| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Roots of unity with trigonometric identities |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on roots of unity that guides students through standard techniques: finding fifth roots of unity (routine), factorizing the cyclotomic polynomial (bookwork), manipulating factors with conjugate pairs, and using a substitution to find an exact trigonometric value. While it requires multiple steps and some algebraic manipulation, each part follows predictable patterns taught in FP2, with clear signposting between parts. The final result is a well-known identity, making this moderately above average but not requiring novel insight. |
| Spec | 4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers4.05b Transform equations: substitution for new roots |
8
\begin{enumerate}[label=(\alph*)]
\item Write down the five roots of the equation $z ^ { 5 } = 1$, giving your answers in the form $\mathrm { e } ^ { \mathrm { i } \theta }$, where $- \pi < \theta \leqslant \pi$.
\item Hence find the four linear factors of
$$z ^ { 4 } + z ^ { 3 } + z ^ { 2 } + z + 1$$
\item Deduce that
$$z ^ { 2 } + z + 1 + z ^ { - 1 } + z ^ { - 2 } = \left( z - 2 \cos \frac { 2 \pi } { 5 } + z ^ { - 1 } \right) \left( z - 2 \cos \frac { 4 \pi } { 5 } + z ^ { - 1 } \right)$$
\item Use the substitution $z + z ^ { - 1 } = w$ to show that $\cos \frac { 2 \pi } { 5 } = \frac { \sqrt { 5 } - 1 } { 4 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2012 Q8 [14]}}