AQA FP2 2007 June — Question 5 9 marks

Exam BoardAQA
ModuleFP2 (Further Pure Mathematics 2)
Year2007
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Argand & Loci
TypeIntersection of two loci
DifficultyStandard +0.3 This is a standard FP2 loci question with straightforward components: (a) requires recognizing that multiplication by i represents 90° rotation; (b) involves drawing standard loci (perpendicular bisector and a ray); (c) requires solving simultaneous equations using the locus definitions. While it involves multiple parts and Further Maths content, each step follows well-established techniques without requiring novel insight or complex problem-solving.
Spec4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines

5 The sketch shows an Argand diagram. The points \(A\) and \(B\) represent the complex numbers \(z _ { 1 }\) and \(z _ { 2 }\) respectively. The angle \(A O B = 90 ^ { \circ }\) and \(O A = O B\). \includegraphics[max width=\textwidth, alt={}, center]{847295e3-d806-43b1-8d25-688c5558bfe1-3_533_869_852_632}
  1. Explain why \(z _ { 2 } = \mathrm { i } z _ { 1 }\).
  2. On a single copy of the diagram, draw:
    1. the locus \(L _ { 1 }\) of points satisfying \(\left| z - z _ { 2 } \right| = \left| z - z _ { 1 } \right|\);
    2. the locus \(L _ { 2 }\) of points satisfying \(\arg \left( z - z _ { 2 } \right) = \arg z _ { 1 }\).
  3. Find, in terms of \(z _ { 1 }\), the complex number representing the point of intersection of \(L _ { 1 }\) and \(L _ { 2 }\).

Question 5:
Part (a)
AnswerMarks Guidance
WorkingMarks Guidance
ExplanationE2,1,0 E1 for \(i = e^{\frac{\pi i}{2}}\) or \(iz_1 = -y_1 + ix_1\)
Part (b)(i)
AnswerMarks Guidance
WorkingMarks Guidance
Perpendicular bisector of \(AB\)B1
through \(O\)B1
Part (b)(ii)
AnswerMarks Guidance
WorkingMarks Guidance
half-lineB1 If \(L_2\) is taken to be the line \(AB\) give B0
from \(B\)B1
parallel to \(OA\)B1
Part (c)
AnswerMarks Guidance
WorkingMarks Guidance
\((1+i)z_1\)M1A1 ft if \(L_2\) taken as line \(AB\)
## Question 5:

### Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| Explanation | E2,1,0 | E1 for $i = e^{\frac{\pi i}{2}}$ or $iz_1 = -y_1 + ix_1$ |

### Part (b)(i)
| Working | Marks | Guidance |
|---------|-------|----------|
| Perpendicular bisector of $AB$ | B1 | |
| through $O$ | B1 | |

### Part (b)(ii)
| Working | Marks | Guidance |
|---------|-------|----------|
| half-line | B1 | If $L_2$ is taken to be the line $AB$ give B0 |
| from $B$ | B1 | |
| parallel to $OA$ | B1 | |

### Part (c)
| Working | Marks | Guidance |
|---------|-------|----------|
| $(1+i)z_1$ | M1A1 | ft if $L_2$ taken as line $AB$ |

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5 The sketch shows an Argand diagram. The points $A$ and $B$ represent the complex numbers $z _ { 1 }$ and $z _ { 2 }$ respectively. The angle $A O B = 90 ^ { \circ }$ and $O A = O B$.\\
\includegraphics[max width=\textwidth, alt={}, center]{847295e3-d806-43b1-8d25-688c5558bfe1-3_533_869_852_632}
\begin{enumerate}[label=(\alph*)]
\item Explain why $z _ { 2 } = \mathrm { i } z _ { 1 }$.
\item On a single copy of the diagram, draw:
\begin{enumerate}[label=(\roman*)]
\item the locus $L _ { 1 }$ of points satisfying $\left| z - z _ { 2 } \right| = \left| z - z _ { 1 } \right|$;
\item the locus $L _ { 2 }$ of points satisfying $\arg \left( z - z _ { 2 } \right) = \arg z _ { 1 }$.
\end{enumerate}\item Find, in terms of $z _ { 1 }$, the complex number representing the point of intersection of $L _ { 1 }$ and $L _ { 2 }$.
\end{enumerate}

\hfill \mbox{\textit{AQA FP2 2007 Q5 [9]}}