| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| 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 FP2 complex transformation question requiring substitution z = x + iy, algebraic manipulation to find the image of a circle under a Möbius transformation, and sketching. While it involves multiple steps and careful algebra, it follows a well-established technique taught explicitly in Further Maths courses with no novel insight required. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.02m Geometrical effects: multiplication and division |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(w = \frac{3iz-2}{z+i} \Rightarrow z = \frac{2+wi}{3i-w}\) | M1 | 2.1 |
| \(\ | z+i\ | =1 \Rightarrow \left\ |
| \(\left\ | \frac{2+wi-3-iw}{3i-w}\right\ | =1 \Rightarrow \left\ |
| \(\ | w-3i\ | =1 \Rightarrow u^2+(v-3)^2=1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(w = \frac{3iz-2}{z+i} \Rightarrow w(z+i)=3i(z+i)+3-2\); attempts to isolate \(z+i\) terms | M1 | 2.1 |
| \((z+i)(w-3i)=1 \Rightarrow \ | (z+i)(w-3i)\ | =1 \Rightarrow \ |
| As main scheme | A1 | 1.1b |
| As main scheme | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(z = \frac{2+wi}{3i-w}\) as main scheme | M1 | 2.1 |
| \(x+yi = \frac{2-v+ui}{-u-(v-3)i} \times \frac{-u+(v-3)i}{-u+(v-3)i} = \cdots = \frac{u-(u^2+v^2-5v+6)i}{u^2+(v-3)^2}\) | ||
| \(\Rightarrow \left(\frac{u}{u^2+(v-3)^2}\right)^2 + \left(\frac{-(u^2+v^2-5v+6)}{u^2+(v-3)^2}+1\right)^2=1\); applies Cartesian coordinates, extracts \(x\) and \(y\) and applies \(x^2+(y+1)^2=1\) | M1 | 2.1 |
| \(\left(\frac{u}{u^2+(v-3)^2}\right)^2+\left(\frac{3-v}{u^2+(v-3)^2}\right)^2=1\); correct expression with \(y+1\) combined and simplified | A1 | 1.1b |
| \(\Rightarrow u^2+(v-3)^2=1\) | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(u+iv = \frac{3i(x+iy)-2}{x+iy+i} \times \frac{x-(y+1)i}{x-(y+1)i} = \frac{f(x,y)+g(x,y)i}{x^2+(y+1)^2}\); applies Cartesian coordinates and uses complex conjugate of denominator | M1 | 2.1 |
| \(x^2+(y+1)^2=1 \Rightarrow u+iv = x+(3x^2+3y^2+5y+2)i\); \(\Rightarrow u=x\) and \(v=3x^2+3(y+1)^2-y-1 = a+by\); uses \(x^2+(y+1)^2=1\) and extracts \(u\) and \(v\) as linear terms in \(x\) and \(y\) | M1 | 2.1 |
| \(u=x\) and \(v=2-y\); correct \(u\) and \(v\) | A1 | 1.1b |
| \(\Rightarrow u^2+(2-v+1)^2=1 \Rightarrow u^2+(v-3)^2=1\); uses \(x^2+(y+1)^2=1\) again to find correct equation | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Circle \(C\) correctly positioned, passing through origin, coordinates of centre labelled (accept as coordinates or marked on axes); centre at \(-1\) on real axis shown | B1 | 1.1b |
| Circle \(D\) correctly positioned with centre correctly labelled (accept as coordinates or marked on axes); centre at \(3i\) shown | B1ft | 1.1b |
# Question 7(a):
| Working | Mark | Guidance |
|---------|------|----------|
| $w = \frac{3iz-2}{z+i} \Rightarrow z = \frac{2+wi}{3i-w}$ | M1 | 2.1 |
| $\|z+i\|=1 \Rightarrow \left\|\frac{2+wi}{3i-w}+i\right\|=1$ | M1 | 2.1 |
