| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| 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 structured Möbius transformation question with clear guidance through parts (a) and (b). While it requires algebraic manipulation with complex numbers and substitution of y=x, the steps are methodical and the 'show that' format provides target expressions to work towards, reducing problem-solving demands compared to open-ended circle mapping questions. |
| Spec | 4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02k Argand diagrams: geometric interpretation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(w = \frac{4(1-i)(x+xi)-8i}{2(i-1)(x+xi)-i}\) | M1 | Substitutes for \(z\) |
| \(w = \frac{4(x+xi-xi+x)-8i}{2(xi-x-x-xi)-i}\) | M1A1 | M1: Attempt to expand numerator and denominator; A1: Correct expression |
| \(\frac{8i-8x}{4x+i} \cdot \frac{4x-i}{4x-i}\) | M1A1 | M1: Multiplies numerator and denominator by conjugate of their denom. No expansion needed; A1: Uses correct conjugate (not ft) |
| \(= \frac{-32x^2+40xi+8}{16x^2+1}\) | B1 | Award only if final answer is correct and follows correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(w = \frac{(1-i)z-8i}{2(-1+i)-i} = \frac{4(1-i)(x+iy)-8i}{2(-1+i)(x+iy)-i}\) | M1 | Substitutes for \(z\) |
| \(= \frac{4(1-i)x+4(1-i)iy-8i}{2(-1+i)x+2(-1+i)iy-i}\) | M1 | Attempt to expand numerator and denominator |
| \(= \frac{4x+4y+(4y-4x-8)i}{-2x-2y+(2x-2y-1)i}\) | A1 | Correct expression |
| \(= \frac{4x+4y+(4y-4x-8)i}{-2x-2y+(2x-2y-1)i} \times \frac{-2x-2y-(2x-2y-1)i}{-2x-2y-(2x-2y-1)i}\) | M1A1 | M1: Multiplies by conjugate of denom, no expansion needed; A1: Uses correct conjugate (not ft) |
| \(= \frac{-16x^2-16y^2+12y-12x+8+(20x+20y)i}{8x^2+8y^2-4x+4y+1}\) | — | — |
| \(= \frac{-32x^2+40xi+8}{16x^2+1}\) | B1 | cso; correct answer using \(y=x\); award only if final answer is correct and follows correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(u = \frac{-32x^2+8}{16x^2+1}\), \(v = \frac{40x}{16x^2+1}\) | M1 | Identifies \(u\) and \(v\) (real and imaginary parts); may be implied by working; may be in terms of \(x\) and \(y\) |
| \(\left(\frac{8-32x^2}{16x^2+1}-3\right)^2 + \left(\frac{40x}{16x^2+1}\right)^2\) | dM1 | Substitutes for their \(u\) and \(v\) in given equation; may be in terms of \(x\) and \(y\) |
| \(= \left(\frac{8-32x^2-48x^2-3}{16x^2+1}\right)^2 + \left(\frac{40x}{16x^2+1}\right)^2\) | — | — |
| \(= \frac{(5-80x^2)^2}{(16x^2+1)^2} + \frac{1600x^2}{(16x^2+1)^2}\) | — | — |
| \(= \frac{6400x^4+800x^2+25}{(16x^2+1)^2}\) | A1 | Combines to form a single correct fraction |
| \(= \frac{25(16x^2+1)^2}{(16x^2+1)^2} = 25\) | A1 | \(k=5\) or \(k^2=25\) may (but need not) be seen explicitly |
## Question 6(a):
**Method 1 (Substituting $z = x + xi$ at the start):**
| Working | Mark | Guidance |
|---------|------|----------|
