| Exam Board | OCR |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Intersection of two loci |
| Difficulty | Standard +0.3 This is a straightforward Further Maths AS question testing standard loci concepts. Part (a) requires writing a circle equation in set notation (routine). Part (b) asks to sketch the perpendicular bisector of two points (standard technique). Part (c) involves finding the intersection of a half-line and vertical line (simple coordinate geometry). While it's Further Maths content, these are foundational loci skills with no novel problem-solving required. |
| Spec | 4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \( | z-(4-2\mathbf{i}) | \) |
| \(\{z : z \in \mathbb{C},\ | z-(4-2\mathbf{i}) | =3\}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| (need points equidistant from) \(\mathbf{i}\) and \(-2\) | B1 | \(\mathbf{i}\) and \(-2\) both clearly identified either on the sketch or in words. Ignore other points. |
| Their two identified points joined and perpendicularly bisected by a single, straight, solid line labelled \(L\) or otherwise unambiguously indicated | B1FT | NB equation of this line (need not be shown) is \(y=-2x-\frac{3}{2}\) so \(x\)- & \(y\)-intercepts are \(-\frac{3}{4}\) and \(-1\frac{1}{2}\) respectively. These need not be shown but if shown must be correct. If points not explicitly identified then 1st B1 can be implied by either correct equation of line or both intercepts of line given. Line must be indicated as perpendicular or implied to be perpendicular. SC If B0 for not labelling the points but the points are located correctly, and line indicated appears to be the perpendicular bisector then allow B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Either \(x=3\) or \(\tan^{-1}\left(\frac{y}{x-1}\right)=\frac{1}{4}\pi\) stated or indicated | M1 | Understanding of one of the conditions. Could be shown on a diagram (e.g. the line \(x=3\) drawn or 3 as a clear, special label on the real axis or the half-line with gradient 1 drawn from \((1,0)\), etc). \(y=x-1\) implies M1; answer of \(3+b\mathbf{i}\) implies M1 |
| \(\tan^{-1}\left(\frac{y}{x-1}\right)=\frac{1}{4}\pi \Rightarrow \frac{y}{3-1}=1 \Rightarrow y=2\) so the number is \(3+2\mathbf{i}\) | A1 | Could be shown on a diagram. Find — might not be much (or even any) working shown |
# Question 4:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $|z-(4-2\mathbf{i})|$ | M1 | Any solution which involves $|z-(4-2\mathbf{i})|$ or $|z-4+2\mathbf{i}|$. M1 for $(x-4)^2+(y+2)^2=9$ |
| $\{z : z \in \mathbb{C},\ |z-(4-2\mathbf{i})|=3\}$ | A1 | Must show the $\{\}$ brackets and correct set notation used, i.e. the form shown but BOD lack of comma. Or $\left\{z=x+y\mathbf{i}: x,y\in\mathbb{R},\ (x-4)^2+(y+2)^2=9\ (\text{or } 3^2)\right\}$ |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| (need points equidistant from) $\mathbf{i}$ and $-2$ | B1 | $\mathbf{i}$ and $-2$ both clearly identified either on the sketch or in words. Ignore other points. |
| Their two identified points joined and perpendicularly bisected by a single, straight, solid line labelled $L$ or otherwise unambiguously indicated | B1FT | NB equation of this line (need not be shown) is $y=-2x-\frac{3}{2}$ so $x$- & $y$-intercepts are $-\frac{3}{4}$ and $-1\frac{1}{2}$ respectively. These need not be shown but if shown must be correct. If points not explicitly identified then 1st B1 can be implied by either correct equation of line or both intercepts of line given. Line must be indicated as perpendicular or implied to be perpendicular. **SC** If B0 for not labelling the points but the points are located correctly, and line indicated appears to be the perpendicular bisector then allow B1 |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Either $x=3$ or $\tan^{-1}\left(\frac{y}{x-1}\right)=\frac{1}{4}\pi$ stated or indicated | M1 | Understanding of one of the conditions. Could be shown on a diagram (e.g. the line $x=3$ drawn or 3 as a clear, special label on the real axis or the half-line with gradient 1 drawn from $(1,0)$, etc). $y=x-1$ implies M1; answer of $3+b\mathbf{i}$ implies M1 |
| $\tan^{-1}\left(\frac{y}{x-1}\right)=\frac{1}{4}\pi \Rightarrow \frac{y}{3-1}=1 \Rightarrow y=2$ so the number is $3+2\mathbf{i}$ | A1 | Could be shown on a diagram. Find — might not be much (or even any) working shown |
---4 The Argand diagram shows a circle of radius 3. The centre of the circle is the point which represents the complex number $4 - 2 \mathrm { i }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{4159328b-475e-4f29-91f2-f2f343573251-3_417_775_349_644}
\begin{enumerate}[label=(\alph*)]
\item Use set notation to define the locus of complex numbers, $z$, represented by points which lie on the circle.
The locus $L$ is defined by $\mathrm { L } = \{ \mathrm { z } : \mathrm { z } \in \mathbb { C } , | \mathrm { z } - \mathrm { i } | = | \mathrm { z } + 2 | \}$.
\item On the Argand diagram in the Printed Answer Booklet, sketch and label the locus $L$.
You are given that the locus $\left\{ z : z \in \mathbb { C } , \arg ( z - 1 ) = \frac { 1 } { 4 } \pi , \operatorname { Re } ( z ) = 3 \right\}$ contains only one number.
\item Find this number.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core AS 2024 Q4 [6]}}