OCR Further Pure Core 1 2019 June — Question 9 12 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2019
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeRoots of unity with trigonometric identities
DifficultyChallenging +1.2 This is a structured Further Maths question on roots of unity that guides students through each step systematically. Parts (a)-(c) are routine applications of De Moivre's theorem and geometric series, (d) requires algebraic manipulation using ω^5=1, and (e) involves solving a quadratic. While it requires multiple techniques and careful algebra, the scaffolding makes it more accessible than typical Further Maths proof questions, placing it moderately above average difficulty.
Spec4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers
9 In this question you must show detailed reasoning.
You are given the complex number \(\omega = \cos \frac { 2 } { 5 } \pi + \mathrm { i } \sin \frac { 2 } { 5 } \pi\) and the equation \(z ^ { 5 } = 1\).
  1. Show that \(\omega\) is a root of the equation.
  2. Write down the other four roots of the equation.
  3. Show that \(\omega + \omega ^ { 2 } + \omega ^ { 3 } + \omega ^ { 4 } = - 1\).
  4. Hence show that \(\left( \omega + \frac { 1 } { \omega } \right) ^ { 2 } + \left( \omega + \frac { 1 } { \omega } \right) - 1 = 0\).
  5. Hence determine the value of \(\cos \frac { 2 } { 5 } \pi\) in the form \(a + b \sqrt { c }\) where \(a , b\) and \(c\) are rational numbers to be found.
Question 9:
AnswerMarks Guidance
9(a) DR
2π 2π
ω=cos +isin
5 5
 2π 2π 5
⇒ω5 = cos +isin  =cos2π+isin2π=1+0i=1
AnswerMarks
 5 5 M1
A1
AnswerMarks
[2]2.1
1.1Finding ω5
AGUse of exponentials is
satisfactory
Could be argued
backwards
AnswerMarks Guidance
(b)ω2,ω3,ω4,1 B1
[1]1.1 Alternative:
2kπ 2kπ
Roots are cos +isin
5 5
for k =2,3,4
AnswerMarks
and 1( or k = 5)Exponentials
satisfactory
AnswerMarks
(c)DR
ω5 −1=0
⇒(ω−1 )( ω4 +ω3 +ω2 +ω+1 ) =0
⇒ω4 +ω3 +ω2 +ω=−1
Alternatively:
1−ω5 0
1+ω+ω2+ω3+ω4 = = M1
1−ω 1−ω
1−ω4 ω−ω5 ω−1
or ω+ω2+ω3+ω4 =ω =  = =−1 A1
 1−ω  1−ω  1−ω
Alternatively:
b
sum of roots = =− where b=0 M1 – needs
a
AnswerMarks
explanation – i.e. coefficient of z4 term = 0M1
A1
AnswerMarks
[2]1.1a
2.1Use equation and ω
AG
AnswerMarks
(d)AG
2
 1   1  1 1
ω+ + ω+ −1=ω2 +2+ +ω+ −1
   
 ω  ω ω2 ω
1 ( )
= ω4 +ω2 +1+ω3+ω =0
ω2
1
Since ≠0,ω4 +ω2 +1+ω3+ω=0
ω2
AnswerMarks
or from part (c)M1
A1
AnswerMarks
A12.1
1.1
AnswerMarks
2.2aMultiply out
Alternatively:
ω4 +ω3+ω2 +ω+1=0
 1 1 
⇒ω2  ω2 +ω+1+ +  =0 M1
 ω ω2 
 1   1  
⇒ω2  ω2 +2+  +  ω+  +1−2 =0 A1
 ω2   ω 
 2 
 1   1 
⇒ω2 ω+ + ω+ −1=0
  
  ω  ω 
 
2
 1   1 
Since ω2 ≠0,  ω+  +  ω+  −1=0 A1
AnswerMarks Guidance
 ω  ω[3] For extraction of ω2
For dealing with the 2
AnswerMarks
(e)1 2π 2π
=cos −isin
ω 5 5
 1  2π
⇒ ω+ =2cos
 
