OCR MEI Further Pure Core AS 2019 June — Question 7 12 marks

Exam BoardOCR MEI
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2019
SessionJune
Marks12
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 a circle and a half-line in the complex plane. While sketching the loci is routine, finding exact coordinates requires setting up and solving a system involving both Cartesian and polar forms, demanding careful algebraic manipulation and geometric understanding—moderately challenging but within standard Further Maths scope.
Spec4.02k Argand diagrams: geometric interpretation4.02o Loci in Argand diagram: circles, half-lines

7
  1. Sketch on a single Argand diagram
    1. the set of points for which \(| z - 1 - 3 i | = 3\),
    2. the set of points for which \(\arg ( z + 4 ) = \frac { 1 } { 4 } \pi\).
  2. Find, in exact form, the two values of \(z\) for which \(| z - 1 - 3 i | = 3\) and \(\arg ( z + 4 ) = \frac { 1 } { 4 } \pi\).

Question 7:
AnswerMarks Guidance
7(a) (i)
A1
A1
AnswerMarks
[3]3.1a
1.1
AnswerMarks
1.1circle
centre 1 + 3i indicated
touching real axis
AnswerMarks Guidance
7(a) (ii)
B1
B1
AnswerMarks
[3]3.1a
1.1
AnswerMarks
1.1line at 45° to real axis
through −4
AnswerMarks Guidance
correct half-line indicatedwith evidence
7(b) circle is (x−1)2 +(y−3)2 =9
line is y = x+4
⇒(x−1)2 +(x+1)2 =9
⇒2x2 =7⇒ x=± 7
2
⇒ y =4± 7
2
( ) ( )
so z =− 7 + 4− 7 i or 7 + 4+ 7 i
AnswerMarks
2 2 2 2B1ft
B1ft
M1
A1
A1
A1
AnswerMarks
[6]3.1a
3.1a
1.1
1.1
1.1
AnswerMarks
3.2aft their centre
ft their −4
eliminating y (or x)
Question 7:
7 | (a) | (i) | 4i | M1
A1
A1
[3] | 3.1a
1.1
1.1 | circle
centre 1 + 3i indicated
touching real axis
7 | (a) | (ii) | B1
B1
B1
[3] | 3.1a
1.1
1.1 | line at 45° to real axis
through −4
correct half-line indicated | with evidence
7 | (b) | circle is (x−1)2 +(y−3)2 =9
line is y = x+4
⇒(x−1)2 +(x+1)2 =9
⇒2x2 =7⇒ x=± 7
2
⇒ y =4± 7
2
( ) ( )
so z =− 7 + 4− 7 i or 7 + 4+ 7 i
2 2 2 2 | B1ft
B1ft
M1
A1
A1
A1
[6] | 3.1a
3.1a
1.1
1.1
1.1
3.2a | ft their centre
ft their −4
eliminating y (or x)
7
\begin{enumerate}[label=(\alph*)]
\item Sketch on a single Argand diagram
\begin{enumerate}[label=(\roman*)]
\item the set of points for which $| z - 1 - 3 i | = 3$,
\item the set of points for which $\arg ( z + 4 ) = \frac { 1 } { 4 } \pi$.
\end{enumerate}\item Find, in exact form, the two values of $z$ for which $| z - 1 - 3 i | = 3$ and $\arg ( z + 4 ) = \frac { 1 } { 4 } \pi$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2019 Q7 [12]}}