OCR MEI Further Pure Core 2022 June — Question 11 8 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2022
SessionJune
Marks8
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Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeDirect nth roots: roots with geometric or algebraic follow-up
DifficultyStandard +0.3 This is a straightforward Further Maths question on cube roots of complex numbers. Part (a) involves basic rotation by multiplying by exponentials and showing three equally-spaced roots sum to zero (standard result). Part (b) requires finding cube roots of 8i using modulus-argument form, which is a direct application of de Moivre's theorem with minimal problem-solving. While it's Further Maths content, the execution is routine and methodical, making it slightly easier than average overall.
Spec4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers
11 An Argand diagram with the point A representing a complex number \(z _ { 1 }\) is shown below. \includegraphics[max width=\textwidth, alt={}, center]{b57a2590-84e8-4998-9633-902db861f23a-4_716_778_932_239} The complex numbers \(z _ { 2 }\) and \(z _ { 3 }\) are \(z _ { 1 } \mathrm { e } ^ { \frac { 2 } { 3 } \mathrm { i } \pi }\) and \(z _ { 1 } \mathrm { e } ^ { \frac { 4 } { 3 } \mathrm { i } \pi }\) respectively.
    1. On the copy of the Argand diagram in the Printed Answer Booklet, mark the points B and C representing the complex numbers \(z _ { 2 }\) and \(z _ { 3 }\).
    2. Show that \(z _ { 1 } + z _ { 2 } + z _ { 3 } = 0\).
  2. Given now that \(z _ { 1 } , z _ { 2 }\) and \(z _ { 3 }\) are roots of the equation \(z ^ { 3 } = 8 \mathrm { i }\), find these three roots, giving your answers in the form \(\mathrm { a } + \mathrm { ib }\), where \(a\) and \(b\) are real and exact.
Question 11:
AnswerMarks Guidance
11(a) (i)
B
z2
A
z1
O
Re
z3
AnswerMarks
CM1
A1
AnswerMarks
[2]1.1
1.1on a circle centre O
form an approximate equilateral triangle
B and C must be labelled
AnswerMarks Guidance
11(a) (ii)
1 2 3 1
1βˆ’π‘’2π‘–πœ‹
= 𝑧 = 0
AnswerMarks
11βˆ’π‘’2π‘–πœ‹/3M1
A12.1
2.2asum of GP used or other correct method
Alternative solution
𝑧 +𝑧 +𝑧 =
1 2 3
2πœ‹ 4πœ‹ 2πœ‹ 4πœ‹
𝑧 [1+cos +cos +𝑖(sin +sin )]
1 3 3 3 3
M1
1 1 √3 √3
= 𝑧 [1βˆ’ βˆ’ +𝑖( βˆ’ )] = 0
AnswerMarks
1 2 2 2 2A1
A1
[2]
AnswerMarks Guidance
11(b) 1 1
𝑧3 = 8(cos πœ‹+𝑖sin πœ‹)
2 2
= 8π‘’π‘–πœ‹/2
AnswerMarks
𝑧= 2
𝑧 = 2π‘’π‘–πœ‹/6, 2𝑒5π‘–πœ‹/6, 2𝑒3π‘–πœ‹/2
AnswerMarks
= √3+𝑖, βˆ’βˆš3+𝑖, βˆ’2𝑖B1
B1
B1B1
AnswerMarks
[4]3.1a
1.1
AnswerMarks
1.1Exponential or modulus argument form soi
Soi
B1B0 if two out of three roots given
Question 11:
11 | (a) | (i) | Im
B
z2
A
z1
O
Re
z3
C | M1
A1
[2] | 1.1
1.1 | on a circle centre O
form an approximate equilateral triangle
B and C must be labelled
11 | (a) | (ii) | 𝑧 +𝑧 +𝑧 = 𝑧 (1+𝑒2π‘–πœ‹/3+𝑒4π‘–πœ‹/3 )
1 2 3 1
1βˆ’π‘’2π‘–πœ‹
= 𝑧 = 0
11βˆ’π‘’2π‘–πœ‹/3 | M1
A1 | 2.1
2.2a | sum of GP used or other correct method
Alternative solution
𝑧 +𝑧 +𝑧 =
1 2 3
2πœ‹ 4πœ‹ 2πœ‹ 4πœ‹
𝑧 [1+cos +cos +𝑖(sin +sin )]
1 3 3 3 3
M1
1 1 √3 √3
= 𝑧 [1βˆ’ βˆ’ +𝑖( βˆ’ )] = 0
1 2 2 2 2 | A1
A1
[2]
11 | (b) | 1 1
𝑧3 = 8(cos πœ‹+𝑖sin πœ‹)
2 2
= 8π‘’π‘–πœ‹/2
|𝑧| = 2
𝑧 = 2π‘’π‘–πœ‹/6, 2𝑒5π‘–πœ‹/6, 2𝑒3π‘–πœ‹/2
= √3+𝑖, βˆ’βˆš3+𝑖, βˆ’2𝑖 | B1
B1
B1B1
[4] | 3.1a
1.1
1.1 | Exponential or modulus argument form soi
Soi
B1B0 if two out of three roots given
11 An Argand diagram with the point A representing a complex number $z _ { 1 }$ is shown below.\\
\includegraphics[max width=\textwidth, alt={}, center]{b57a2590-84e8-4998-9633-902db861f23a-4_716_778_932_239}

The complex numbers $z _ { 2 }$ and $z _ { 3 }$ are $z _ { 1 } \mathrm { e } ^ { \frac { 2 } { 3 } \mathrm { i } \pi }$ and $z _ { 1 } \mathrm { e } ^ { \frac { 4 } { 3 } \mathrm { i } \pi }$ respectively.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item On the copy of the Argand diagram in the Printed Answer Booklet, mark the points B and C representing the complex numbers $z _ { 2 }$ and $z _ { 3 }$.
\item Show that $z _ { 1 } + z _ { 2 } + z _ { 3 } = 0$.
\end{enumerate}\item Given now that $z _ { 1 } , z _ { 2 }$ and $z _ { 3 }$ are roots of the equation $z ^ { 3 } = 8 \mathrm { i }$, find these three roots, giving your answers in the form $\mathrm { a } + \mathrm { ib }$, where $a$ and $b$ are real and exact.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2022 Q11 [8]}}
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