4.02q De Moivre's theorem: multiple angle formulae

The Argand diagram shows the solutions to the equation \(z^5 = 1\) \includegraphics{figure_3}

1. Solve the equation $$z^5 = 1$$ giving your answers in the form \(z = \cos\theta + i\sin\theta\), where \(0 \leq \theta < 2\pi\) [2 marks]
2. Explain why the points on an Argand diagram which represent the solutions found in part (a) are the vertices of a regular pentagon. [2 marks]
3. Show that if \(c = \cos\theta\), where \(z = \cos\theta + i\sin\theta\) is a solution to the equation \(z^5 = 1\), then \(c\) satisfies the equation $$16c^5 - 20c^3 + 5c - 1 = 0$$ [5 marks]
4. The Argand diagram on page 22 is repeated below. \includegraphics{figure_4} Explain, with reference to the Argand diagram, why the expression $$16c^5 - 20c^3 + 5c - 1$$ has a repeated quadratic factor. [3 marks]
5. \(O\) is the centre of a regular pentagon \(ABCDE\) such that \(OA = OB = OC = OD = OE = 1\) unit. The distance from \(O\) to \(AB\) is \(h\) By solving the equation \(16c^5 - 20c^3 + 5c - 1 = 0\), show that $$h = \frac{\sqrt{5} + 1}{4}$$ [5 marks]

The complex numbers \(z\) and \(w\) are defined by $$z = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4}$$ and $$w = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6}$$ By evaluating the product \(zw\), show that $$\tan\frac{5\pi}{12} = 2 + \sqrt{3}$$ [6 marks]

Find the smallest value \(\theta\) of for which \((\cos \theta + i \sin \theta)^5 = \frac{1}{\sqrt{2}}(1 - i)\) \(\{\theta \in \mathbb{R} : \theta > 0\}\) [4 marks]

Use de Moivre's theorem to find the constants \(A\), \(B\) and \(C\) in the identity \(\sin^3 \theta \equiv A \sin \theta + B \sin 3\theta + C \sin 5\theta\). [4]

1. Given that \(z^n + \frac{1}{z^n} = 2\cos n\theta\), where \(z = \cos\theta + \mathrm{i}\sin\theta\), express \(16\cos^4\theta\) in the form $$a\cos 4\theta + b\cos 2\theta + c,$$ where \(a\), \(b\), \(c\) are integers whose values are to be determined. [5]

The diagram below shows a sketch of the curve C with polar equation $$r = 3 - 4\cos^2\theta, \quad \text{where } \frac{\pi}{6} \leq \theta \leq \frac{5\pi}{6}.$$ \includegraphics{figure_4}

1. Calculate the area of the region enclosed by the curve C. [8]
2. Find the exact polar coordinates of the points on C at which the tangent is perpendicular to the initial line. [8]

Let \(C = \sum_{r=0}^{20} \binom{20}{r} \cos(r\theta)\). Show that \(C = 2^{20} \cos^{20}\left(\frac{1}{2}\theta\right) \cos(10\theta)\). [8]

In this question you must show detailed reasoning. The complex number \(-4 + i\sqrt{48}\) is denoted by \(z\).

1. Determine the cube roots of \(z\), giving the roots in exponential form. [6]

The points which represent the cube roots of \(z\) are denoted by \(A\), \(B\) and \(C\) and these form a triangle in an Argand diagram.

1. Write down the angles that any lines of symmetry of triangle \(ABC\) make with the positive real axis, justifying your answer. [3]

Use de Moivre's theorem to express \(\cos 3\theta\) in terms of powers of \(\cos \theta\) only, and deduce the identity \(\cos 6x \equiv \cos 2x(2\cos 4x - 1)\). [5]

Given that \(w_n = 3^{-n} \cos 2n\theta\) for \(n = 1, 2, 3, \ldots\), use de Moivre's theorem to show that $$1 + w_1 + w_2 + w_3 + \ldots + w_{N-1} = \frac{9 - 3\cos 2\theta + 3^{-N+1} \cos 2(N-1)\theta - 3^{-N+2} \cos 2N\theta}{10 - 6\cos 2\theta}.$$ [7] Hence show that the infinite series $$1 + w_1 + w_2 + w_3 + \ldots$$ is convergent for all values of \(\theta\), and find the sum to infinity. [2]

11 Answer only one of the following two alternatives. EITHER State the fifth roots of unity in the form \(\cos \theta + \mathrm { i } \sin \theta\), where \(- \pi < \theta \leqslant \pi\). Simplify $$\left( x - \left[ \cos \frac { 2 } { 5 } \pi + i \sin \frac { 2 } { 5 } \pi \right] \right) \left( x - \left[ \cos \frac { 2 } { 5 } \pi - i \sin \frac { 2 } { 5 } \pi \right] \right) .$$ Hence find the real factors of $$x ^ { 5 } - 1$$ Express the six roots of the equation $$x ^ { 6 } - x ^ { 3 } + 1 = 0$$ as three conjugate pairs, in the form \(\cos \theta \pm \mathrm { i } \sin \theta\). Hence find the real factors of $$x ^ { 6 } - x ^ { 3 } + 1$$ OR Given that $$y ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 6 y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 3 y ^ { 3 } = 25 \mathrm { e } ^ { - 2 x }$$ and that \(v = y ^ { 3 }\), show that $$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } - 6 \frac { \mathrm {~d} v } { \mathrm {~d} x } + 9 v = 75 \mathrm { e } ^ { - 2 x }$$ Find the particular solution for \(y\) in terms of \(x\), given that when \(x = 0 , y = 2\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\). \end{document}