 ω 5
 1  −1± 5
From (iii) solving quadratic:  ω+  =
 ω 2
2π 5−1 2π 5−1
⇒2cos = ⇒cos =
5 2 5 4
1 5 1 1
=− + or − + 5 or −0.25+0.25 5
AnswerMarks
4 4 4 4B1
B1
M1
AnswerMarks
A13.1a
3.1a
2.2a
AnswerMarks
2.31
ω+ may be seen in (d)
ω
BC
Equating
For taking the valid value and
presenting in correct form
AnswerMarks
oeNo other forms
acceptable
[4]
Question 9:
9 | (a) | DR
2π 2π
ω=cos +isin
5 5
 2π 2π 5
⇒ω5 = cos +isin  =cos2π+isin2π=1+0i=1
 5 5  | M1
A1
[2] | 2.1
1.1 | Finding ω5
AG | Use of exponentials is
satisfactory
Could be argued
backwards
(b) | ω2,ω3,ω4,1 | B1
[1] | 1.1 | Alternative:
2kπ 2kπ
Roots are cos +isin
5 5
for k =2,3,4
and 1( or k = 5) | Exponentials
satisfactory
(c) | DR
ω5 −1=0
⇒(ω−1 )( ω4 +ω3 +ω2 +ω+1 ) =0
⇒ω4 +ω3 +ω2 +ω=−1
Alternatively:
1−ω5 0
1+ω+ω2+ω3+ω4 = = M1
1−ω 1−ω
1−ω4 ω−ω5 ω−1
or ω+ω2+ω3+ω4 =ω =  = =−1 A1
 1−ω  1−ω  1−ω
Alternatively:
b
sum of roots = =− where b=0 M1 – needs
a
explanation – i.e. coefficient of z4 term = 0 | M1
A1
[2] | 1.1a
2.1 | Use equation and ω
AG
(d) | AG
2
 1   1  1 1
ω+ + ω+ −1=ω2 +2+ +ω+ −1
   
 ω  ω ω2 ω
1 ( )
= ω4 +ω2 +1+ω3+ω =0
ω2
1
Since ≠0,ω4 +ω2 +1+ω3+ω=0
ω2
or from part (c) | M1
A1
A1 | 2.1
1.1
2.2a | Multiply out
Alternatively:
ω4 +ω3+ω2 +ω+1=0
 1 1 
⇒ω2  ω2 +ω+1+ +  =0 M1
 ω ω2 
 1   1  
⇒ω2  ω2 +2+  +  ω+  +1−2 =0 A1
 ω2   ω 
 2 
 1   1 
⇒ω2 ω+ + ω+ −1=0
  
  ω  ω 
 
2
 1   1 
Since ω2 ≠0,  ω+  +  ω+  −1=0 A1
 ω  ω | [3] | For extraction of ω2
For dealing with the 2
(e) | 1 2π 2π
=cos −isin
ω 5 5
 1  2π
⇒ ω+ =2cos
 
 ω 5
 1  −1± 5
From (iii) solving quadratic:  ω+  =
 ω 2
2π 5−1 2π 5−1
⇒2cos = ⇒cos =
5 2 5 4
1 5 1 1
=− + or − + 5 or −0.25+0.25 5
4 4 4 4 | B1
B1
M1
A1 | 3.1a
3.1a
2.2a
2.3 | 1
ω+ may be seen in (d)
ω
BC
Equating
For taking the valid value and
presenting in correct form
oe | No other forms
acceptable
[4]
9 In this question you must show detailed reasoning.\\
You are given the complex number $\omega = \cos \frac { 2 } { 5 } \pi + \mathrm { i } \sin \frac { 2 } { 5 } \pi$ and the equation $z ^ { 5 } = 1$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\omega$ is a root of the equation.
\item Write down the other four roots of the equation.
\item Show that $\omega + \omega ^ { 2 } + \omega ^ { 3 } + \omega ^ { 4 } = - 1$.
\item Hence show that $\left( \omega + \frac { 1 } { \omega } \right) ^ { 2 } + \left( \omega + \frac { 1 } { \omega } \right) - 1 = 0$.
\item Hence determine the value of $\cos \frac { 2 } { 5 } \pi$ in the form $a + b \sqrt { c }$ where $a , b$ and $c$ are rational numbers to be found.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 1 2019 Q9 [12]}}
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