| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Cartesian equation from argument condition |
| Difficulty | Challenging +1.2 This is a Further Maths FP2 question on complex loci requiring multiple steps: finding the circle's center from given constraints, determining the unknown constant a, and writing the locus as an argument condition. While it involves several techniques (circle geometry, argument form), each step follows standard FP2 methods without requiring novel insight. The constraints guide the solution path clearly, making it moderately above average difficulty but well within typical FP2 scope. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Centre lies on perpendicular bisector of \(z_2\) and \(z_3\), which is vertical as same imaginary components. Hence \(\frac{a+1}{2} = -1 \Rightarrow a = \ldots\) Or \((a-(-1))^2 + (4-b)^2 = (1-(-1))^2 + (4-b)^2 \Rightarrow a = \ldots\) \((a-(-1))^2 = (1-(-1))^2 \Rightarrow a = \ldots\) Or \(-1-a = 1-(-1) \Rightarrow a = \ldots\) | M1 | Forms a correct strategy to find the value of \(a\) |
| \(a = -3\) | A1 | Correct value |
| Alternative: \((x-(-1))^2 + (y-b)^2 = r^2\); \((1+1)^2 + (4-b)^2 = r^2\) and \((3+1)^2 + (2-b)^2 = r^2 \Rightarrow 4+(4-b)^2 = 16+(2-b)^2 \Rightarrow b = \ldots\{0\}\); \(r^2 = 4+(4-"0")^2 = \ldots\{20\}\); \((a+1)^2 + (4-"0")^2 = "20" \Rightarrow a = \ldots\) | M1 | Forms a correct strategy |
| \(a = -3\) only, the other root must be rejected | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Equation has form \(\arg\left(\frac{z-z_1}{z-z_2}\right) = \theta\) | B1 | Recalls correct form for equation of an arc, with any angle or \(\theta\). Allow if \(z_1\) and \(z_2\) are the other way round |
| \(\arg\left(\frac{1+4\mathrm{i}-(3+2\mathrm{i})}{1+4\mathrm{i}-("-3"+4\mathrm{i})}\right) = \arg(-2+2\mathrm{i}) - \arg(4) = \ldots\left(=\frac{3\pi}{4}\right)\) Or \(\arg\left(\frac{1+4\mathrm{i}-3-2\mathrm{i}}{1+4\mathrm{i}-a-4\mathrm{i}}\right) = \arg\left(\frac{-2+2\mathrm{i}}{4}\right) = \arg\left(-\frac{1}{2}+\frac{1}{2}\mathrm{i}\right) = \frac{3\pi}{4}\) Or \(\overrightarrow{z_3 z_2} = \begin{pmatrix}-4\\0\end{pmatrix}\) and \(\overrightarrow{z_3 z_1} = \begin{pmatrix}2\\-2\end{pmatrix}\); \(\cos\theta = \frac{\begin{pmatrix}-4\\0\end{pmatrix}\cdot\begin{pmatrix}2\\-2\end{pmatrix}}{4\sqrt{2^2+(-2)^2}} \Rightarrow \theta = \ldots\) Or \(\cos\theta = \frac{4^2+8-40}{2\times4\times\sqrt{8}} \Rightarrow \theta = \ldots\) Or Using trigonometry | M1 | Any correct full method to find the value of \(\theta\) |
| Equation is \(\arg\left(\frac{z-3-2\mathrm{i}}{z+3-4\mathrm{i}}\right) = \frac{3\pi}{4}\) o.e.e. Or \(\arg(z-3-2\mathrm{i}) - \arg(z+3-4\mathrm{i}) = \frac{3\pi}{4}\) o.e.e. | A1 | Correct equation. Correct answer implies full marks |
## Question 9:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Centre lies on perpendicular bisector of $z_2$ and $z_3$, which is vertical as same imaginary components. Hence $\frac{a+1}{2} = -1 \Rightarrow a = \ldots$ **Or** $(a-(-1))^2 + (4-b)^2 = (1-(-1))^2 + (4-b)^2 \Rightarrow a = \ldots$ $(a-(-1))^2 = (1-(-1))^2 \Rightarrow a = \ldots$ **Or** $-1-a = 1-(-1) \Rightarrow a = \ldots$ | M1 | Forms a correct strategy to find the value of $a$ |
| $a = -3$ | A1 | Correct value |
| **Alternative:** $(x-(-1))^2 + (y-b)^2 = r^2$; $(1+1)^2 + (4-b)^2 = r^2$ and $(3+1)^2 + (2-b)^2 = r^2 \Rightarrow 4+(4-b)^2 = 16+(2-b)^2 \Rightarrow b = \ldots\{0\}$; $r^2 = 4+(4-"0")^2 = \ldots\{20\}$; $(a+1)^2 + (4-"0")^2 = "20" \Rightarrow a = \ldots$ | M1 | Forms a correct strategy |
| $a = -3$ only, the other root must be rejected | A1 | Correct value |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Equation has form $\arg\left(\frac{z-z_1}{z-z_2}\right) = \theta$ | B1 | Recalls correct form for equation of an arc, with any angle or $\theta$. Allow if $z_1$ and $z_2$ are the other way round |
| $\arg\left(\frac{1+4\mathrm{i}-(3+2\mathrm{i})}{1+4\mathrm{i}-("-3"+4\mathrm{i})}\right) = \arg(-2+2\mathrm{i}) - \arg(4) = \ldots\left(=\frac{3\pi}{4}\right)$ **Or** $\arg\left(\frac{1+4\mathrm{i}-3-2\mathrm{i}}{1+4\mathrm{i}-a-4\mathrm{i}}\right) = \arg\left(\frac{-2+2\mathrm{i}}{4}\right) = \arg\left(-\frac{1}{2}+\frac{1}{2}\mathrm{i}\right) = \frac{3\pi}{4}$ **Or** $\overrightarrow{z_3 z_2} = \begin{pmatrix}-4\\0\end{pmatrix}$ and $\overrightarrow{z_3 z_1} = \begin{pmatrix}2\\-2\end{pmatrix}$; $\cos\theta = \frac{\begin{pmatrix}-4\\0\end{pmatrix}\cdot\begin{pmatrix}2\\-2\end{pmatrix}}{4\sqrt{2^2+(-2)^2}} \Rightarrow \theta = \ldots$ **Or** $\cos\theta = \frac{4^2+8-40}{2\times4\times\sqrt{8}} \Rightarrow \theta = \ldots$ **Or** Using trigonometry | M1 | Any correct full method to find the value of $\theta$ |
| Equation is $\arg\left(\frac{z-3-2\mathrm{i}}{z+3-4\mathrm{i}}\right) = \frac{3\pi}{4}$ o.e.e. **Or** $\arg(z-3-2\mathrm{i}) - \arg(z+3-4\mathrm{i}) = \frac{3\pi}{4}$ o.e.e. | A1 | Correct equation. Correct answer implies full marks |
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{78543314-72b7-4366-98a1-dbb6b852632f-30_312_634_278_717}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a locus in the complex plane.\\
The locus is an arc of a circle from the point represented by $z _ { 1 } = 3 + 2 i$ to the point represented by $z _ { 2 } = a + 4 \mathrm { i }$, where $a$ is a constant, $a \neq 1$
Given that
\begin{itemize}
\item the point $z _ { 3 } = 1 + 4 \mathrm { i }$ also lies on the locus
\item the centre of the circle has real part equal to - 1
\begin{enumerate}[label=(\alph*)]
\item determine the value of $a$.
\item Hence determine a complex equation for the locus, giving any angles in the equation as positive values.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2023 Q9 [5]}}