| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Geometric properties in Argand diagram |
| Difficulty | Standard +0.3 This is a standard FP3 question on cube roots of unity and their geometric interpretation. Part (i) is routine algebraic manipulation, part (ii) requires knowing that ω represents 120° rotation (core FP3 knowledge), and part (iii) is straightforward algebra using the given relation and 1+ω+ω²=0. While it involves multiple parts and geometric reasoning, these are well-practiced techniques in Further Maths with clear signposting, making it slightly easier than average overall. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02b Express complex numbers: cartesian and modulus-argument forms4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| EITHER \(1 + \omega + \omega^2\) | M1 | For result shown by any correct method AG |
| A1 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Multiplication by \(\omega \Rightarrow\) rotation through \(\frac{2}{3}\pi\) \(\angle \text{ to }\) (⊐) | B1 | For correct interpretation of \(\times\) by \(\omega\) (allow \(120°\) and omission of, or error in, ⊐) |
| \(z_1 - z_3 = \overrightarrow{CA}, \quad z_3 - z_2 = \overrightarrow{BC}\) | B1 | For identification of vectors soi (ignore direction errors) |
| \(\overrightarrow{BC}\) rotates through \(\frac{2}{3}\pi\) to direction of \(\overrightarrow{CA}\) | M1 | For linking \(BC\) and \(CA\) by rotation of \(\frac{2}{3}\pi\) OR \(\omega\) |
| \(\triangle ABC\) has \(BC = CA\), hence result | A1 4 | For stating equal magnitudes \(\Rightarrow\) AG |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \(\Rightarrow z_1 + \omega z_2 - (1 + \omega)z_3 = 0\) | M1 | For using \(1 + \omega + \omega^2 = 0\) in (ii) |
| \(1 + \omega + \omega^2 = 0 \Rightarrow z_1 + \omega z_2 + \omega^2z_3 = 0\) | A1 2 | For obtaining AG |
## (i)
EITHER $1 + \omega + \omega^2$ | M1 | For result shown by any correct method **AG**
| A1 2 |
$= $ sum of roots of $(z^3 - 1 = 0) = 0$
OR $\omega^3 = 1 \Rightarrow (\omega - 1)(\omega^2 + \omega + 1) = 0$
$\Rightarrow 1 + \omega + \omega^2 = 0$ (for $\omega \neq 1$)
OR sum of G.P.
$1 + \omega + \omega^2 = \frac{1 - \omega^3}{1 - \omega} = \frac{0}{1-\omega} = 0$
OR shown on Argand diagram or explained in terms of vectors
OR $1 + \cis\frac{2}{3}\pi + \cis\frac{4}{3}\pi = 1 + (-\frac{1}{2} + \frac{\sqrt{3}}{2}i) + (-\frac{1}{2} - \frac{\sqrt{3}}{2}i) = 0$
## (ii)
Multiplication by $\omega \Rightarrow$ rotation through $\frac{2}{3}\pi$ $\angle \text{ to }$ (⊐) | B1 | For correct interpretation of $\times$ by $\omega$ (allow $120°$ and omission of, or error in, ⊐)
$z_1 - z_3 = \overrightarrow{CA}, \quad z_3 - z_2 = \overrightarrow{BC}$ | B1 | For identification of vectors soi (ignore direction errors)
$\overrightarrow{BC}$ rotates through $\frac{2}{3}\pi$ to direction of $\overrightarrow{CA}$ | M1 | For linking $BC$ and $CA$ by rotation of $\frac{2}{3}\pi$ OR $\omega$
$\triangle ABC$ has $BC = CA$, hence result | A1 4 | For stating equal magnitudes $\Rightarrow$ **AG**
## (iii)
(ii) $\Rightarrow z_1 + \omega z_2 - (1 + \omega)z_3 = 0$ | M1 | For using $1 + \omega + \omega^2 = 0$ in (ii)
$1 + \omega + \omega^2 = 0 \Rightarrow z_1 + \omega z_2 + \omega^2z_3 = 0$ | A1 2 | For obtaining **AG**
---The cube roots of 1 are denoted by $1$, $\omega$ and $\omega^2$, where the imaginary part of $\omega$ is positive.
\begin{enumerate}[label=(\roman*)]
\item Show that $1 + \omega + \omega^2 = 0$. [2]
\end{enumerate}
\includegraphics{figure_1}
In the diagram, $ABC$ is an equilateral triangle, labelled anticlockwise. The points $A$, $B$ and $C$ represent the complex numbers $z_1$, $z_2$ and $z_3$ respectively.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item State the geometrical effect of multiplication by $\omega$ and hence explain why $z_1 - z_3 = \omega(z_3 - z_2)$. [4]
\item Hence show that $z_1 + \omega z_2 + \omega^2 z_3 = 0$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2011 Q4 [8]}}