M1 | Deduce circle centre $O$ |
| Centre $O$ | A1 | |

**Alternative 2:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\|z - \mathrm{i}\| = \|z\|$ | 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 | |
| $\|w - \mathrm{i} + \mathrm{i}^2 w\| = \|w\|$ | A1 | |
| $\Rightarrow \|w\| = 1$ | M1 | |
| $u^2 + v^2 = 1$ | M1 | |
| Centre $O$ | A1 | (6) |

**Alternative 3:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = \dfrac{w}{1-\mathrm{i}w} = \dfrac{u+\mathrm{i}v}{1-(u+\mathrm{i}v)\mathrm{i}}$ | M1 | |
| Realise the denominator | M1 | |
| Correct result | A1 | |
| Set imaginary part $= \frac{1}{2}$ and simplify | M1 | |
| $u^2 + v^2 = 1$ | M1 | |
| Centre $O$ | A1 | (6) |
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

$$w = \frac { z } { i z + 1 } , \quad z \neq i$$

The transformation $T$ maps the line $l$ in the $z$-plane onto the line with equation $v = - 1$ in the $w$-plane.
\begin{enumerate}[label=(\alph*)]
\item Find a cartesian equation of $l$ in terms of $x$ and $y$.

The transformation $T$ maps the line with equation $y = \frac { 1 } { 2 }$ in the $z$-plane onto the curve $C$ in the $w$-plane.
\item \begin{enumerate}[label=(\roman*)]
\item Show that $C$ is a circle with centre the origin.
\item Write down a cartesian equation of $C$ in terms of $u$ and $v$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{Edexcel FP2 2014 Q6 [11]}}
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