OCR Further Pure Core 1 2023 June — Question 3 5 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2023
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
Typenth roots with preliminary simplification
DifficultyStandard +0.3 This is a standard Further Maths question on finding nth roots of complex numbers. Part (a) is routine verification of converting to exponential form (finding modulus and argument). Part (b) applies the standard formula for fifth roots with straightforward arithmetic. While it requires knowledge beyond A-level Core, it's a textbook exercise within Further Maths with no novel problem-solving required.
Spec4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02r nth roots: of complex numbers
3
  1. Show that \(\frac { - 3 + \sqrt { 3 } \mathrm { i } } { 2 } = \sqrt { 3 } \mathrm { e } ^ { \frac { 5 } { 6 } \pi \mathrm { i } }\).
  2. Hence determine the exact roots of the equation \(z ^ { 5 } = \frac { 9 ( - 3 + \sqrt { 3 } \mathrm { i } ) } { 2 }\), giving the roots in the form \(r \mathrm { e } ^ { \mathrm { i } \theta }\) where \(r > 0\) and \(0 \leqslant \theta < 2 \pi\).
Question 3:
AnswerMarks Guidance
3(a) r = 1 ( − 3 ) 2 + ( 3 ) 2 = 1 1 2 = 3
2 2
3   
a r c t a n − = −
3 6
5  
   = − + =
AnswerMarks
6 6B1
B1AG so must show use of z = a2 +b2
√3
AG. Or 𝜃 = 𝜋−arctan( ), may be indicated on a diagram, but clear
3
reasoning must be shown (eg.finding complementary angle, or use of
Pythagoras’ theorem and then arcsin or arccos)
AnswerMarks
Alternative method:M1
A1AG Clearly shown
56 i 5 5       
3 e 3 c o s i s i n   = +
6 6
3 1 3 i 3   −
3 i = − + =
2 2 2
[2]
AnswerMarks
(b)15
15  52 
( )
r = 9 3 = 3 = 3
1 5   
2 r ( 5 1 2 r ) f o r r 0 , 1 , 2 , 3 , 4    = + = + =
5 6 3 0
1 17 29 41 53
i i i i i
AnswerMarks
 z = 3e6 , 3e30 , 3e30 , 3e30 , 3e30B1
M1
AnswerMarks
A1For 𝑟 = √3 oe (including 1.73....)
5
For their 𝜋+2𝜋𝑛 from (a) divided by 5 (either in terms of n, or for at
6
least two values of n).
1
(5+12𝑛)𝜋i
Allow √3e30 for n = 0, 1, 2, 3, 4.
1
( )
Accept only r = 3 or 3 2
For last two marks, If M0 then SC B1 for all five roots
[3]
M1
A1
AG Clearly shown
Question 3:
3 | (a) | r = 1 ( − 3 ) 2 + ( 3 ) 2 = 1 1 2 = 3
2 2
3   
a r c t a n − = −
3 6
5  
   = − + =
6 6 | B1
B1 | AG so must show use of z = a2 +b2
√3
AG. Or 𝜃 = 𝜋−arctan( ), may be indicated on a diagram, but clear
3
reasoning must be shown (eg.finding complementary angle, or use of
Pythagoras’ theorem and then arcsin or arccos)
Alternative method: | M1
A1 | AG Clearly shown
56 i 5 5       
3 e 3 c o s i s i n   = +
6 6
3 1 3 i 3   −
3 i = − + =
2 2 2
[2]
(b) | 15
15  52 
( )
r = 9 3 = 3 = 3
1 5   
2 r ( 5 1 2 r ) f o r r 0 , 1 , 2 , 3 , 4    = + = + =
5 6 3 0
1 17 29 41 53
i i i i i
 z = 3e6 , 3e30 , 3e30 , 3e30 , 3e30 | B1
M1
A1 | For 𝑟 = √3 oe (including 1.73....)
5
For their 𝜋+2𝜋𝑛 from (a) divided by 5 (either in terms of n, or for at
6
least two values of n).
1
(5+12𝑛)𝜋i
Allow √3e30 for n = 0, 1, 2, 3, 4.
1
( )
Accept only r = 3 or 3 2
For last two marks, If M0 then SC B1 for all five roots
[3]
M1
A1
AG Clearly shown
3
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { - 3 + \sqrt { 3 } \mathrm { i } } { 2 } = \sqrt { 3 } \mathrm { e } ^ { \frac { 5 } { 6 } \pi \mathrm { i } }$.
\item Hence determine the exact roots of the equation $z ^ { 5 } = \frac { 9 ( - 3 + \sqrt { 3 } \mathrm { i } ) } { 2 }$, giving the roots in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$ where $r > 0$ and $0 \leqslant \theta < 2 \pi$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 1 2023 Q3 [5]}}
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