Question 5 - A-Level Maths

Edexcel F2 2016 June — Question 5 7 marks

Exam Board	Edexcel
Module	F2 (Further Pure Mathematics 2)
Year	2016
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Complex numbers 2
Type	Complex transformations (Möbius)
Difficulty	Challenging +1.8 This is a Möbius transformation question requiring students to find the image of a circle under the transformation. It demands understanding of complex transformations, algebraic manipulation to find the image equation, and completing the square. While the technique is standard for Further Maths F2, it requires multiple sophisticated steps and careful algebraic work, making it significantly harder than average A-level questions but not exceptionally difficult for Further Maths students who have practiced this topic.
Spec	4.02k Argand diagrams: geometric interpretation

5. The transformation $T$ from the $z$-plane to the $w$-plane is given by $$w = \frac { 2 z - 1 } { z + 3 } , \quad z \neq - 3$$ The circle in the $z$-plane with equation $x ^ { 2 } + y ^ { 2 } = 1$, where $z = x + \mathrm { i } y$, is mapped by $T$ onto the circle $C$ in the $w$-plane. Find the centre and the radius of $C$.

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Question 5:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$z = \frac{3w+1}{2-w}$	M1A1	M1: Attempt to make $z$ the subject. A1: Correct equation
\(\	z\	=1 \Rightarrow \left\
$(3u+1)^2+(3v)^2=(u-2)^2+v^2$	M1	Correct use of Pythagoras; condone missing brackets if intention clear; no $i$'s
$u^2+v^2+\frac{10}{8}u-\frac{3}{8}=0$	—	—
$\left(u+\frac{5}{8}\right)^2-\frac{25}{64}+v^2=\frac{3}{8}$	ddddM1	Attempt to complete the square on equation of a circle where coefficients of $u^2$ and $v^2$ are equal; dependent on all previous M marks
$\left(u+\frac{5}{8}\right)^2+v^2=\frac{49}{64}$	—	—
Centre $\left(-\frac{5}{8},0\right)$	A1	Correct centre
Radius $\frac{7}{8}$	A1	Correct radius

Total: (7)

# Question 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = \frac{3w+1}{2-w}$ | M1A1 | M1: Attempt to make $z$ the subject. A1: Correct equation |
| $\|z\|=1 \Rightarrow \left\|\frac{3w+1}{2-w}\right\|=1 \Rightarrow \left\|\frac{3(u+iv)+1}{2-(u+iv)}\right\|=1$ | M1 | Uses $\|z\|=1$ and introduces $w=u+iv$ |
| $(3u+1)^2+(3v)^2=(u-2)^2+v^2$ | M1 | Correct use of Pythagoras; condone missing brackets if intention clear; no $i$'s |
| $u^2+v^2+\frac{10}{8}u-\frac{3}{8}=0$ | — | — |
| $\left(u+\frac{5}{8}\right)^2-\frac{25}{64}+v^2=\frac{3}{8}$ | ddddM1 | Attempt to complete the square on equation of a circle where coefficients of $u^2$ and $v^2$ are equal; dependent on all previous M marks |
| $\left(u+\frac{5}{8}\right)^2+v^2=\frac{49}{64}$ | — | — |
| Centre $\left(-\frac{5}{8},0\right)$ | A1 | Correct centre |
| Radius $\frac{7}{8}$ | A1 | Correct radius |

**Total: (7)**

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5. The transformation $T$ from the $z$-plane to the $w$-plane is given by

$$w = \frac { 2 z - 1 } { z + 3 } , \quad z \neq - 3$$

The circle in the $z$-plane with equation $x ^ { 2 } + y ^ { 2 } = 1$, where $z = x + \mathrm { i } y$, is mapped by $T$ onto the circle $C$ in the $w$-plane.

Find the centre and the radius of $C$.

\hfill \mbox{\textit{Edexcel F2 2016 Q5 [7]}}

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Q1 6 Q2 6 Q3 6 Q4 8 Q5 7 Q6 14 Q7 11 Q8 10 Q9 7