| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Complex transformations and mappings |
| Difficulty | Challenging +1.2 This is a multi-part FP2 question on complex loci and transformations. Part (a) requires recognizing that arg((z-a)/(z-b))=π/2 gives a semicircular arc, which is standard FP2 knowledge. Part (b) follows directly from geometric observation. Part (c) involves showing a transformation maps a circle to a line, requiring algebraic manipulation but following standard techniques. While it requires multiple concepts and careful execution, these are well-practiced FP2 topics without requiring novel insight, placing it moderately above average difficulty. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines |
| Answer | Marks | Guidance |
|---|---|---|
| Relating lines and angle (generous) | M1 | |
| [angle between \(\pm 2i\) to \(P\) and \(\pm 2\) to \(P\)] | A1 | |
| Angle between correct lines is \(\frac{\pi}{2}\) | M1, A1 | 4 |
| Answer | Marks |
|---|---|
| Selecting correct ("top half") semi-circle |
| Answer | Marks |
|---|---|
| Method for finding Cartesian equation | M1 |
| Correct equation, any form, \(\Rightarrow x(x + 2) + y(y - 2) = 0\) | A1 |
| Sketch: showing circle | M1 |
| Correct circle [centre \((-1, 1)\)], choosing only "top half" | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \( | z + 1 - i | \) is radius; \(= \sqrt{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(z = \frac{2(1 + i) - 2\omega}{\omega} \left(= \frac{2(1 + i)}{\omega} - 2\right)\) | M1 | |
| \(\frac{z - 2i}{z + 2} = \frac{2(1 + i) - 2(1 + i)\omega}{2(1 + i)} (= -\omega)\) | M1, A1 | |
| \(\text{Arg}(1 - \omega) = \frac{\pi}{2}\) is line segment, passing through \((1,0)\) | A1, A1 | |
| [Diagram showing horizontal line from 0 to 1 on u-axis with vertical arrow pointing down] | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(u + iv = \frac{2 + 2i}{(x + 2) + iy} = \frac{(2x + 2y + 4) + i(x + 2 - y)}{(x + 2)^2 + y^2}\) | M1 | |
| \(x = -1 + \sqrt{2}\cos\theta, y = 1 + \sqrt{2}\sin\theta\) | M1 | |
| \(\Rightarrow w = \frac{(2\sqrt{2}\cos\theta + 2\sqrt{2}\sin\theta + 4) + i(...)}{(2\sqrt{2}\cos\theta + 2\sqrt{2}\sin\theta + 4)} = | = 1 + i f(\theta) | \) |
| \(\Rightarrow\) part of line \(u = 1\), show lower "half" of line | A1, A1 |
**Part (a)**
| Relating lines and angle (generous) | M1 | |
| [angle between $\pm 2i$ to $P$ and $\pm 2$ to $P$] | A1 | |
| Angle between correct lines is $\frac{\pi}{2}$ | M1, A1 | 4 |
**Circle**
| Selecting correct ("top half") semi-circle | | |
**[If algebraic approach:]**
| Method for finding Cartesian equation | M1 | |
| Correct equation, any form, $\Rightarrow x(x + 2) + y(y - 2) = 0$ | A1 | |
| Sketch: showing circle | M1 | |
| Correct circle [centre $(-1, 1)$], choosing only "top half" | A1 | |
**Part (b)**
| $|z + 1 - i|$ is radius; $= \sqrt{2}$ | M1, A1 | 2 |
**Part (c)**
| $z = \frac{2(1 + i) - 2\omega}{\omega} \left(= \frac{2(1 + i)}{\omega} - 2\right)$ | M1 | |
| $\frac{z - 2i}{z + 2} = \frac{2(1 + i) - 2(1 + i)\omega}{2(1 + i)} (= -\omega)$ | M1, A1 | |
| $\text{Arg}(1 - \omega) = \frac{\pi}{2}$ is line segment, passing through $(1,0)$ | A1, A1 | |
| [Diagram showing horizontal line from 0 to 1 on u-axis with vertical arrow pointing down] | | A1 | 6 |
[12]
**Alt (c):**
| $u + iv = \frac{2 + 2i}{(x + 2) + iy} = \frac{(2x + 2y + 4) + i(x + 2 - y)}{(x + 2)^2 + y^2}$ | M1 | |
| $x = -1 + \sqrt{2}\cos\theta, y = 1 + \sqrt{2}\sin\theta$ | M1 | |
| $\Rightarrow w = \frac{(2\sqrt{2}\cos\theta + 2\sqrt{2}\sin\theta + 4) + i(...)}{(2\sqrt{2}\cos\theta + 2\sqrt{2}\sin\theta + 4)} = |= 1 + i f(\theta)|$ | A1 | |
| $\Rightarrow$ part of line $u = 1$, show lower "half" of line | A1, A1 | |8. In the Argand diagram the point $P$ represents the complex number $z$.
Given that arg $\left( \frac { z - 2 \mathrm { i } } { z + 2 } \right) = \frac { \pi } { 2 }$,
\begin{enumerate}[label=(\alph*)]
\item sketch the locus of $P$,
\item deduce the value of $| \mathrm { z } + 1 - \mathrm { i } |$.
The transformation $T$ from the $z$-plane to the $w$-plane is defined by
$$w = \frac { 2 ( 1 + \mathrm { i } ) } { z + 2 } , \quad z \neq - 2$$
\item Show that the locus of $P$ in the $z$-plane is mapped to part of a straight line in the $w$-plane, and show this in an Argand diagram.\\
(6)(Total 12 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2006 Q8 [12]}}