Question 4 - A-Level Maths

Edexcel F2 2018 Specimen — Question 4 9 marks

Exam Board	Edexcel
Module	F2 (Further Pure Mathematics 2)
Year	2018
Session	Specimen
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Complex numbers 2
Type	Complex transformations (Möbius)
Difficulty	Challenging +1.2 This is a standard Further Maths Möbius transformation question requiring substitution of z = x + iy, algebraic manipulation to find the image circle's equation, and determining which region maps where. While it involves several steps and complex number manipulation, it follows a well-established procedure taught in F2 with no novel insight required—moderately above average difficulty due to the algebraic complexity and being a Further Maths topic.
Spec	4.02k Argand diagrams: geometric interpretation 4.02l Geometrical effects: conjugate, addition, subtraction

A transformation $T$ from the $z$-plane to the $w$-plane is given by

$$w = \frac { z - 1 } { z + 1 } , \quad z \neq - 1$$ The line in the $z$-plane with equation $y = 2 x$ is mapped by $T$ onto the curve $C$ in the $w$-plane.

Show that $C$ is a circle and find its centre and radius. The region $y < 2 x$ in the $z$-plane is mapped by $T$ onto the region $R$ in the $w$-plane.
Sketch circle $C$ on an Argand diagram and shade and label region $R$.
VIIIV SIHI NI IIIYM ION OC VIIV SIHI NI JIIIM ION OC VEXV SIHIL NI JIIIM ION OO

Show mark scheme Show mark scheme source

Question 4:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$w = \frac{z-1}{z+1} \Rightarrow wz + w = z - 1 \Rightarrow z = ...$	M1	Attempt to make $z$ the subject
$z = \frac{w+1}{1-w}$	A1	Correct expression in terms of $w$
$= \frac{u+iv+1}{1-u-iv} \times \frac{1-u+iv}{1-u+iv}$	M1	Introduces "$u+iv$" and multiplies top and bottom by the complex conjugate of the bottom
$x = \frac{-u^2 - v^2 + 1}{....}$, $y = \frac{2v}{....}$
$y = 2x \Rightarrow 2v = -2u^2 - 2v^2 + 2$	M1	Uses real and imaginary parts and $y = 2x$ to obtain an equation connecting "$u$" and "$v$". Can have the 2 on the wrong side
$u^2 + \left(v + \frac{1}{2}\right)^2 - \frac{1}{4} = 1$	M1	Processes their equation to a form recognisable as a circle, ie coefficients of $u^2$ and $v^2$ are the same and no $uv$ terms
Centre $\left(0, -\frac{1}{2}\right)$, radius $\frac{\sqrt{5}}{2}$	A1, A1	A1: Correct centre (allow $-\frac{1}{2}i$); A1: Correct radius

Total: (7)

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Circle correctly positioned with shading inside	B1ft	Their circle correctly positioned provided their equation does give a circle
Completely correct sketch and shading	B1	Completely correct sketch and shading

Total: (2)

# Question 4:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $w = \frac{z-1}{z+1} \Rightarrow wz + w = z - 1 \Rightarrow z = ...$ | M1 | Attempt to make $z$ the subject |
| $z = \frac{w+1}{1-w}$ | A1 | Correct expression in terms of $w$ |
| $= \frac{u+iv+1}{1-u-iv} \times \frac{1-u+iv}{1-u+iv}$ | M1 | Introduces "$u+iv$" and multiplies top and bottom by the complex conjugate of the bottom |
| $x = \frac{-u^2 - v^2 + 1}{....}$, $y = \frac{2v}{....}$ | | |
| $y = 2x \Rightarrow 2v = -2u^2 - 2v^2 + 2$ | M1 | Uses real and imaginary parts and $y = 2x$ to obtain an equation connecting "$u$" and "$v$". Can have the 2 on the wrong side |
| $u^2 + \left(v + \frac{1}{2}\right)^2 - \frac{1}{4} = 1$ | M1 | Processes their equation to a form recognisable as a circle, ie coefficients of $u^2$ and $v^2$ are the same and no $uv$ terms |
| Centre $\left(0, -\frac{1}{2}\right)$, radius $\frac{\sqrt{5}}{2}$ | A1, A1 | A1: Correct centre (allow $-\frac{1}{2}i$); A1: Correct radius |

**Total: (7)**

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Circle correctly positioned with shading inside | B1ft | Their circle correctly positioned provided their equation does give a circle |
| Completely correct sketch and shading | B1 | Completely correct sketch and shading |

**Total: (2)**

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Show LaTeX source

\begin{enumerate}
  \item A transformation $T$ from the $z$-plane to the $w$-plane is given by
\end{enumerate}

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

The line in the $z$-plane with equation $y = 2 x$ is mapped by $T$ onto the curve $C$ in the $w$-plane.\\
(a) Show that $C$ is a circle and find its centre and radius.

The region $y < 2 x$ in the $z$-plane is mapped by $T$ onto the region $R$ in the $w$-plane.\\
(b) Sketch circle $C$ on an Argand diagram and shade and label region $R$.

\begin{center}
\begin{tabular}{|l|l|l|}
\hline
VIIIV SIHI NI IIIYM ION OC & VIIV SIHI NI JIIIM ION OC & VEXV SIHIL NI JIIIM ION OO \\
\hline
\end{tabular}
\end{center}

\hfill \mbox{\textit{Edexcel F2 2018 Q4 [9]}}

This paper (7 questions)

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Q2 5 Q3 10 Q4 9 Q5 9 Q6 13 Q7 8 Q8 14

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(w = \frac{z-1}{z+1} \Rightarrow wz + w = z - 1 \Rightarrow z = ...\)	M1	Attempt to make \(z\) the subject
\(z = \frac{w+1}{1-w}\)	A1	Correct expression in terms of \(w\)
\(= \frac{u+iv+1}{1-u-iv} \times \frac{1-u+iv}{1-u+iv}\)	M1	Introduces "\(u+iv\)" and multiplies top and bottom by the complex conjugate of the bottom
\(x = \frac{-u^2 - v^2 + 1}{....}\), \(y = \frac{2v}{....}\)
\(y = 2x \Rightarrow 2v = -2u^2 - 2v^2 + 2\)	M1	Uses real and imaginary parts and \(y = 2x\) to obtain an equation connecting "\(u\)" and "\(v\)". Can have the 2 on the wrong side
\(u^2 + \left(v + \frac{1}{2}\right)^2 - \frac{1}{4} = 1\)	M1	Processes their equation to a form recognisable as a circle, ie coefficients of \(u^2\) and \(v^2\) are the same and no \(uv\) terms
Centre \(\left(0, -\frac{1}{2}\right)\), radius \(\frac{\sqrt{5}}{2}\)	A1, A1	A1: Correct centre (allow \(-\frac{1}{2}i\)); A1: Correct radius