| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Topic | Complex numbers 2 |
| Type | Roots of unity with trigonometric identities |
| Difficulty | Challenging +1.2 This is a structured roots of unity question with clear scaffolding through multiple parts. Part (i) involves routine verification of standard results (ω^5=1 is immediate, sum of roots is a standard geometric series result, and pairing conjugate roots for cosine is a well-known technique). Part (ii) requires forming a quadratic from sum and product of roots, which is a standard Further Maths technique. While it involves multiple steps and requires familiarity with roots of unity, the heavy scaffolding and standard nature of each component makes this easier than average for a Further Maths question, though still requiring more sophistication than typical single-maths content. |
| Spec | 4.02d Exponential form: re^(i*theta)4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02r nth roots: of complex numbers |
**(i)(a)** Either $\omega^5 = e^{2\pi i} = 1$ or $\omega$ a root of $z^5-1=0 \Rightarrow \omega^5=1$ **B1** [1]
**(b)** $0 = z^5-1 = (\omega-1)(\omega^4+\omega^3+\omega^2+\omega+1)$ **M1** or via trig. work with values
Since $\omega\neq 1$, $\omega^4+\omega^3+\omega^2+\omega = -1$ **A1** [2]
**(c)** $\omega+\omega^4 = \omega+\frac{1}{\omega} = \left(\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}\right)+\left(\cos\frac{2\pi}{5}-i\sin\frac{2\pi}{5}\right) = 2\cos\frac{2\pi}{5}$ **M1 A1**
or via $\left(\cos\frac{8\pi}{5}+i\sin\frac{8\pi}{5}\right)$
Similarly, $\omega^2+\omega^3 = 2\cos\frac{4\pi}{5}$ **B1** Ignore any spurious reasoning [3]
**(ii)** Then $\cos\frac{2\pi}{5}+\cos\frac{4\pi}{5} = \frac{1}{2}(\omega+\omega^4+\omega^2+\omega^3) = -\frac{1}{2}$ **B1**
And $\cos\frac{2\pi}{5}\times\cos\frac{4\pi}{5} = \frac{1}{2}(\omega+\omega^4)\frac{1}{2}(\omega^2+\omega^3) = \frac{1}{4}(\omega^3+\omega^4+\omega^6+\omega^7)$
$= \frac{1}{4}(\omega^3+\omega^4+\omega^6+\omega^7) = \frac{1}{4}(\omega^3+\omega^4+\omega+\omega^2) = -\frac{1}{4}$ **M1 A1**
Use of $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$ **M1** Allow $\pm$ (sum of roots)
$x^2+\frac{1}{2}x-\frac{1}{4}=0$ or $4x^2+2x-1=0$ **A1** With/out integers [5]10 One root of the equation $z ^ { 5 } - 1 = 0$ is the complex number $\omega = \mathrm { e } ^ { \frac { 2 } { 5 } \pi \mathrm { i } }$.\\
(i) Show that
\begin{enumerate}[label=(\alph*)]
\item $\quad \omega ^ { 5 } = 1$,
\item $\quad \omega + \omega ^ { 2 } + \omega ^ { 3 } + \omega ^ { 4 } = - 1$,
\item $\quad \omega + \omega ^ { 4 } = 2 \cos \frac { 2 } { 5 } \pi$, and write down a similar expression for $\omega ^ { 2 } + \omega ^ { 3 }$.\\
(ii) Using these results, find the values of $\cos \frac { 2 } { 5 } \pi + \cos \frac { 4 } { 5 } \pi$ and $\cos \frac { 2 } { 5 } \pi \times \cos \frac { 4 } { 5 } \pi$, and deduce a quadratic equation, with integer coefficients, which has roots
$$\cos \frac { 2 } { 5 } \pi \quad \text { and } \quad \cos \frac { 4 } { 5 } \pi$$
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2010 Q10 [11]}}