OCR Further Pure Core AS 2022 June — Question 7 7 marks

Exam BoardOCR
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2022
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Argand & Loci
TypeIntersection of two loci
DifficultyStandard +0.8 This is a Further Maths question requiring students to find the intersection of two argument loci (half-lines from different points). It demands visualization on an Argand diagram, converting arguments to geometric rays, finding their intersection point algebraically, and showing detailed reasoning. While the individual concepts are standard, combining them requires solid geometric insight and multi-step problem-solving beyond typical A-level pure maths.
Spec4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines

7 In this question you must show detailed reasoning.
Two loci, \(C _ { 1 }\) and \(C _ { 2 }\), are defined as follows. \(\mathrm { C } _ { 1 } = \left\{ \mathrm { z } : \arg ( \mathrm { z } + 2 - \mathrm { i } ) = \frac { 1 } { 4 } \pi \right\}\) and \(\mathrm { C } _ { 2 } = \left\{ \mathrm { z } : \arg ( \mathrm { z } - 2 - \sqrt { 3 } - 2 \mathrm { i } ) = \frac { 2 } { 3 } \pi \right\}\) By considering the representations of \(C _ { 1 }\) and \(C _ { 2 }\) on an Argand diagram, determine the locus \(C _ { 1 } \cap C _ { 2 }\).

Question 7:
AnswerMarks Guidance
AnswerMarks Guidance
\(m_1=[\tan(\pi/4)]=1\) and \(m_2=[\tan(2\pi/3)]=-\sqrt{3}\)B1 3.1a Must be connect to the gradients of the lines (soi)
\(C_1\): \(z+2-i=z-(-2+i)\) so equation of (half-)line is \(y-1=x--2\) or \(y=x+3\)M1 3.1a Identifying a point on the extended line (condone sign errors) and using it and their gradient to form the equation of a line. Don't need to see equation in \(z\) first. Gradient must have come from considering angle of \((\pi/4)\)
\(C_2\): \(z-2-\sqrt{3}-2i = z-(2+\sqrt{3}+2i)\) so equation is \(y-2=-\sqrt{3}(x-(2+\sqrt{3}))\) or \(y=-\sqrt{3}x+2\sqrt{3}+5\)M1 1.1 Identifying a point on the extended line (condone sign errors) and using it and their gradient to form the equation of a line. Don't need to see equation in \(z\) first. Gradient must have come from considering angle of \((2\pi/3)\) or \((\pi/3)\)
\(x+3=-\sqrt{3}x+2\sqrt{3}+5 \Rightarrow (1+\sqrt{3})x=2(1+\sqrt{3}) \Rightarrow x=2\)M1 2.1 Eliminating one unknown and solving for the other
\(\Rightarrow y=2+3=5\)A1 1.1
Sketch showing \(C_1\) and \(C_2\) as half lines with correct start points clearly indicated, PoI lies on both half lines in approximately the right positions, angles approximately correct, lines need to intersectB1 2.3 Or: \(C_1\): Need \(x>-2\). \(C_2\): Need \(x<2+\sqrt{3}\), therefore solution with \(x=2\) is valid. Or: \(C_1\): Need \(y>1\). \(C_2\) Need \(y>2\), so solution with \(y=5\) is valid
\(C_1\cap C_2 = \{2+5i\}\) (or in words: "so the required locus contains only the number \(2+5i\)")A1 [7] 3.2a A1 can be awarded if answer represented unambiguously on an Argand Diagram either as \(2+5i\), or \(2\) and \(5i\) marked on axes. Not just \((2,5)\) or \((2,5i)\). Needs to be an indication that the locus is a single point, expressed as a complex number. Cannot be expressed as coordinates
## Question 7:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $m_1=[\tan(\pi/4)]=1$ and $m_2=[\tan(2\pi/3)]=-\sqrt{3}$ | B1 3.1a | Must be connect to the gradients of the lines (soi) |
| $C_1$: $z+2-i=z-(-2+i)$ so equation of (half-)line is $y-1=x--2$ or $y=x+3$ | M1 3.1a | Identifying a point on the extended line (condone sign errors) and using it and their gradient to form the equation of a line. Don't need to see equation in $z$ first. Gradient must have come from considering angle of $(\pi/4)$ |
| $C_2$: $z-2-\sqrt{3}-2i = z-(2+\sqrt{3}+2i)$ so equation is $y-2=-\sqrt{3}(x-(2+\sqrt{3}))$ or $y=-\sqrt{3}x+2\sqrt{3}+5$ | M1 1.1 | Identifying a point on the extended line (condone sign errors) and using it and their gradient to form the equation of a line. Don't need to see equation in $z$ first. Gradient must have come from considering angle of $(2\pi/3)$ or $(\pi/3)$ |
| $x+3=-\sqrt{3}x+2\sqrt{3}+5 \Rightarrow (1+\sqrt{3})x=2(1+\sqrt{3}) \Rightarrow x=2$ | M1 2.1 | Eliminating one unknown and solving for the other |
| $\Rightarrow y=2+3=5$ | A1 1.1 | |
| Sketch showing $C_1$ and $C_2$ as half lines with correct start points clearly indicated, PoI lies on both half lines in approximately the right positions, angles approximately correct, lines need to intersect | B1 2.3 | Or: $C_1$: Need $x>-2$. $C_2$: Need $x<2+\sqrt{3}$, therefore solution with $x=2$ is valid. Or: $C_1$: Need $y>1$. $C_2$ Need $y>2$, so solution with $y=5$ is valid |
| $C_1\cap C_2 = \{2+5i\}$ (or in words: "so the required locus contains only the number $2+5i$") | A1 [7] 3.2a | A1 can be awarded if answer represented unambiguously on an Argand Diagram either as $2+5i$, or $2$ and $5i$ marked on axes. Not just $(2,5)$ or $(2,5i)$. Needs to be an indication that the locus is a single point, expressed as a complex number. Cannot be expressed as coordinates |
7 In this question you must show detailed reasoning.\\
Two loci, $C _ { 1 }$ and $C _ { 2 }$, are defined as follows.\\
$\mathrm { C } _ { 1 } = \left\{ \mathrm { z } : \arg ( \mathrm { z } + 2 - \mathrm { i } ) = \frac { 1 } { 4 } \pi \right\}$ and $\mathrm { C } _ { 2 } = \left\{ \mathrm { z } : \arg ( \mathrm { z } - 2 - \sqrt { 3 } - 2 \mathrm { i } ) = \frac { 2 } { 3 } \pi \right\}$\\
By considering the representations of $C _ { 1 }$ and $C _ { 2 }$ on an Argand diagram, determine the locus $C _ { 1 } \cap C _ { 2 }$.

\hfill \mbox{\textit{OCR Further Pure Core AS 2022 Q7 [7]}}