| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| 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 students to show a circle maps to a circle and find its properties. While it involves multiple steps (substituting z = x+iy, algebraic manipulation, completing the square), these are well-practiced techniques in FP2. The shading part requires understanding of image regions but follows directly from part (a). More routine than the average Further Maths question but still requires solid algebraic facility. |
| Spec | 4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02k Argand diagrams: geometric interpretation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(w(z+3i) = z\), rearrange to \(z = \frac{3iw}{1-w}\) or \(\frac{-3iw}{w-1}\) | M1A1 | Re-arrange to \(z = \ldots\); A1 for correct result |
| \( | z | = 2 \Rightarrow \left |
| \(w = u+iv\): \(9(u^2+v^2) = 4((1-u)^2+v^2)\) | ddM1A1 | Dep on both M marks; use \(w=u+iv\), find moduli; moduli contain no \(i\) |
| \(9u^2+9v^2 = 4(1-2u+u^2+v^2)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(5u^2+5v^2+8u-4=0\) | dddM1 | Dep on all previous M marks; re-arrange to circle form with same coeffs for squared terms |
| \(\left(u+\frac{4}{5}\right)^2 + v^2 = \frac{36}{25}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Circle, Centre \(\left(-\frac{4}{5}, 0\right)\), Radius \(\frac{6}{5}\) | A1A1 (8) | Deduct 1 for each error or omission |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Circle drawn on Argand diagram in correct position for their centre and radius | B1ft | No numbers needed but circle must be in correct region/on correct axis; consistent with centre and radius |
| Region inside correct circle shaded | B1 (2)[10] | No ft; region inside correct circle shaded |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Let \(z = x+iy\); rationalise: \(w = \frac{(x+iy)(x-i(y+3))}{(x+i(y+3))(x-i(y+3))}\) | M1 | Must use conjugate of denominator |
| \(= \frac{x^2+y^2+3y-3ix}{x^2+y^2+6y+9}\) | A1 | Expand brackets, obtain correct numerator and denominator |
| Using \( | z | =2 \Rightarrow x^2+y^2=4\): \(\frac{3y+4-3ix}{6y+13}\) |
| \(w=u+iv\): \(u=\frac{3y+4}{6y+13}\), \(v=\frac{-3x}{6y+13}\) | ddM1 A1 | Equating real and imaginary parts; A1 for correct expressions |
| \(u^2+v^2 = \frac{9y^2+24y+16+9x^2}{(6y+13)^2} = \frac{24y+52}{(6y+13)^2} = \frac{4}{6y+13} = \frac{8}{5}\left(\frac{1}{2}-u\right)\) | dddM1 | Uses \(u^2+v^2=\ldots\) to eliminate \(x\) and \(y\); obtain equation of circle |
| \(\therefore 5u^2+5v^2+8u=4\); then as main scheme: Circle, centre, radius | A1A1 (8) | As main scheme |
## Question 5:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $w(z+3i) = z$, rearrange to $z = \frac{3iw}{1-w}$ or $\frac{-3iw}{w-1}$ | M1A1 | Re-arrange to $z = \ldots$; A1 for correct result |
| $|z| = 2 \Rightarrow \left|\frac{3iw}{1-w}\right| = 2$, so $|3iw| = 2|1-w|$ | dM1 | Dep on first M1, using $|z|=2$ with previous result |
| $w = u+iv$: $9(u^2+v^2) = 4((1-u)^2+v^2)$ | ddM1A1 | Dep on both M marks; use $w=u+iv$, find moduli; moduli contain no $i$ |
| $9u^2+9v^2 = 4(1-2u+u^2+v^2)$ | | |
### Part (a)(i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $5u^2+5v^2+8u-4=0$ | dddM1 | Dep on all previous M marks; re-arrange to circle form with same coeffs for squared terms |
| $\left(u+\frac{4}{5}\right)^2 + v^2 = \frac{36}{25}$ | | |
### Part (a)(ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| Circle, Centre $\left(-\frac{4}{5}, 0\right)$, Radius $\frac{6}{5}$ | A1A1 (8) | Deduct 1 for each error or omission |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| Circle drawn on Argand diagram in correct position for their centre and radius | B1ft | No numbers needed but circle must be in correct region/on correct axis; consistent with centre and radius |
| Region inside correct circle shaded | B1 (2)[10] | No ft; region inside **correct** circle shaded |
---
### Alternative for 5(a):
| Working | Marks | Guidance |
|---------|-------|----------|
| Let $z = x+iy$; rationalise: $w = \frac{(x+iy)(x-i(y+3))}{(x+i(y+3))(x-i(y+3))}$ | M1 | Must use conjugate of denominator |
| $= \frac{x^2+y^2+3y-3ix}{x^2+y^2+6y+9}$ | A1 | Expand brackets, obtain correct numerator and denominator |
| Using $|z|=2 \Rightarrow x^2+y^2=4$: $\frac{3y+4-3ix}{6y+13}$ | dM1 | Use $x^2+y^2=4$ to remove squares |
| $w=u+iv$: $u=\frac{3y+4}{6y+13}$, $v=\frac{-3x}{6y+13}$ | ddM1 A1 | Equating real and imaginary parts; A1 for correct expressions |
| $u^2+v^2 = \frac{9y^2+24y+16+9x^2}{(6y+13)^2} = \frac{24y+52}{(6y+13)^2} = \frac{4}{6y+13} = \frac{8}{5}\left(\frac{1}{2}-u\right)$ | dddM1 | Uses $u^2+v^2=\ldots$ to eliminate $x$ and $y$; obtain equation of circle |
| $\therefore 5u^2+5v^2+8u=4$; then as main scheme: Circle, centre, radius | A1A1 (8) | As main scheme |
---\begin{enumerate}
\item A transformation $T$ from the $z$-plane to the $w$-plane is given by
\end{enumerate}
$$w = \frac { z } { z + 3 \mathrm { i } } , \quad z \neq - 3 \mathrm { i }$$
The circle with equation $| z | = 2$ is mapped by $T$ onto the curve $C$.\\
(a) (i) Show that $C$ is a circle.\\
(ii) Find the centre and radius of $C$.
The region $| z | \leqslant 2$ 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 2015 Q5 [10]}}