| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Complex transformations and mappings |
| Difficulty | Challenging +1.2 Part (a) is a standard perpendicular bisector locus requiring basic geometric interpretation. Part (b) involves a Möbius transformation mapping, which is a Further Maths topic, but the algebraic manipulation follows a well-established method (rearranging w = iz/(z-2i) to find z in terms of w, substituting into the locus equation, and simplifying). This is a typical F2 examination question requiring multiple steps but following standard techniques without requiring novel insight. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.02l Geometrical effects: conjugate, addition, subtraction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| A straight line through the origin, not vertical or horizontal | M1 | May be solid or dotted. Clear "V" shapes score M0 |
| Line with positive gradient in quadrants 1, 3 and 4 | A1 | Ignore intercepts correct or incorrect. If there are other lines that are clearly "construction" lines they can be ignored. If there are clearly several lines score A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(w = \frac{iz}{z-2i} \Rightarrow z = \frac{2wi}{w-i}\) | M1 | Attempts to make \(z\) the subject. Must obtain form \(\frac{awi}{bw+ci}\), \(a,b,c\) real and non-zero |
| \(z = \frac{2(u+iv)i}{u+iv-i}\), multiplied by \(\frac{u-(v-1)i}{u-(v-1)i}\) | dM1 | Introduces \(w=u+iv\) and attempts to multiply by complex conjugate of denominator |
| \(z = \frac{-2u}{u^2+(v-1)^2} + \frac{2u^2+2v(v-1)}{u^2+(v-1)^2}i\) | A1 | Correct expression for \(z\) with real and imaginary parts identified |
| \(\frac{2u^2+2v(v-1)}{u^2+(v-1)^2}-1 = \frac{3}{2}\left(\frac{-2u}{u^2+(v-1)^2}-\frac{3}{2}\right)\) | ddM1 | Attempts Cartesian equation of locus of \(z\), substitutes for \(x\) and \(y\); must be linear with non-zero constant term |
| \(13u^2+13v^2+12u-18v+5=0 \Rightarrow \left(u+\frac{6}{13}\right)^2+\left(v-\frac{9}{13}\right)^2=\frac{4}{13}\) | dddM1 | Attempts to complete the square where \(u^2\) and \(v^2\) have same coefficient |
| \(\left\ | w-\left(-\frac{6}{13}+\frac{9}{13}i\right)\right\ | =\frac{2}{\sqrt{13}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(w = \frac{iz}{z-2i} \Rightarrow z = \frac{2wi}{w-i}\) | M1 | Must obtain form \(\frac{awi}{bw+ci}\) |
| \(\left\ | \frac{2wi}{w-i}-2i\right\ | = \left\ |
| \(\left\ | \frac{-2}{w-i}\right\ | = \left\ |
| \( | 2(u+iv)i - 3(u+iv)+3i | = 2 \Rightarrow (3u+2v)^2+(3v-2u-3)^2=4\) |
| \(\left(u+\frac{6}{13}\right)^2+\left(v-\frac{9}{13}\right)^2=\frac{4}{13}\) | dddM1 | Attempts to complete the square |
| \(\left\ | w-\left(-\frac{6}{13}+\frac{9}{13}i\right)\right\ | =\frac{2}{\sqrt{13}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(w = \frac{iz}{z-2i} \Rightarrow z = \frac{2wi}{w-i}\) | M1 | Must obtain form \(\frac{awi}{bw+ci}\) |
| \(\left\ | \frac{2wi-2wi-2}{w-i}\right\ | = \left\ |
| \( | -2 | = |
| \( | w(2i-3)+3i | = |
| \(\sqrt{13}\left\ | w-\left(-\frac{6}{13}+\frac{9}{13}i\right)\right\ | =2 \Rightarrow \left\ |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| A straight line through the origin, not vertical or horizontal | M1 | May be solid or dotted. Clear "V" shapes score M0 |
| Line with positive gradient in quadrants 1, 3 and 4 | A1 | Ignore intercepts correct or incorrect. If there are other lines that are clearly "construction" lines they can be ignored. If there are clearly several lines score A0 |
