CAIE FP1 2013 November — Question 16 EITHER

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionNovember
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeBasic roots of unity properties
DifficultyStandard +0.3 This is a structured multi-part question on roots of unity that follows standard Further Maths patterns. While it requires knowledge of De Moivre's theorem and complex conjugate pairs, each part guides students through familiar techniques: listing roots of unity, multiplying conjugate pairs to get real quadratics, and factorizing polynomials. The final part requires recognizing that x^6 - x^3 + 1 relates to 9th roots of unity, which is slightly less routine but still a standard FP1 exercise. Overall, this is slightly easier than average for A-level due to its structured, guided nature.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem4.02j Cubic/quartic equations: conjugate pairs and factor theorem4.02r nth roots: of complex numbers

State the fifth roots of unity in the form \(\cos \theta + \mathrm { i } \sin \theta\), where \(- \pi < \theta \leqslant \pi\). Simplify $$\left( x - \left[ \cos \frac { 2 } { 5 } \pi + i \sin \frac { 2 } { 5 } \pi \right] \right) \left( x - \left[ \cos \frac { 2 } { 5 } \pi - i \sin \frac { 2 } { 5 } \pi \right] \right)$$ Hence find the real factors of $$x ^ { 5 } - 1$$ Express the six roots of the equation $$x ^ { 6 } - x ^ { 3 } + 1 = 0$$ as three conjugate pairs, in the form \(\cos \theta \pm \mathrm { i } \sin \theta\). Hence find the real factors of $$x ^ { 6 } - x ^ { 3 } + 1$$

State the fifth roots of unity in the form $\cos \theta + \mathrm { i } \sin \theta$, where $- \pi < \theta \leqslant \pi$.

Simplify

$$\left( x - \left[ \cos \frac { 2 } { 5 } \pi + i \sin \frac { 2 } { 5 } \pi \right] \right) \left( x - \left[ \cos \frac { 2 } { 5 } \pi - i \sin \frac { 2 } { 5 } \pi \right] \right)$$

Hence find the real factors of

$$x ^ { 5 } - 1$$

Express the six roots of the equation

$$x ^ { 6 } - x ^ { 3 } + 1 = 0$$

as three conjugate pairs, in the form $\cos \theta \pm \mathrm { i } \sin \theta$.

Hence find the real factors of

$$x ^ { 6 } - x ^ { 3 } + 1$$

\hfill \mbox{\textit{CAIE FP1 2013 Q16 EITHER}}