OCR MEI Further Pure Core AS 2021 November — Question 7 9 marks

Exam BoardOCR MEI
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2021
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Argand & Loci
TypeArgument calculations and identities
DifficultyChallenging +1.2 This is a multi-step Further Maths question requiring conversion between forms, complex multiplication, and using De Moivre's theorem to find an exact trigonometric value. While it involves several techniques, each step follows standard procedures with clear signposting. The final trigonometric extraction requires some insight but is a well-known application type in Further Maths complex numbers.
Spec4.02b Express complex numbers: cartesian and modulus-argument forms4.02f Convert between forms: cartesian and modulus-argument4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)

7
    1. Find the modulus and argument of \(z _ { 1 }\), where \(z _ { 1 } = 1 + \mathrm { i }\).
    2. Given that \(\left| z _ { 2 } \right| = 2\) and \(\arg \left( z _ { 2 } \right) = \frac { 1 } { 6 } \pi\), express \(z _ { 2 }\) in a + bi form, where \(a\) and \(b\) are exact real numbers.
  1. Using these results, find the exact value of \(\sin \frac { 5 } { 12 } \pi\), giving the answer in the form \(\frac { \sqrt { m } + \sqrt { n } } { p }\), where \(m , n\) and \(p\) are integers.

Question 7:
AnswerMarks Guidance
7(a) (i)
1
arg(z )= 1π
AnswerMarks
1 4B1
B1
AnswerMarks
[2]1.1
1.11.41 or better
allow 45°
AnswerMarks Guidance
7(a) (ii)
2 6 6
AnswerMarks
= 3+iM1
A1
AnswerMarks Guidance
[2]1.1
1.11.73 or better
7(b) z z =(1+i)( 3+i)
1 2
= 3−1+( 3+1)i
z z =2 2 ( cos 5π+isin 5π )
1 2 12 12
3+1
So sin 5π=
12 2 2
2( 3+1) 6+ 2
= =
AnswerMarks
4 4M1
A1
M1
A1
A1
AnswerMarks
[5]3.1a
1.1
3.1a
1.1
AnswerMarks
3.2aFinding z z (using cartesian form)
1 2
Use of arg(z z ) = arg z + argz
1 2 1 2
Question 7:
7 | (a) | (i) | | z |= 2
1
arg(z )= 1π
1 4 | B1
B1
[2] | 1.1
1.1 | 1.41 or better
allow 45°
7 | (a) | (ii) | z =2 ( cos1π+isin1π )
2 6 6
= 3+i | M1
A1
[2] | 1.1
1.1 | 1.73 or better
7 | (b) | z z =(1+i)( 3+i)
1 2
= 3−1+( 3+1)i
z z =2 2 ( cos 5π+isin 5π )
1 2 12 12
3+1
So sin 5π=
12 2 2
2( 3+1) 6+ 2
= =
4 4 | M1
A1
M1
A1
A1
[5] | 3.1a
1.1
3.1a
1.1
3.2a | Finding z z (using cartesian form)
1 2
Use of arg(z z ) = arg z + argz
1 2 1 2
7
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the modulus and argument of $z _ { 1 }$, where $z _ { 1 } = 1 + \mathrm { i }$.
\item Given that $\left| z _ { 2 } \right| = 2$ and $\arg \left( z _ { 2 } \right) = \frac { 1 } { 6 } \pi$, express $z _ { 2 }$ in a + bi form, where $a$ and $b$ are exact real numbers.
\end{enumerate}\item Using these results, find the exact value of $\sin \frac { 5 } { 12 } \pi$, giving the answer in the form $\frac { \sqrt { m } + \sqrt { n } } { p }$, where $m , n$ and $p$ are integers.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2021 Q7 [9]}}