| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Intersection of two loci |
| Difficulty | Standard +0.8 This is a Further Maths question requiring understanding of loci in the Argand diagram. Part (a) involves finding a point on an argument locus given its real part (straightforward substitution). Part (b) requires finding the second intersection point of a circle and a ray, which demands geometric insight about the circle's center and radius, then solving simultaneous equations. The multi-step nature and need for geometric visualization place it moderately above average difficulty. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (a) | line is y = 7 − x |
| Answer | Marks |
|---|---|
| so w = 1 + 6i | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 3.2a | 3 b |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (b) | 1 + 6 i + 3 − 9 i = k k2 = 42 + (−3)2 |
| Answer | Marks |
|---|---|
| other complex number is −6 + 13i | M1 |
| Answer | Marks |
|---|---|
| [7] | 1.1 |
| Answer | Marks |
|---|---|
| 3.2a | finding k using modulus or substituting their x = 1, y = 6 into |
Question 8:
8 | (a) | line is y = 7 − x
when x = 1, y = 6
so w = 1 + 6i | M1
A1
[2] | 3.1a
3.2a | 3 b
oe, e.g. w = 1 + bi, arg(−6+bi)= =−1
4 −6
8 | (b) | 1 + 6 i + 3 − 9 i = k k2 = 42 + (−3)2
k = 5
Circle equation is (x + 3)2 + (y − 9)2 = k2
(x +3)2 + (7 − x − 9)2 = 25
x = [1 and] −6
when x = −6, y = 13
other complex number is −6 + 13i | M1
A1
M1
M1
A1
A1
A1
[7] | 1.1
1.1
3.1a
1.1
1.1
1.1
3.2a | finding k using modulus or substituting their x = 1, y = 6 into
circle equation (see below)
oe condone sign errors or k for k2 (but not both)
solving simultaneously with y = 7 − x
or (eliminating x) y = [6 and] 13
or when y = 13, x = −68 In an Argand diagram, the point P representing the complex number $w$ lies on the locus defined by $\left\{ z : \arg ( z - 7 ) = \frac { 3 } { 4 } \pi \right\}$. You are given that $\operatorname { Re } ( w ) = 1$.
\begin{enumerate}[label=(\alph*)]
\item Find $w$.
The point P also lies on the locus defined by $\{ \mathrm { z } : | \mathrm { z } + 3 - 9 \mathrm { i } | = \mathrm { k } \}$, where $k$ is a constant.
\item Find the complex number represented by the other point of intersection of the loci defined by
$$\{ z : | z + 3 - 9 i | = k \} \text { and } \left\{ z : \arg ( z - 7 ) = \frac { 3 } { 4 } \pi \right\} .$$
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2024 Q8 [9]}}