| $w = \frac{4(1-i)(x+xi)-8i}{2(i-1)(x+xi)-i}$ | M1 | Substitutes for $z$ |
| $w = \frac{4(x+xi-xi+x)-8i}{2(xi-x-x-xi)-i}$ | M1A1 | M1: Attempt to expand numerator and denominator; A1: Correct expression |
| $\frac{8i-8x}{4x+i} \cdot \frac{4x-i}{4x-i}$ | M1A1 | M1: Multiplies numerator and denominator by conjugate of their denom. No expansion needed; A1: Uses correct conjugate (not ft) |
| $= \frac{-32x^2+40xi+8}{16x^2+1}$ | B1 | Award only if final answer is correct and follows correct working |
**Method 2 (without $y = x$ substitution):**
| Working | Mark | Guidance |
|---------|------|----------|
| $w = \frac{(1-i)z-8i}{2(-1+i)-i} = \frac{4(1-i)(x+iy)-8i}{2(-1+i)(x+iy)-i}$ | M1 | Substitutes for $z$ |
| $= \frac{4(1-i)x+4(1-i)iy-8i}{2(-1+i)x+2(-1+i)iy-i}$ | M1 | Attempt to expand numerator and denominator |
| $= \frac{4x+4y+(4y-4x-8)i}{-2x-2y+(2x-2y-1)i}$ | A1 | Correct expression |
| $= \frac{4x+4y+(4y-4x-8)i}{-2x-2y+(2x-2y-1)i} \times \frac{-2x-2y-(2x-2y-1)i}{-2x-2y-(2x-2y-1)i}$ | M1A1 | M1: Multiplies by conjugate of denom, no expansion needed; A1: Uses correct conjugate (not ft) |
| $= \frac{-16x^2-16y^2+12y-12x+8+(20x+20y)i}{8x^2+8y^2-4x+4y+1}$ | — | — |
| $= \frac{-32x^2+40xi+8}{16x^2+1}$ | B1 | cso; correct answer using $y=x$; award only if final answer is correct and follows correct working |
---
## Question 6(b):
| Working | Mark | Guidance |
|---------|------|----------|
| $u = \frac{-32x^2+8}{16x^2+1}$, $v = \frac{40x}{16x^2+1}$ | M1 | Identifies $u$ and $v$ (real and imaginary parts); may be implied by working; may be in terms of $x$ and $y$ |
| $\left(\frac{8-32x^2}{16x^2+1}-3\right)^2 + \left(\frac{40x}{16x^2+1}\right)^2$ | dM1 | Substitutes for their $u$ and $v$ in given equation; may be in terms of $x$ and $y$ |
| $= \left(\frac{8-32x^2-48x^2-3}{16x^2+1}\right)^2 + \left(\frac{40x}{16x^2+1}\right)^2$ | — | — |
| $= \frac{(5-80x^2)^2}{(16x^2+1)^2} + \frac{1600x^2}{(16x^2+1)^2}$ | — | — |
| $= \frac{6400x^4+800x^2+25}{(16x^2+1)^2}$ | A1 | Combines to form a single correct fraction |
| $= \frac{25(16x^2+1)^2}{(16x^2+1)^2} = 25$ | A1 | $k=5$ or $k^2=25$ may (but need not) be seen explicitly |
**Total: 10 marks**
---6. 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
$$w = \frac { 4 ( 1 - \mathrm { i } ) z - 8 \mathrm { i } } { 2 ( - 1 + \mathrm { i } ) z - \mathrm { i } } , \quad z \neq \frac { 1 } { 4 } - \frac { 1 } { 4 } \mathrm { i }$$
The transformation $T$ maps the points on the line $l$ with equation $y = x$ in the $z$-plane to a circle $C$ in the $w$-plane.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$w = \frac { a x ^ { 2 } + b x i + c } { 16 x ^ { 2 } + 1 }$$
where $a$, $b$ and $c$ are real constants to be found.
\item Hence show that the circle $C$ has equation
$$( u - 3 ) ^ { 2 } + v ^ { 2 } = k ^ { 2 }$$
where $k$ is a constant to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2014 Q6 [10]}}