FP2 June 2013 Q4

EdexcelOld spec7 marksDe Moivre's Theorem Induction

(a) Given that \[z = r(\cos\theta + \mathrm{i}\sin\theta), \qquad r \in \mathbb{R}\] prove, by induction, that \(z^n = r^n(\cos n\theta + \mathrm{i}\sin n\theta)\), \(\quad n \in \mathbb{Z}^{+}\) (5)

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

(b) Find the exact value of \(w^5\), giving your answer in the form \(a + \mathrm{i}b\), where \(a, b \in \mathbb{R}\). (2)

Mark scheme (a)

Scheme	Marks
Assume true for \(n = k\): \(z^k = r^k(\cos k\theta + \mathrm{i}\sin k\theta)\)
\(n = k + 1:\ \ z^{k+1} = \left(z^k \times z =\right) r^k(\cos k\theta + \mathrm{i}\sin k\theta) \times r(\cos\theta + \mathrm{i}\sin\theta)\)	M1
\(= r^{k+1}\left(\cos k\theta\cos\theta - \sin k\theta\sin\theta + \mathrm{i}(\sin k\theta\cos\theta + \cos k\theta\sin\theta)\right)\)	M1
\(= r^{k+1}\left(\cos(k + 1)\theta + \mathrm{i}\sin(k + 1)\theta\right)\)	M1depA1cso
\(\therefore\) if true for \(n = k\), also true for \(n = k + 1\)
\(k = 1\quad \underline{z^1 = r^1(\cos\theta + \mathrm{i}\sin\theta)}\); True for \(n = 1\) \(\therefore\) true for all \(n\)	A1cso
Alternative: See notes for use of \(r\mathrm{e}^{\mathrm{i}\theta}\) form
	(5)

Notes

NB: Allow each mark if \(n\), \(n + 1\) used instead of \(k\), \(k + 1\)

M1 for using the result for \(n = k\) to write \(z^{k+1}\left(= z^k \times z\right) = r^k(\cos k\theta + \mathrm{i}\sin k\theta) \times r(\cos\theta + \mathrm{i}\sin\theta)\)

M1 for multiplying out and collecting real and imaginary parts, using \(\mathrm{i}^2 = -1\)
OR using sum of arguments and product of moduli to get \(r^{k+1}\left(\cos(k\theta + \theta) + \mathrm{i}\sin(k\theta + \theta)\right)\)

M1dep for using the addition formulae to obtain single cos and sin terms
OR factorise the argument \(r^{k+1}\left(\cos\theta(k + 1) + \mathrm{i}\sin\theta(k + 1)\right)\)
Dependent on the second M mark.

A1cso for \(r^{k+1}\left(\cos(k + 1)\theta + \mathrm{i}\sin(k + 1)\theta\right)\) Only give this mark if all previous steps are fully correct.

A1cso All 5 underlined statements must be seen

Alternative: Using Euler's form

Scheme	Marks
\(z = r(\cos\theta + \mathrm{i}\sin\theta) = r\mathrm{e}^{\mathrm{i}\theta}\)	M1 May not be seen explicitly
\(z^{k+1} = z^k \times z = \left(r\mathrm{e}^{\mathrm{i}\theta}\right)^k \times r\mathrm{e}^{\mathrm{i}\theta} = r^k\mathrm{e}^{\mathrm{i}k\theta} \times r\mathrm{e}^{\mathrm{i}\theta}\)	M1
\(= r^{k+1}\mathrm{e}^{\mathrm{i}(k+1)\theta}\)	M1dep on 2^nd M mark
\(= r^{k+1}\left(\cos(k + 1)\theta + \mathrm{i}\sin(k + 1)\theta\right)\)	A1cso
\(k = 1\quad z^1 = r^1(\cos\theta + \mathrm{i}\sin\theta)\)
True for \(n = 1\) \(\therefore\) true for all \(n\) etc	A1 cso All 5 underlined statements must be seen

Mark scheme (b)

Scheme	Marks
\(w = 3\left(\cos\dfrac{3\pi}{4} + \mathrm{i}\sin\dfrac{3\pi}{4}\right)\)
\(w^5 = 3^5\left(\cos\dfrac{15\pi}{4} + \mathrm{i}\sin\dfrac{15\pi}{4}\right)\)	M1
\(w^5 = 243\left(\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}\mathrm{i}\right)\) \(\left[= \dfrac{243\sqrt{2}}{2} - \dfrac{243\sqrt{2}}{2}\mathrm{i} \text{ or}\right]\) oe	A1
	(2)
	(7 marks)

Notes

M1 for attempting to apply de Moivre to \(w\) or attempting to expand \(w^5\) and collecting real and imaginary parts, but no need to simplify these.

A1cao for \(243\left(\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}\mathrm{i}\right)\) \(\left[= \dfrac{243\sqrt{2}}{2} - \dfrac{243\sqrt{2}}{2}\mathrm{i}\right]\) (oe eg \(3^5\) instead of 243)