OCR MEI Further Pure Core 2021 November — Question 3 6 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2021
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Argand & Loci
TypeModulus-argument form conversion
DifficultyModerate -0.3 This is a straightforward Further Maths question requiring standard conversions between Cartesian and modulus-argument forms, followed by division using the quotient rule. While it involves multiple steps and exact values (requiring knowledge of special angles), these are routine techniques for Further Maths students with no problem-solving or novel insight required. Slightly easier than average due to its mechanical nature.
Spec4.02b Express complex numbers: cartesian and modulus-argument forms4.02c Complex notation: z, z*, Re(z), Im(z), |z|, arg(z)4.02e Arithmetic of complex numbers: add, subtract, multiply, divide

3 In this question you must show detailed reasoning.
The complex numbers \(z _ { 1 }\) and \(z _ { 2 }\) are given by \(z _ { 1 } = - 2 + 2 i\) and \(z _ { 2 } = 2 \left( \cos \frac { 1 } { 6 } \pi + i \sin \frac { 1 } { 6 } \pi \right)\).
  1. Find the modulus and argument of \(z _ { 1 }\).
  2. Hence express \(\frac { z _ { 1 } } { z _ { 2 } }\) in exact modulus-argument form.

Question 3:
AnswerMarks Guidance
3(a) DR
z = 8
1
3 
a r g ( z ) =
1
AnswerMarks
4B1
E1
AnswerMarks Guidance
[2]1.1
1.1Must see some reasoning for arg(z)
3(b) DR
z 8
1 = = 2
z 2
2
( z ) 3 7   
a r g 1 = − =
z 4 6 1 2
2
z ( 7 7)
so 1 = 2 cos +isin
z 12 12
AnswerMarks
2B1
M1
A1
B1
AnswerMarks
[4]1.1
1.1
1.1
AnswerMarks
1.1( z )
a r g 1 = a r g ( z ) − a r g ( z ) used
1 2
z
2
Question 3:
3 | (a) | DR
z = 8
1
3 
a r g ( z ) =
1
4 | B1
E1
[2] | 1.1
1.1 | Must see some reasoning for arg(z)
3 | (b) | DR
z 8
1 = = 2
z 2
2
( z ) 3 7   
a r g 1 = − =
z 4 6 1 2
2
z ( 7 7)
so 1 = 2 cos +isin
z 12 12
2 | B1
M1
A1
B1
[4] | 1.1
1.1
1.1
1.1 | ( z )
a r g 1 = a r g ( z ) − a r g ( z ) used
1 2
z
2
3 In this question you must show detailed reasoning.\\
The complex numbers $z _ { 1 }$ and $z _ { 2 }$ are given by $z _ { 1 } = - 2 + 2 i$ and $z _ { 2 } = 2 \left( \cos \frac { 1 } { 6 } \pi + i \sin \frac { 1 } { 6 } \pi \right)$.
\begin{enumerate}[label=(\alph*)]
\item Find the modulus and argument of $z _ { 1 }$.
\item Hence express $\frac { z _ { 1 } } { z _ { 2 } }$ in exact modulus-argument form.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2021 Q3 [6]}}