Edexcel CP2 2024 June — Question 7 9 marks

Exam BoardEdexcel
ModuleCP2 (Core Pure 2)
Year2024
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Arithmetic
TypeRoots of unity and special equations
DifficultyModerate -0.8 This is a structured, multi-part question on roots of unity with clear scaffolding. Parts (a) and (b) are standard textbook exercises requiring direct application of De Moivre's theorem. Part (c) is straightforward verification. Part (d) uses the hint from (c) to transform the equation into the form from (a). While it covers several techniques, each step is routine for Core Pure 2 students with no novel problem-solving required.
Spec4.02k Argand diagrams: geometric interpretation4.02r nth roots: of complex numbers

  1. (a) Determine the roots of the equation
$$z ^ { 6 } = 1$$ giving your answers in the form \(\mathrm { e } ^ { \mathrm { i } \theta }\) where \(0 \leqslant \theta < 2 \pi\) (b) Show the roots of the equation in part (a) on a single Argand diagram.
(c) Show that $$( \sqrt { 3 } + i ) ^ { 6 } = - 64$$ (d) Hence, or otherwise, solve the equation $$z ^ { 6 } + 64 = 0$$ giving your answers in the form \(r \mathrm { e } ^ { \mathrm { i } \theta }\) where \(0 \leqslant \theta < 2 \pi\)

Question 7:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(z = e^{\frac{k\pi}{3}i}\), \(k = 0, 1, 2, 3, 4, 5\)M1 For sight of \(e^{\frac{k\pi}{3}i}\), accept any value of \(k\)
All six roots fully definedA1 All six roots listed with \(\theta\) in given range, no incorrect/extra values; condone \(e^0\) and/or \(-1\) for \(e^{\pi i}\)
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
6 points forming a hexagon with point on positive real axis and point on negative real axis, one point in each quadrantB1 Points form hexagon; do not concern about distance from centre
Hexagon centred at origin, axes as lines of symmetrydB1 Centre at origin; axes need not be labelled
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\((\sqrt{3}+i)^6 = \left(2e^{\frac{\pi}{6}i}\right)^6 = 64e^{i\pi} = -64\)M1 Valid method shown
\(-64\)A1* Correct answer with no errors, printed answer confirmed
Part (d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(r = 2\)B1 Correct modulus
\(z = 2e^{\frac{\pi}{6}i} \times e^{\frac{k\pi}{3}i}\), \(k = 0, 1, 2, 3, 4, 5\)M1 Multiplies their roots from (a) by \(2e^{\frac{\pi}{6}i}\)
\(z = 2e^{\left(\frac{\pi}{6}+\frac{k\pi}{3}\right)i}\), \(k = 0, 1, 2, 3, 4, 5\)A1 All six roots correct
Question 7 (Complex Numbers - parts c and d):
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Converts \(\sqrt{3}+i\) to polar form \(re^{i\theta}\) with \(r=2\) or \(\theta=\frac{\pi}{6}\), applies power of 6 to obtain \(r^6e^{6\theta i}\)M1
Obtains given answer; minimum: \(2^6e^{\frac{6\pi i}{6}}=-64\) or \(2^6e^{\pi i}=-64\)A1* If \(r=-2\) seen in workings, withhold this mark
OR: Converts to modulus-argument form \(r(\cos\theta+i\sin\theta)\) with \(r=2\) or \(\theta=\frac{\pi}{6}\), applies power of 6 to obtain \(r^6(\cos 6\theta+i\sin 6\theta)\)M1
OR: Attempts full expansion of \((\sqrt{3}+i)^6\) using binomial expansion with 7 terms, correct coefficients, \(a=\sqrt{3}\), \(b=i\), \(n=6\)M1
Obtains given answer with at least one intermediate lineA1* No brackets, no irrational numbers, no \(i\) terms in simplified answer
Part (d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Deduces \(r=2\) (only)B1
Obtains at least one value of \(z\) in form \(re^{i\theta}\) with consistent \(r\), and \(\theta\) taking one of \(\left\{\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{7\pi}{6},\frac{3\pi}{2},\frac{11\pi}{6}\right\}\)M1
\(2e^{\frac{\pi}{6}i},\ 2e^{\frac{\pi}{2}i},\ 2e^{\frac{5\pi}{6}i},\ 2e^{\frac{7\pi}{6}i},\ 2e^{\frac{3\pi}{2}i},\ 2e^{\frac{11\pi}{6}i}\) with no incorrect or extra valuesA1 Accept unsimplified arguments e.g. \(2e^{\frac{9\pi}{6}i}\); ensure \(i\) and \(\pi\) present; accept \(2e^{\frac{\pi}{2}i}\) as \(2i\) and \(2e^{\frac{3\pi}{2}i}\) as \(-2i\)
# Question 7:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = e^{\frac{k\pi}{3}i}$, $k = 0, 1, 2, 3, 4, 5$ | M1 | For sight of $e^{\frac{k\pi}{3}i}$, accept any value of $k$ |
| All six roots fully defined | A1 | All six roots listed with $\theta$ in given range, no incorrect/extra values; condone $e^0$ and/or $-1$ for $e^{\pi i}$ |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| 6 points forming a hexagon with point on positive real axis and point on negative real axis, one point in each quadrant | B1 | Points form hexagon; do not concern about distance from centre |
| Hexagon centred at origin, axes as lines of symmetry | dB1 | Centre at origin; axes need not be labelled |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\sqrt{3}+i)^6 = \left(2e^{\frac{\pi}{6}i}\right)^6 = 64e^{i\pi} = -64$ | M1 | Valid method shown |
| $-64$ | A1* | Correct answer with no errors, printed answer confirmed |

## Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = 2$ | B1 | Correct modulus |
| $z = 2e^{\frac{\pi}{6}i} \times e^{\frac{k\pi}{3}i}$, $k = 0, 1, 2, 3, 4, 5$ | M1 | Multiplies their roots from (a) by $2e^{\frac{\pi}{6}i}$ |
| $z = 2e^{\left(\frac{\pi}{6}+\frac{k\pi}{3}\right)i}$, $k = 0, 1, 2, 3, 4, 5$ | A1 | All six roots correct |

# Question 7 (Complex Numbers - parts c and d):

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Converts $\sqrt{3}+i$ to polar form $re^{i\theta}$ with $r=2$ or $\theta=\frac{\pi}{6}$, applies power of 6 to obtain $r^6e^{6\theta i}$ | M1 | |
| Obtains given answer; minimum: $2^6e^{\frac{6\pi i}{6}}=-64$ or $2^6e^{\pi i}=-64$ | A1* | If $r=-2$ seen in workings, withhold this mark |
| **OR:** Converts to modulus-argument form $r(\cos\theta+i\sin\theta)$ with $r=2$ or $\theta=\frac{\pi}{6}$, applies power of 6 to obtain $r^6(\cos 6\theta+i\sin 6\theta)$ | M1 | |
| **OR:** Attempts full expansion of $(\sqrt{3}+i)^6$ using binomial expansion with 7 terms, correct coefficients, $a=\sqrt{3}$, $b=i$, $n=6$ | M1 | |
| Obtains given answer with at least one intermediate line | A1* | No brackets, no irrational numbers, no $i$ terms in simplified answer |

## Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces $r=2$ (only) | B1 | |
| Obtains at least one value of $z$ in form $re^{i\theta}$ with consistent $r$, and $\theta$ taking one of $\left\{\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{7\pi}{6},\frac{3\pi}{2},\frac{11\pi}{6}\right\}$ | M1 | |
| $2e^{\frac{\pi}{6}i},\ 2e^{\frac{\pi}{2}i},\ 2e^{\frac{5\pi}{6}i},\ 2e^{\frac{7\pi}{6}i},\ 2e^{\frac{3\pi}{2}i},\ 2e^{\frac{11\pi}{6}i}$ with no incorrect or extra values | A1 | Accept unsimplified arguments e.g. $2e^{\frac{9\pi}{6}i}$; ensure $i$ and $\pi$ present; accept $2e^{\frac{\pi}{2}i}$ as $2i$ and $2e^{\frac{3\pi}{2}i}$ as $-2i$ |

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\begin{enumerate}
  \item (a) Determine the roots of the equation
\end{enumerate}

$$z ^ { 6 } = 1$$

giving your answers in the form $\mathrm { e } ^ { \mathrm { i } \theta }$ where $0 \leqslant \theta < 2 \pi$\\
(b) Show the roots of the equation in part (a) on a single Argand diagram.\\
(c) Show that

$$( \sqrt { 3 } + i ) ^ { 6 } = - 64$$

(d) Hence, or otherwise, solve the equation

$$z ^ { 6 } + 64 = 0$$

giving your answers in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$ where $0 \leqslant \theta < 2 \pi$

\hfill \mbox{\textit{Edexcel CP2 2024 Q7 [9]}}