| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| 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 line maps to a circle and sketch the result. While it involves Further Maths content (Möbius transformations), the technique is routine: substitute z = x + iy with y = 2x, manipulate to find |w - a| = r form, and test a point for the region. The algebraic manipulation is moderately involved but follows a well-practiced method with no novel insight required. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02k Argand diagrams: geometric interpretation4.02l Geometrical effects: conjugate, addition, subtraction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(w = \frac{z-1}{z+1} \Rightarrow wz + w = z - 1 \Rightarrow z = \ldots\) | M1 | Attempt to make \(z\) the subject |
| \(z = \frac{w+1}{1-w}\) | A1 | Correct expression in terms of \(w\) |
| \(= \frac{u+iv+1}{1-u-iv} \times \frac{1-u+iv}{1-u+iv}\) | M1 | Introduces \(u+iv\) and multiplies top and bottom by the complex conjugate of the bottom |
| \(x = \frac{-u^2 - v^2 + 1}{\ldots},\quad y = \frac{2v}{\ldots}\) | ||
| \(y = 2x \Rightarrow 2v = -2u^2 - 2v^2 + 2\) | M1 | Uses real and imaginary parts and \(y=2x\) to obtain equation connecting \(u\) and \(v\). Can have the 2 on the wrong side |
| \(u^2 + \left(v + \frac{1}{2}\right)^2 - \frac{1}{4} = 1\) | M1 | Processes their equation to a form recognisable as a circle, i.e. coefficients of \(u^2\) and \(v^2\) are the same and no \(uv\) terms |
| Centre \(\left(0, -\frac{1}{2}\right)\), radius \(\frac{\sqrt{5}}{2}\) | A1, A1 | A1: Correct centre (allow \(-\frac{1}{2}i\)). A1: Correct radius |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(w = \frac{x+iy-1}{x+iy+1} = \frac{(x-1)+2xi}{(x+1)+2xi} \times \frac{(x+1)-2xi}{(x+1)-2xi}\) | M1 | Rationalise the denominator, may have \(2x\) or \(y\) |
| \(= \frac{(x^2-1)+4x^2 + 2xi(x+1-(x-1))}{(x+1)^2+4x^2}\) | A1 | Correct result in terms of \(x\) only. Must have rational denominator shown, but no other simplification needed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| [Circle sketch with shading in region R] | B1ft | B1ft: Their circle correctly positioned provided their equation does give a circle |
| [Completely correct sketch and shading] | B1 | B1: Completely correct sketch and shading |
## Question 4:
$$w = \frac{z-1}{z+1}$$
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $w = \frac{z-1}{z+1} \Rightarrow wz + w = z - 1 \Rightarrow z = \ldots$ | M1 | Attempt to make $z$ the subject |
| $z = \frac{w+1}{1-w}$ | A1 | Correct expression in terms of $w$ |
| $= \frac{u+iv+1}{1-u-iv} \times \frac{1-u+iv}{1-u+iv}$ | M1 | Introduces $u+iv$ and multiplies top and bottom by the complex conjugate of the bottom |
| $x = \frac{-u^2 - v^2 + 1}{\ldots},\quad y = \frac{2v}{\ldots}$ | | |
| $y = 2x \Rightarrow 2v = -2u^2 - 2v^2 + 2$ | M1 | Uses real and imaginary parts and $y=2x$ to obtain equation connecting $u$ and $v$. Can have the 2 on the wrong side |
| $u^2 + \left(v + \frac{1}{2}\right)^2 - \frac{1}{4} = 1$ | M1 | Processes their equation to a form recognisable as a circle, i.e. coefficients of $u^2$ and $v^2$ are the same and no $uv$ terms |
| Centre $\left(0, -\frac{1}{2}\right)$, radius $\frac{\sqrt{5}}{2}$ | A1, A1 | A1: Correct centre (allow $-\frac{1}{2}i$). A1: Correct radius |
**Special Case:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $w = \frac{x+iy-1}{x+iy+1} = \frac{(x-1)+2xi}{(x+1)+2xi} \times \frac{(x+1)-2xi}{(x+1)-2xi}$ | M1 | Rationalise the denominator, may have $2x$ or $y$ |
| $= \frac{(x^2-1)+4x^2 + 2xi(x+1-(x-1))}{(x+1)^2+4x^2}$ | A1 | Correct result in terms of $x$ only. Must have rational denominator shown, but no other simplification needed |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| [Circle sketch with shading in region R] | B1ft | B1ft: Their circle correctly positioned provided their equation does give a circle |
| [Completely correct sketch and shading] | B1 | B1: Completely correct sketch and shading |\begin{enumerate}
\item A transformation $T$ from the $z$-plane to the $w$-plane is given by
\end{enumerate}
$$w = \frac { z - 1 } { z + 1 } , \quad z \neq - 1$$
The line in the $z$-plane with equation $y = 2 x$ is mapped by $T$ onto the curve $C$ in the $w$-plane.\\
(a) Show that $C$ is a circle and find its centre and radius.
The region $y < 2 x$ in the $z$-plane is mapped by $T$ onto the region $R$ in the $w$-plane.\\
(b) Sketch circle $C$ on an Argand diagram and shade and label region $R$.
\hfill \mbox{\textit{Edexcel F2 2015 Q4 [9]}}