OCR FP3 2007 June — Question 7 10 marks

Exam BoardOCR
ModuleFP3 (Further Pure Mathematics 3)
Year2007
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeBasic roots of unity properties
DifficultyStandard +0.3 This is a structured question on roots of unity with clear steps: (i) is algebraic verification using Euler's formula, (ii) is standard recall of 7th roots of unity, and (iii) applies part (i) to factor the polynomial by pairing conjugate roots. While it requires understanding of complex exponentials and systematic application across multiple parts, each step follows a well-established procedure taught in FP3 with no novel insight required.
Spec4.02k Argand diagrams: geometric interpretation4.02q De Moivre's theorem: multiple angle formulae

  1. Show that \(\left( z - \mathrm { e } ^ { \mathrm { i } \phi } \right) \left( z - \mathrm { e } ^ { - \mathrm { i } \phi } \right) \equiv z ^ { 2 } - ( 2 \cos \phi ) z + 1\).
  2. Write down the seven roots of the equation \(z ^ { 7 } = 1\) in the form \(\mathrm { e } ^ { \mathrm { i } \theta }\) and show their positions in an Argand diagram.
  3. Hence express \(z ^ { 7 } - 1\) as the product of one real linear factor and three real quadratic factors.

AnswerMarks Guidance
(i) \((z - e^{i\theta})(z - e^{-i\theta}) = z^2 - (2) \frac{e^{i\theta} + e^{-i\theta}}{2} + 1\)B1 1 For correct justification AG
\(= z^2 - (2\cos\theta)z + 1\)
AnswerMarks Guidance
(ii) \(z = e^{\frac{2}{7}\pi i}\)B1 For general form OR any one non-real root
for \(k = 0, 1, 2, 3, 4, 5, 6\) OR \(0, \pm 1, \pm 2, \pm 3\)B1 For other roots specified (\(k=0\) may be seen in any form, eg \(1\), \(e^{2\pi i}\))
For answers in form \(\cos\theta + i\sin\theta\) allow maximum B1 B0B1 For answers in form \(\cos\theta + i\sin\theta\) allow maximum B1 B0
B1For any 7 points equally spaced round unit circle (circumference need not be shown)
B1 4For 1 point on \(+\)ve real axis, and other points in correct quadrants
(iii) \((z^7 - 1) = (z-1)(z-e^{\frac{2\pi i}{7}})(z - e^{\frac{4\pi i}{7}})\)M1 For using linear factors from (ii), seen or implied
\((z-e^{\frac{6\pi i}{7}})(z - e^{\frac{-6\pi i}{7}})(z-e^{\frac{-4\pi i}{7}})(z-e^{\frac{-2\pi i}{7}})\)
\((z-e^{\frac{2\pi i}{7}})(z-e^{\frac{-2\pi i}{7}})(z-e^{\frac{4\pi i}{7}})(z-e^{\frac{-4\pi i}{7}})\)
\((z-e^{\frac{6\pi i}{7}})(z-e^{\frac{-6\pi i}{7}}) \times\)
AnswerMarks Guidance
\(\times (z-1)\)M1 For identifying at least one pair of complex conjugate factors
\(= (z^2 - (2\cos \frac{2}{7}\pi)z + 1) \times\)B1 For linear factor seen
\(\times (z^2 - (2\cos \frac{4}{7}\pi)z + 1) \times (z^2 - (2\cos \frac{6}{7}\pi)z + 1) \times (z-1)\)A1 5 For the other 2 quadratic factors and expression written as product of 4 factors
**(i)** $(z - e^{i\theta})(z - e^{-i\theta}) = z^2 - (2) \frac{e^{i\theta} + e^{-i\theta}}{2} + 1$ | B1 1 | For correct justification AG

$= z^2 - (2\cos\theta)z + 1$

**(ii)** $z = e^{\frac{2}{7}\pi i}$ | B1 | For general form OR any one non-real root

for $k = 0, 1, 2, 3, 4, 5, 6$ OR $0, \pm 1, \pm 2, \pm 3$ | B1 | For other roots specified ($k=0$ may be seen in any form, eg $1$, $e^{2\pi i}$)

For answers in form $\cos\theta + i\sin\theta$ allow maximum B1 B0 | B1 | For answers in form $\cos\theta + i\sin\theta$ allow maximum B1 B0

| B1 | For any 7 points equally spaced round unit circle (circumference need not be shown)

| B1 4 | For 1 point on $+$ve real axis, and other points in correct quadrants

**(iii)** $(z^7 - 1) = (z-1)(z-e^{\frac{2\pi i}{7}})(z - e^{\frac{4\pi i}{7}})$ | M1 | For using linear factors from (ii), seen or implied

$(z-e^{\frac{6\pi i}{7}})(z - e^{\frac{-6\pi i}{7}})(z-e^{\frac{-4\pi i}{7}})(z-e^{\frac{-2\pi i}{7}})$

$(z-e^{\frac{2\pi i}{7}})(z-e^{\frac{-2\pi i}{7}})(z-e^{\frac{4\pi i}{7}})(z-e^{\frac{-4\pi i}{7}})$

$(z-e^{\frac{6\pi i}{7}})(z-e^{\frac{-6\pi i}{7}}) \times$

$\times (z-1)$ | M1 | For identifying at least one pair of complex conjugate factors

$= (z^2 - (2\cos \frac{2}{7}\pi)z + 1) \times$ | B1 | For linear factor seen

$\times (z^2 - (2\cos \frac{4}{7}\pi)z + 1) \times (z^2 - (2\cos \frac{6}{7}\pi)z + 1) \times (z-1)$ | A1 5 | For the other 2 quadratic factors and expression written as product of 4 factors
(i) Show that $\left( z - \mathrm { e } ^ { \mathrm { i } \phi } \right) \left( z - \mathrm { e } ^ { - \mathrm { i } \phi } \right) \equiv z ^ { 2 } - ( 2 \cos \phi ) z + 1$.\\
(ii) Write down the seven roots of the equation $z ^ { 7 } = 1$ in the form $\mathrm { e } ^ { \mathrm { i } \theta }$ and show their positions in an Argand diagram.\\
(iii) Hence express $z ^ { 7 } - 1$ as the product of one real linear factor and three real quadratic factors.

\hfill \mbox{\textit{OCR FP3 2007 Q7 [10]}}