| $\left\|\frac{2+wi-3-iw}{3i-w}\right\|=1 \Rightarrow \left\|\frac{1}{w-3i}\right\|=1$ | A1 | 1.1b |
| $\|w-3i\|=1 \Rightarrow u^2+(v-3)^2=1$ | A1 | 2.2a |
**ALT 1:**
| Working | Mark | Guidance |
|---------|------|----------|
| $w = \frac{3iz-2}{z+i} \Rightarrow w(z+i)=3i(z+i)+3-2$; attempts to isolate $z+i$ terms | M1 | 2.1 |
| $(z+i)(w-3i)=1 \Rightarrow \|(z+i)(w-3i)\|=1 \Rightarrow \|w-3i\|=1$; gathers $z+i$ terms and applies $\|z+i\|=1$ | M1 | 2.1 |
| As main scheme | A1 | 1.1b |
| As main scheme | A1 | 2.2a |
**ALT 2:**
| Working | Mark | Guidance |
|---------|------|----------|
| $z = \frac{2+wi}{3i-w}$ as main scheme | M1 | 2.1 |
| $x+yi = \frac{2-v+ui}{-u-(v-3)i} \times \frac{-u+(v-3)i}{-u+(v-3)i} = \cdots = \frac{u-(u^2+v^2-5v+6)i}{u^2+(v-3)^2}$ | | |
| $\Rightarrow \left(\frac{u}{u^2+(v-3)^2}\right)^2 + \left(\frac{-(u^2+v^2-5v+6)}{u^2+(v-3)^2}+1\right)^2=1$; applies Cartesian coordinates, extracts $x$ and $y$ and applies $x^2+(y+1)^2=1$ | M1 | 2.1 |
| $\left(\frac{u}{u^2+(v-3)^2}\right)^2+\left(\frac{3-v}{u^2+(v-3)^2}\right)^2=1$; correct expression with $y+1$ combined and simplified | A1 | 1.1b |
| $\Rightarrow u^2+(v-3)^2=1$ | A1 | 2.2a |
**ALT 3:**
| Working | Mark | Guidance |
|---------|------|----------|
| $u+iv = \frac{3i(x+iy)-2}{x+iy+i} \times \frac{x-(y+1)i}{x-(y+1)i} = \frac{f(x,y)+g(x,y)i}{x^2+(y+1)^2}$; applies Cartesian coordinates and uses complex conjugate of denominator | M1 | 2.1 |
| $x^2+(y+1)^2=1 \Rightarrow u+iv = x+(3x^2+3y^2+5y+2)i$; $\Rightarrow u=x$ and $v=3x^2+3(y+1)^2-y-1 = a+by$; uses $x^2+(y+1)^2=1$ and extracts $u$ and $v$ as linear terms in $x$ and $y$ | M1 | 2.1 |
| $u=x$ and $v=2-y$; correct $u$ and $v$ | A1 | 1.1b |
| $\Rightarrow u^2+(2-v+1)^2=1 \Rightarrow u^2+(v-3)^2=1$; uses $x^2+(y+1)^2=1$ again to find correct equation | A1 | 2.2a |
---
# Question 7(b):
| Working | Mark | Guidance |
|---------|------|----------|
| Circle $C$ correctly positioned, passing through origin, coordinates of centre labelled (accept as coordinates or marked on axes); centre at $-1$ on real axis shown | B1 | 1.1b |
| Circle $D$ correctly positioned with centre correctly labelled (accept as coordinates or marked on axes); centre at $3i$ shown | B1ft | 1.1b |
**Notes:**
- B1: Circle $C$ correctly positioned, passing through origin, centre labelled. Accept coordinates or marked on axes.
- B1ft: Their $D$ correctly positioned with centre correctly labelled.
- Accept both drawn on same diagram.
- Allow S.C. B1B0 for two circles in correct respective positions but with no labelling.\begin{enumerate}
\item A transformation from the $z$-plane to the $w$-plane is given by
\end{enumerate}
$$w = \frac { 3 \mathrm { i } z - 2 } { z + \mathrm { i } } \quad z \neq - \mathrm { i }$$
(a) Show that the circle $C$ with equation $| z + \mathrm { i } | = 1$ in the $z$-plane is mapped to a circle $D$ in the $w$-plane, giving a Cartesian equation for $D$.\\
(b) Sketch $C$ and $D$ on Argand diagrams.\\
\hfill \mbox{\textit{Edexcel FP2 2019 Q7 [6]}}