CAIE FP1 2012 June — Question 6 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeRoots of unity with derived equations
DifficultyChallenging +1.2 This is a structured Further Maths question on roots of unity requiring multiple steps: identifying fifth roots of unity (routine), algebraic manipulation to the given form, and using the relationship between roots of unity and cotangent expressions. While it involves several techniques and is from FP1, the question provides significant scaffolding and follows a standard pattern for this topic, making it moderately above average difficulty but not requiring novel insight.
Spec4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02r nth roots: of complex numbers
6 Write down the values of \(\theta\), in the interval \(0 \leqslant \theta < 2 \pi\), for which \(\cos \theta + \mathrm { i } \sin \theta\) is a fifth root of unity. By writing the equation \(( z + 1 ) ^ { 5 } = z ^ { 5 }\) in the form $$\left( \frac { z + 1 } { z } \right) ^ { 5 } = 1$$ show that its roots are $$- \frac { 1 } { 2 } \left\{ 1 + \mathrm { i } \cot \left( \frac { k \pi } { 5 } \right) \right\} , \quad k = 1,2,3,4$$
Question 6:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(1 = \cos(2k\pi) + i\sin(2k\pi)\), \(k = 0,1,2,3,4\)M1 Obtains fifth roots of unity by de Moivre's Theorem
\(\theta = \left(\frac{2k\pi}{5}\right)\), \(k=0,1,2,3,4\)A1
\((z+1)^5 = z^5 \Rightarrow \frac{(z+1)^5}{z^5} = 1 \Rightarrow \left(\frac{z+1}{z}\right)^5 = 1\)M1 Rewrites
\(\frac{z+1}{z} = \text{cis}\!\left(\frac{2k\pi}{5}\right) \Rightarrow z+1 = z\,\text{cis}\!\left(\frac{2k\pi}{5}\right)\)
\(\Rightarrow z\!\left(1 - \text{cis}\!\left(\frac{2k\pi}{5}\right)\right) = -1\)A1 and factorises
\(\Rightarrow z = \frac{-1}{1-\text{cis}\!\left(\frac{2k\pi}{5}\right)} = \frac{-\left(\text{cis}\!\left(-\frac{k\pi}{5}\right)\right)}{\text{cis}\!\left(-\frac{k\pi}{5}\right)-\text{cis}\!\left(\frac{k\pi}{5}\right)}\)M1A1 Isolates \(z\)
\(= \frac{-\cos\!\left(\frac{k\pi}{5}\right)+i\sin\!\left(\frac{k\pi}{5}\right)}{-2i\sin\!\left(\frac{k\pi}{5}\right)} = -\frac{1}{2} + \frac{1}{2i}\cot\!\left(\frac{k\pi}{5}\right)\), \(k=1,2,3\)A1 Obtains purely imaginary denominator
\(= -\frac{1}{2}\!\left(1 + i\cot\!\left(\frac{k\pi}{5}\right)\right)\), \(k=1,2,3,4\) (AG)A1 and obtains result
Observes original equation is a quartic with real coefficients, so roots occur in conjugate pairs and \(k=0\) must be rejectedB1
Total: [9]
## Question 6:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 = \cos(2k\pi) + i\sin(2k\pi)$, $k = 0,1,2,3,4$ | M1 | Obtains fifth roots of unity by de Moivre's Theorem |
| $\theta = \left(\frac{2k\pi}{5}\right)$, $k=0,1,2,3,4$ | A1 | |
| $(z+1)^5 = z^5 \Rightarrow \frac{(z+1)^5}{z^5} = 1 \Rightarrow \left(\frac{z+1}{z}\right)^5 = 1$ | M1 | Rewrites |
| $\frac{z+1}{z} = \text{cis}\!\left(\frac{2k\pi}{5}\right) \Rightarrow z+1 = z\,\text{cis}\!\left(\frac{2k\pi}{5}\right)$ | | |
| $\Rightarrow z\!\left(1 - \text{cis}\!\left(\frac{2k\pi}{5}\right)\right) = -1$ | A1 | and factorises |
| $\Rightarrow z = \frac{-1}{1-\text{cis}\!\left(\frac{2k\pi}{5}\right)} = \frac{-\left(\text{cis}\!\left(-\frac{k\pi}{5}\right)\right)}{\text{cis}\!\left(-\frac{k\pi}{5}\right)-\text{cis}\!\left(\frac{k\pi}{5}\right)}$ | M1A1 | Isolates $z$ |
| $= \frac{-\cos\!\left(\frac{k\pi}{5}\right)+i\sin\!\left(\frac{k\pi}{5}\right)}{-2i\sin\!\left(\frac{k\pi}{5}\right)} = -\frac{1}{2} + \frac{1}{2i}\cot\!\left(\frac{k\pi}{5}\right)$, $k=1,2,3$ | A1 | Obtains purely imaginary denominator |
| $= -\frac{1}{2}\!\left(1 + i\cot\!\left(\frac{k\pi}{5}\right)\right)$, $k=1,2,3,4$ (AG) | A1 | and obtains result |
| Observes original equation is a quartic with real coefficients, so roots occur in conjugate pairs and $k=0$ must be rejected | B1 | |

**Total: [9]**
6 Write down the values of $\theta$, in the interval $0 \leqslant \theta < 2 \pi$, for which $\cos \theta + \mathrm { i } \sin \theta$ is a fifth root of unity.

By writing the equation $( z + 1 ) ^ { 5 } = z ^ { 5 }$ in the form

$$\left( \frac { z + 1 } { z } \right) ^ { 5 } = 1$$

show that its roots are

$$- \frac { 1 } { 2 } \left\{ 1 + \mathrm { i } \cot \left( \frac { k \pi } { 5 } \right) \right\} , \quad k = 1,2,3,4$$

\hfill \mbox{\textit{CAIE FP1 2012 Q6 [9]}}
This paper (12 questions)
View full paper
Q1 5 Q2 5 Q3 6 Q4 9 Q5 9 Q6 9 Q7 10 Q8 11 Q9 11 Q10 11 Q11 EITHER Q11 OR