Edexcel FP2 2013 June — Question 4 7 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve de Moivre's theorem
DifficultyStandard +0.8 This is a standard Further Maths proof by induction of de Moivre's theorem followed by a routine application. Part (a) requires formal inductive proof structure with compound angle formulae, which is more demanding than typical A-level proofs. Part (b) is straightforward application. Overall slightly above average difficulty due to the proof component and Further Maths context.
Spec4.01a Mathematical induction: construct proofs4.02q De Moivre's theorem: multiple angle formulae

4. (a) Given that $$z = r ( \cos \theta + \mathrm { i } \sin \theta ) , \quad r \in \mathbb { R }$$ prove, by induction, that \(z ^ { n } = r ^ { n } ( \cos n \theta + \mathrm { i } \sin n \theta ) , \quad n \in \mathbb { Z } ^ { + }\) $$w = 3 \left( \cos \frac { 3 \pi } { 4 } + i \sin \frac { 3 \pi } { 4 } \right)$$ (b) Find the exact value of \(w ^ { 5 }\), giving your answer in the form \(a + \mathrm { i } b\), where \(a , b \in \mathbb { R }\).

Question 4:
Part (a):
AnswerMarks Guidance
AnswerMark Guidance
Assume true for \(n=k\): \(z^k = r^k(\cos k\theta + \mathrm{i}\sin k\theta)\)
\(n=k+1\): \(z^{k+1} = (z^k \times z =) r^k(\cos k\theta + \mathrm{i}\sin k\theta) \times r(\cos\theta + \mathrm{i}\sin\theta)\)M1
\(= r^{k+1}(\cos k\theta\cos\theta - \sin k\theta\sin\theta + \mathrm{i}(\sin k\theta\cos\theta + \cos k\theta\sin\theta))\)M1
\(= r^{k+1}(\cos(k+1)\theta + \mathrm{i}\sin(k+1)\theta)\)M1dep A1 cso
\(\therefore\) if true for \(n=k\), also true for \(n=k+1\)
\(k=1\): \(z^1 = r^1(\cos\theta + \mathrm{i}\sin\theta)\); True for \(n=1\), \(\therefore\) true for all \(n\)A1 cso (5)
Part (b):
AnswerMarks Guidance
AnswerMark Guidance
\(w = 3\left(\cos\dfrac{3\pi}{4} + \mathrm{i}\sin\dfrac{3\pi}{4}\right)\)
\(w^5 = 3^5\left(\cos\dfrac{15\pi}{4} + \mathrm{i}\sin\dfrac{15\pi}{4}\right)\)M1
\(w^5 = 243\left(\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}\mathrm{i}\right)\) \(\left[= \dfrac{243\sqrt{2}}{2} - \dfrac{243\sqrt{2}}{2}\mathrm{i}\right]\) or equivalentA1 (2)
Question 4:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(z^{k+1} = z^k \times z = r^k(\cos k\theta + i\sin k\theta) \times r(\cos\theta + i\sin\theta)\)M1 Using result for \(n=k\) to write \(z^{k+1}\)
Multiplying out and collecting real and imaginary parts using \(i^2 = -1\), OR using sum of arguments and product of moduli to get \(r^{k+1}(\cos(k\theta+\theta)+i\sin(k\theta+\theta))\)M1
Using addition formulae to obtain single \(\cos\) and \(\sin\) terms, OR factorise argument \(r^{k+1}(\cos\theta(k+1)+i\sin\theta(k+1))\)M1dep Dependent on second M mark
\(r^{k+1}(\cos(k+1)\theta + i\sin(k+1)\theta)\)A1cso Only if all previous steps fully correct
All 5 underlined statements seenA1cso
Alternative (Euler's form):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(z = r(\cos\theta + i\sin\theta) = re^{i\theta}\)M1 May not be seen explicitly
\(z^{k+1} = z^k \times z = (re^{i\theta})^k \times re^{i\theta} = r^k e^{ik\theta} \times re^{i\theta}\)M1
\(= r^{k+1}e^{i(k+1)\theta}\)M1dep Dependent on 2nd M mark
\(= r^{k+1}(\cos(k+1)\theta + i\sin(k+1)\theta)\)A1cso
\(k=1\): \(z^1 = r^1(\cos\theta + i\sin\theta)\); True for \(n=1\) \(\therefore\) true for all \(n\) etcA1cso All 5 underlined statements must be seen
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Attempting to apply de Moivre to \(w\) or attempting to expand \(w^5\) and collecting real and imaginary partsM1 No need to simplify
\(243\left(\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}i\right) \left[= \dfrac{243\sqrt{2}}{2} - \dfrac{243\sqrt{2}}{2}i\right]\)A1cao oe e.g. \(3^5\) instead of 243
## Question 4:

### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Assume true for $n=k$: $z^k = r^k(\cos k\theta + \mathrm{i}\sin k\theta)$ | | |
| $n=k+1$: $z^{k+1} = (z^k \times z =) r^k(\cos k\theta + \mathrm{i}\sin k\theta) \times r(\cos\theta + \mathrm{i}\sin\theta)$ | M1 | |
| $= r^{k+1}(\cos k\theta\cos\theta - \sin k\theta\sin\theta + \mathrm{i}(\sin k\theta\cos\theta + \cos k\theta\sin\theta))$ | M1 | |
| $= r^{k+1}(\cos(k+1)\theta + \mathrm{i}\sin(k+1)\theta)$ | M1dep A1 cso | |
| $\therefore$ if true for $n=k$, also true for $n=k+1$ | | |
| $k=1$: $z^1 = r^1(\cos\theta + \mathrm{i}\sin\theta)$; True for $n=1$, $\therefore$ true for all $n$ | A1 cso (5) | |

### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $w = 3\left(\cos\dfrac{3\pi}{4} + \mathrm{i}\sin\dfrac{3\pi}{4}\right)$ | | |
| $w^5 = 3^5\left(\cos\dfrac{15\pi}{4} + \mathrm{i}\sin\dfrac{15\pi}{4}\right)$ | M1 | |
| $w^5 = 243\left(\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}\mathrm{i}\right)$ $\left[= \dfrac{243\sqrt{2}}{2} - \dfrac{243\sqrt{2}}{2}\mathrm{i}\right]$ or equivalent | A1 (2) | |

# Question 4:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $z^{k+1} = z^k \times z = r^k(\cos k\theta + i\sin k\theta) \times r(\cos\theta + i\sin\theta)$ | M1 | Using result for $n=k$ to write $z^{k+1}$ |
| Multiplying out and collecting real and imaginary parts using $i^2 = -1$, OR using sum of arguments and product of moduli to get $r^{k+1}(\cos(k\theta+\theta)+i\sin(k\theta+\theta))$ | M1 | |
| Using addition formulae to obtain single $\cos$ and $\sin$ terms, OR factorise argument $r^{k+1}(\cos\theta(k+1)+i\sin\theta(k+1))$ | M1dep | Dependent on second M mark |
| $r^{k+1}(\cos(k+1)\theta + i\sin(k+1)\theta)$ | A1cso | Only if all previous steps fully correct |
| All 5 underlined statements seen | A1cso | |

**Alternative (Euler's form):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$ | M1 | May not be seen explicitly |
| $z^{k+1} = z^k \times z = (re^{i\theta})^k \times re^{i\theta} = r^k e^{ik\theta} \times re^{i\theta}$ | M1 | |
| $= r^{k+1}e^{i(k+1)\theta}$ | M1dep | Dependent on 2nd M mark |
| $= r^{k+1}(\cos(k+1)\theta + i\sin(k+1)\theta)$ | A1cso | |
| $k=1$: $z^1 = r^1(\cos\theta + i\sin\theta)$; True for $n=1$ $\therefore$ true for all $n$ etc | A1cso | All 5 underlined statements must be seen |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempting to apply de Moivre to $w$ or attempting to expand $w^5$ and collecting real and imaginary parts | M1 | No need to simplify |
| $243\left(\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}i\right) \left[= \dfrac{243\sqrt{2}}{2} - \dfrac{243\sqrt{2}}{2}i\right]$ | A1cao | oe e.g. $3^5$ instead of 243 |

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4. (a) Given that

$$z = r ( \cos \theta + \mathrm { i } \sin \theta ) , \quad r \in \mathbb { R }$$

prove, by induction, that $z ^ { n } = r ^ { n } ( \cos n \theta + \mathrm { i } \sin n \theta ) , \quad n \in \mathbb { Z } ^ { + }$

$$w = 3 \left( \cos \frac { 3 \pi } { 4 } + i \sin \frac { 3 \pi } { 4 } \right)$$

(b) Find the exact value of $w ^ { 5 }$, giving your answer in the form $a + \mathrm { i } b$, where $a , b \in \mathbb { R }$.\\

\hfill \mbox{\textit{Edexcel FP2 2013 Q4 [7]}}