| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| 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 algebraic manipulation to find the image circle's equation, followed by a routine interior/exterior check. While it involves Further Maths content and multiple steps, the technique is well-practiced in FP2 and follows a predictable method without requiring novel geometric insight. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02k Argand diagrams: geometric interpretation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(w(z+i) = z \Rightarrow wz + iw = z \Rightarrow iw = z - wz \Rightarrow iw = z(1-w) \Rightarrow z = \frac{iw}{1-w}\) | M1 | Complete method of rearranging to make \(z\) the subject |
| \(z = \frac{iw}{1-w}\) | A1 aef | |
| \( | z | = 3 \Rightarrow \left |
| \( | iw | = 3 |
| \(\Rightarrow u^2 + v^2 = 9[(u-1)^2 + v^2]\) | A1 | Correct equation |
| \(\Rightarrow 0 = 8u^2 - 18u + 8v^2 + 9 \Rightarrow 0 = u^2 - \frac{9}{4}u + v^2 + \frac{9}{8}\) | dddM1 | Simplifies down to \(u^2 + v^2 \pm \alpha u \pm \beta v \pm \delta = 0\) |
| \(\Rightarrow \left(u - \frac{9}{8}\right)^2 + v^2 = \frac{9}{64}\) | ||
| Circle, centre \(\left(\frac{9}{8}, 0\right)\), radius \(\frac{3}{8}\) | A1, A1 | One of centre or radius correct; Both centre and radius correct |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Circle indicated on Argand diagram in correct position in correct quadrants | B1ft | Ignore plotted coordinates |
| Region outside circle indicated only | B1 |
## Question 6:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $w(z+i) = z \Rightarrow wz + iw = z \Rightarrow iw = z - wz \Rightarrow iw = z(1-w) \Rightarrow z = \frac{iw}{1-w}$ | M1 | Complete method of rearranging to make $z$ the subject |
| $z = \frac{iw}{1-w}$ | A1 **aef** | |
| $|z| = 3 \Rightarrow \left|\frac{iw}{1-w}\right| = 3$ | dM1 | Putting $|z|$ in terms of their $w$, $|w|=3$ |
| $|iw| = 3|1-w| \Rightarrow |w| = 3|w-1| \Rightarrow |w|^2 = 9|w-1|^2 \Rightarrow |u+iv|^2 = 9|u+iv-1|^2$ | ddM1 | Applies $w = u+iv$, uses Pythagoras correctly to get equation in terms of $u$ and $v$ without any $i$'s |
| $\Rightarrow u^2 + v^2 = 9[(u-1)^2 + v^2]$ | A1 | Correct equation |
| $\Rightarrow 0 = 8u^2 - 18u + 8v^2 + 9 \Rightarrow 0 = u^2 - \frac{9}{4}u + v^2 + \frac{9}{8}$ | dddM1 | Simplifies down to $u^2 + v^2 \pm \alpha u \pm \beta v \pm \delta = 0$ |
| $\Rightarrow \left(u - \frac{9}{8}\right)^2 + v^2 = \frac{9}{64}$ | | |
| Circle, centre $\left(\frac{9}{8}, 0\right)$, radius $\frac{3}{8}$ | A1, A1 | One of centre or radius correct; Both centre and radius correct |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| Circle indicated on Argand diagram in correct position in correct quadrants | B1ft | Ignore plotted coordinates |
| Region outside circle indicated only | B1 | |
---\begin{enumerate}
\item A transformation $T$ from the $z$-plane to the $w$-plane is given by
\end{enumerate}
$$w = \frac { z } { z + \mathrm { i } } , \quad z \neq - \mathrm { i }$$
The circle with equation $| z | = 3$ is mapped by $T$ onto the curve $C$.\\
(a) Show that $C$ is a circle and find its centre and radius.
The region $| z | < 3$ in the $z$-plane is mapped by $T$ onto the region $R$ in the $w$-plane.\\
(b) Shade the region $R$ on an Argand diagram.\\
\hfill \mbox{\textit{Edexcel FP2 2009 Q6 [10]}}