| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Complex transformations (Möbius) |
| Difficulty | Challenging +1.2 This is a standard Möbius transformation question requiring systematic application of algebraic manipulation to find images of lines. Part (a) involves substituting z = x + ix and separating real/imaginary parts. Part (b) requires showing a line maps to a circle (typical for Möbius transformations) using similar techniques. Part (c) is routine sketching. While it requires careful algebra and understanding of complex transformations, it follows predictable patterns that FP2 students practice extensively, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02k Argand diagrams: geometric interpretation4.02l Geometrical effects: conjugate, addition, subtraction |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Let \(z = \lambda + 2i\); \(w = \frac{\lambda + (\lambda + 1)i}{\lambda(1 + i)}\) | M1 | |
| \(= \frac{\lambda + (\lambda + 1)i}{\lambda(1+i)} \cdot \frac{1-i}{1-i}\) | M1 | |
| \(u + iv = \frac{(2\lambda + 1) + i}{2\lambda}\) | A1 | |
| \(u = 1 + \frac{1}{2\lambda}\), \(v = \frac{1}{2\lambda}\) | M1 | |
| Eliminating \(\lambda\) gives a line with equation \(v = u - 1\) or equivalent | A15 | |
| (b) Let \(z = \lambda - (\lambda + 1)i\); \(w = \frac{\lambda - \lambda i}{\lambda - (\lambda + 1)i}\) | M1 | |
| \(= \frac{\lambda - \lambda i}{\lambda - (\lambda + 1)i} \times \frac{\lambda + (\lambda + 1)i}{\lambda + (\lambda + 1)i}\) | M1 | |
| \(u + iv = \frac{\lambda(2\lambda + 1) + \lambda i}{2\lambda^2 + 2\lambda + 1}\) | A1 | |
| \(u = \frac{\lambda(2\lambda + 1)}{2\lambda^2 + 2\lambda + 1}\), \(v = \frac{\lambda}{2\lambda^2 + 2\lambda + 1}\) | M1 | |
| \(\frac{u}{v} = 2\lambda + 1\) | ||
| \(v = \frac{2\lambda}{4\lambda^2 + 4\lambda + 2} = \frac{(2\lambda + 1) - 1}{{(2\lambda + 1)}^2 + 1} = \frac{\frac{u}{v} - 1}{\left(\frac{u}{v}\right)^2 + 1}\) | M1 | |
| Reducing to the circle with equation \(u^2 + v^2 - u + v = 0\) * | cso | M1A17 |
| Answer | Marks |
|---|---|
| Let \(z = \lambda - (\lambda + 1)i\); \(u + iv = \frac{\lambda - \lambda i}{\lambda - (\lambda + 1)i}\) | M1 |
| \((u + iv)(\lambda - (\lambda + 1)i) = \lambda - \lambda i\) | M1 |
| \(u\lambda + v(\lambda + 1) + i[v\lambda - u(\lambda + 1)] = \lambda - \lambda i\) | A1 |
| Equating real & imaginary parts: | |
| \(u\lambda + v(\lambda + 1) = \lambda\) (i) | |
| \(v\lambda - u(\lambda + 1) = -\lambda\) (ii) | M1 |
| From (i) \(\lambda = \frac{v}{1 - u - v}\) | From (ii) \(\lambda = \frac{u}{1 - u + v}\) |
| \(\frac{v}{1 - u - v} = \frac{u}{1 - u + v}\) | M1 |
| Reducing to the circle with equation \(u^2 + v^2 - u + v = 0\) * | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(z = \lambda - (\lambda + 1)i\); \(u + iv = \frac{\lambda - \lambda i}{\lambda - (\lambda + 1)i}\) | M1 | |
| \((u + iv)(\lambda - (\lambda + 1)i) = \lambda - \lambda i\) | M1 | |
| \(u\lambda + v(\lambda + 1) + i[v\lambda - u(\lambda + 1)] = \lambda - \lambda i\) | A1 | |
| Equating real & imaginary parts | ||
| \(u\lambda + v(\lambda + 1) = \lambda\) (i) | ||
| \(v\lambda - u(\lambda + 1) = -\lambda\) (ii) | M1 | |
| From (i) \(\lambda = \frac{v}{1 - u - v}\) | From (ii) \(\lambda = \frac{u}{1 - u + v}\) | |
| \(\frac{v}{1 - u - v} = \frac{u}{1 - u + v}\) | M1 | |
