| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| 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 question requiring standard techniques: finding modulus/argument of a given complex number (routine calculation), then sketching two basic loci (a circle and a half-line). While it's FP1 material, both parts are direct applications of definitions with no problem-solving or geometric insight required, making it slightly easier than average overall. |
| 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \( | a | = 2\) |
| \(\arg a = 60°, \frac{\pi}{3}, 1.05\) | B1 [2] | Correct argument |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| B1 | Circle | |
| B1 | Centre \((1, \sqrt{3})\) | |
| B1 | Through origin, centre \((\pm 1, \pm\sqrt{3})\) and another y intercept | |
| B1 | Vertical line | |
| B1* | Through \(a\) or their centre, with +ve gradient | |
| DB1 [6] | Correct half line |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $|a| = 2$ | B1 | Correct modulus |
| $\arg a = 60°, \frac{\pi}{3}, 1.05$ | B1 **[2]** | Correct argument |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| | B1 | Circle |
| | B1 | Centre $(1, \sqrt{3})$ |
| | B1 | Through origin, centre $(\pm 1, \pm\sqrt{3})$ and another y intercept |
| | B1 | Vertical line |
| | B1* | Through $a$ or their centre, with +ve gradient |
| | DB1 **[6]** | Correct half line |
---5 The complex number $1 + \mathrm { i } \sqrt { 3 }$ is denoted by $a$.\\
(i) Find $| a |$ and $\arg a$.\\
(ii) Sketch on a single Argand diagram the loci given by $| z - a | = | a |$ and $\arg ( z - a ) = \frac { 1 } { 2 } \pi$.
\hfill \mbox{\textit{OCR FP1 2011 Q5 [8]}}