| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Complex transformations and mappings |
| Difficulty | Challenging +1.2 This is a standard Further Maths transformation question requiring substitution of z = x + 4ix into the given transformation, algebraic manipulation to separate real and imaginary parts, elimination of the parameter, and completing the square. While it involves several steps and careful algebra with complex numbers, it follows a well-established procedure taught in F2 with no novel insight required. The 'show that' format provides the target equation, making it more routine than open-ended loci problems. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.02l Geometrical effects: conjugate, addition, subtraction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(w = \frac{z+1}{z-3} \Rightarrow wz - 3w = z+1 \Rightarrow z = \frac{3w+1}{w-1}\) | M1 A1 | M1: Attempt to make \(z\) subject; A1: Correct expression |
| \((z=)\frac{(3u+1)+3iv}{(u-1)+iv} \times \frac{(u-1)-iv}{(u-1)-iv} = ...\) | dM1 | Dependent on M1. Replace \(w\) with \(u+iv\), rationalise denominator. Accept \(x+iy\) instead |
| \(x = \frac{3u^2-2u+3v^2-1}{(u-1)^2+v^2}\), \(y = \frac{-4v}{(u-1)^2+v^2}\); \((y=4x \Rightarrow)\; -4v = 12u^2-8u+12v^2-4\) | ddM1 | Dependent on both M marks. Equate real/imaginary parts, use \(y=4x\) to form equation in \(u,v\) only |
| \(\Rightarrow 3u^2+3v^2-2u+v-1=0\)* | A1cso* | Must equal 0. No incorrect statements leading to answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u^2-\frac{2}{3}u+v^2+\frac{1}{3}v-\frac{1}{3}=0 \Rightarrow \left(u-\frac{1}{3}\right)^2+\left(v+\frac{1}{6}\right)^2=\frac{17}{36}\) | B1 B1 | B1: Correct centre or radius (unsimplified fractions allowed); B1: Correct centre and radius (fractions must be simplified) |
| Centre: \(\left(\frac{1}{3},-\frac{1}{6}\right)\), radius: \(\frac{\sqrt{17}}{6}\) | Allow \(x=\frac{1}{3}, y=-\frac{1}{6}\) |
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $w = \frac{z+1}{z-3} \Rightarrow wz - 3w = z+1 \Rightarrow z = \frac{3w+1}{w-1}$ | M1 A1 | M1: Attempt to make $z$ subject; A1: Correct expression |
| $(z=)\frac{(3u+1)+3iv}{(u-1)+iv} \times \frac{(u-1)-iv}{(u-1)-iv} = ...$ | dM1 | Dependent on M1. Replace $w$ with $u+iv$, rationalise denominator. Accept $x+iy$ instead |
| $x = \frac{3u^2-2u+3v^2-1}{(u-1)^2+v^2}$, $y = \frac{-4v}{(u-1)^2+v^2}$; $(y=4x \Rightarrow)\; -4v = 12u^2-8u+12v^2-4$ | ddM1 | Dependent on both M marks. Equate real/imaginary parts, use $y=4x$ to form equation in $u,v$ only |
| $\Rightarrow 3u^2+3v^2-2u+v-1=0$* | A1cso* | Must equal 0. No incorrect statements leading to answer |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u^2-\frac{2}{3}u+v^2+\frac{1}{3}v-\frac{1}{3}=0 \Rightarrow \left(u-\frac{1}{3}\right)^2+\left(v+\frac{1}{6}\right)^2=\frac{17}{36}$ | B1 B1 | B1: Correct centre **or** radius (unsimplified fractions allowed); B1: Correct centre **and** radius (fractions must be simplified) |
| Centre: $\left(\frac{1}{3},-\frac{1}{6}\right)$, radius: $\frac{\sqrt{17}}{6}$ | | Allow $x=\frac{1}{3}, y=-\frac{1}{6}$ |
---\begin{enumerate}
\item The transformation $T$ from the $z$-plane, where $z = x + \mathrm { i } y$, to the $w$-plane, where $w = u + \mathrm { i } v$ is given by
\end{enumerate}
$$w = \frac { z + 1 } { z - 3 } \quad z \neq 3$$
The straight line in the $z$-plane with equation $y = 4 x$ is mapped by $T$ onto the circle $C$ in the $w$-plane.\\
(a) Show that $C$ has equation
$$3 u ^ { 2 } + 3 v ^ { 2 } - 2 u + v + k = 0$$
where $k$ is a constant to be determined.\\
(b) Hence determine\\
(i) the coordinates of the centre of $C$\\
(ii) the radius of $C$
\hfill \mbox{\textit{Edexcel F2 2023 Q5 [7]}}