| Reducing to the circle with equation \(u^2 + v^2 - u + v = 0\) * | M1A1 | |
| (c) | ft their line | B1ft |
| B1 | ||
| Circle through origin, centre in correct quadrant | B13 | |
| Intersection correctly placed |
**(a)** Let $z = \lambda + 2i$; $w = \frac{\lambda + (\lambda + 1)i}{\lambda(1 + i)}$ | M1 |
$= \frac{\lambda + (\lambda + 1)i}{\lambda(1+i)} \cdot \frac{1-i}{1-i}$ | M1 |
$u + iv = \frac{(2\lambda + 1) + i}{2\lambda}$ | A1 |
$u = 1 + \frac{1}{2\lambda}$, $v = \frac{1}{2\lambda}$ | M1 |
Eliminating $\lambda$ gives a line with equation $v = u - 1$ or equivalent | A15 |
**(b)** Let $z = \lambda - (\lambda + 1)i$; $w = \frac{\lambda - \lambda i}{\lambda - (\lambda + 1)i}$ | M1 |
$= \frac{\lambda - \lambda i}{\lambda - (\lambda + 1)i} \times \frac{\lambda + (\lambda + 1)i}{\lambda + (\lambda + 1)i}$ | M1 |
$u + iv = \frac{\lambda(2\lambda + 1) + \lambda i}{2\lambda^2 + 2\lambda + 1}$ | A1 |
$u = \frac{\lambda(2\lambda + 1)}{2\lambda^2 + 2\lambda + 1}$, $v = \frac{\lambda}{2\lambda^2 + 2\lambda + 1}$ | M1 |
$\frac{u}{v} = 2\lambda + 1$ | |
$v = \frac{2\lambda}{4\lambda^2 + 4\lambda + 2} = \frac{(2\lambda + 1) - 1}{{(2\lambda + 1)}^2 + 1} = \frac{\frac{u}{v} - 1}{\left(\frac{u}{v}\right)^2 + 1}$ | M1 |
Reducing to the circle with equation $u^2 + v^2 - u + v = 0$ * | cso | M1A17 |
**Alternative 1:**
Let $z = \lambda - (\lambda + 1)i$; $u + iv = \frac{\lambda - \lambda i}{\lambda - (\lambda + 1)i}$ | M1 |
$(u + iv)(\lambda - (\lambda + 1)i) = \lambda - \lambda i$ | M1 |
$u\lambda + v(\lambda + 1) + i[v\lambda - u(\lambda + 1)] = \lambda - \lambda i$ | A1 |
Equating real & imaginary parts: | |
$u\lambda + v(\lambda + 1) = \lambda$ (i) | |
$v\lambda - u(\lambda + 1) = -\lambda$ (ii) | M1 |
From (i) $\lambda = \frac{v}{1 - u - v}$ | From (ii) $\lambda = \frac{u}{1 - u + v}$ | |
$\frac{v}{1 - u - v} = \frac{u}{1 - u + v}$ | M1 |
Reducing to the circle with equation $u^2 + v^2 - u + v = 0$ * | M1A1 |
**Alternative 2:**
Let $z = \lambda - (\lambda + 1)i$; $u + iv = \frac{\lambda - \lambda i}{\lambda - (\lambda + 1)i}$ | M1 |
$(u + iv)(\lambda - (\lambda + 1)i) = \lambda - \lambda i$ | M1 |
$u\lambda + v(\lambda + 1) + i[v\lambda - u(\lambda + 1)] = \lambda - \lambda i$ | A1 |
Equating real & imaginary parts | |
$u\lambda + v(\lambda + 1) = \lambda$ (i) | |
$v\lambda - u(\lambda + 1) = -\lambda$ (ii) | M1 |
From (i) $\lambda = \frac{v}{1 - u - v}$ | From (ii) $\lambda = \frac{u}{1 - u + v}$ | |
$\frac{v}{1 - u - v} = \frac{u}{1 - u + v}$ | M1 |
Reducing to the circle with equation $u^2 + v^2 - u + v = 0$ * | M1A1 |
**(c)** | ft their line | B1ft |
| | B1 |
Circle through origin, centre in correct quadrant | B13 |
Intersection correctly placed | |
**Total: [15]**\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}
$$w = \frac { z + \mathrm { i } } { \mathrm { z } } , \quad z \neq 0 .$$
(a) The transformation $T$ maps the points on the line with equation $y = x$ in the $z$-plane, other than $( 0,0 )$, to points on a line $l$ in the $w$-plane. Find a cartesian equation of $l$.\\
(b) Show that the image, under $T$, of the line with equation $x + y + 1 = 0$ in the $z$-plane is a circle $C$ in the $w$-plane, where $C$ has cartesian equation
$$u ^ { 2 } + v ^ { 2 } - u + v = 0$$
(c) On the same Argand diagram, sketch $l$ and $C$.
\hfill \mbox{\textit{Edexcel FP2 2007 Q12 [15]}}