### Part (b) — Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $w = \frac{iz}{z-2i} \Rightarrow z = \frac{2wi}{w-i}$ | M1 | Attempts to make $z$ the subject. Must obtain form $\frac{awi}{bw+ci}$, $a,b,c$ real and non-zero |
| $z = \frac{2(u+iv)i}{u+iv-i}$, multiplied by $\frac{u-(v-1)i}{u-(v-1)i}$ | dM1 | Introduces $w=u+iv$ and attempts to multiply by complex conjugate of denominator |
| $z = \frac{-2u}{u^2+(v-1)^2} + \frac{2u^2+2v(v-1)}{u^2+(v-1)^2}i$ | A1 | Correct expression for $z$ with real and imaginary parts identified |
| $\frac{2u^2+2v(v-1)}{u^2+(v-1)^2}-1 = \frac{3}{2}\left(\frac{-2u}{u^2+(v-1)^2}-\frac{3}{2}\right)$ | ddM1 | Attempts Cartesian equation of locus of $z$, substitutes for $x$ and $y$; must be linear with non-zero constant term |
| $13u^2+13v^2+12u-18v+5=0 \Rightarrow \left(u+\frac{6}{13}\right)^2+\left(v-\frac{9}{13}\right)^2=\frac{4}{13}$ | dddM1 | Attempts to complete the square where $u^2$ and $v^2$ have same coefficient |
| $\left\|w-\left(-\frac{6}{13}+\frac{9}{13}i\right)\right\|=\frac{2}{\sqrt{13}}$ | A1 | Correct equation in required form |
### Part (b) — Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $w = \frac{iz}{z-2i} \Rightarrow z = \frac{2wi}{w-i}$ | M1 | Must obtain form $\frac{awi}{bw+ci}$ |
| $\left\|\frac{2wi}{w-i}-2i\right\| = \left\|\frac{2wi}{w-i}-3\right\| \Rightarrow \left\|\frac{2wi-2wi-2}{w-i}\right\| = \left\|\frac{2wi-3w+3i}{w-i}\right\|$ | dM1 | Introduces $z$ in terms of $w$ into locus and attempts to combine terms |
| $\left\|\frac{-2}{w-i}\right\| = \left\|\frac{2wi-3w+3i}{w-i}\right\| \Rightarrow |-2| = |2wi-3w+3i|$ | A1 | Correct equation with fractions removed |
| $|2(u+iv)i - 3(u+iv)+3i| = 2 \Rightarrow (3u+2v)^2+(3v-2u-3)^2=4$ | ddM1 | Introduces $w=u+iv$ and applies Pythagoras correctly |
| $\left(u+\frac{6}{13}\right)^2+\left(v-\frac{9}{13}\right)^2=\frac{4}{13}$ | dddM1 | Attempts to complete the square |
| $\left\|w-\left(-\frac{6}{13}+\frac{9}{13}i\right)\right\|=\frac{2}{\sqrt{13}}$ | A1 | Correct equation in required form |
### Part (b) — Way 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $w = \frac{iz}{z-2i} \Rightarrow z = \frac{2wi}{w-i}$ | M1 | Must obtain form $\frac{awi}{bw+ci}$ |
| $\left\|\frac{2wi-2wi-2}{w-i}\right\| = \left\|\frac{2wi-3w+3i}{w-i}\right\|$ | dM1 | Introduces $z$ and attempts to combine terms |
| $|-2|=|2wi-3w+3i|$ | A1 | Correct equation with fractions removed |
| $|w(2i-3)+3i| = |(2i-3)|\left\|w+\frac{6-9i}{13}\right\| = 2$ | ddM1 | Attempts to isolate $w$ and rationalise denominator |
| $\sqrt{13}\left\|w-\left(-\frac{6}{13}+\frac{9}{13}i\right)\right\|=2 \Rightarrow \left\|w-\left(-\frac{6}{13}+\frac{9}{13}i\right)\right\|=\frac{2}{\sqrt{13}}$ | dddM1, A1 | M1: divides by $|2i-3|$; A1: correct equation in required form |
---\begin{enumerate}
\item A complex number $z$ is represented by the point $P$ in an Argand diagram.
\end{enumerate}
Given that
$$| z - 2 i | = | z - 3 |$$
(a) sketch the locus of $P$. You do not need to find the coordinates of any intercepts.
The transformation $T$ from the $z$-plane to the $w$-plane is given by
$$w = \frac { \mathrm { i } z } { z - 2 \mathrm { i } } \quad z \neq 2 \mathrm { i }$$
Given that $T$ maps $| z - 2 i | = | z - 3 |$ to a circle $C$ in the $w$-plane,\\
(b) find the equation of $C$, giving your answer in the form
$$| w - ( p + q \mathrm { i } ) | = r$$
where $p , q$ and $r$ are real numbers to be determined.
\hfill \mbox{\textit{Edexcel F2 2023 Q6 [8]}}