Complex numbers - infinite Series 1

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Submitted by roamingfree on Fri, 06/25/2021 - 16:35
skills
Move between exponential and trigonometric representations
Question

Let $\omega = e^{5i\theta}$, express $(1 +\frac{1}{3}\omega)(1 + \frac{1}{3}\overline{\omega})$ in the form $a + b\cos(5\theta)$

Let the sequence $C$ be defined as $$ C = 2\cos(3\theta) - \frac{2}{3}\cos(8\theta) + \frac{2}{9}\cos(13\theta) + \dots $$ Show that $C$ can be rewritten as $$ C = \frac{9\cos(3\theta) + 3\cos(2\theta)}{5 + 3\cos(5\theta)} $$

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$ \frac{9\cos(3\theta) + 3\cos(2\theta)}{5 + 3\cos(5\theta)}$
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The key to this problem is being able to move between exponential and trigonometric representations using complex numbers.

$\omega = e^{5i\theta}$. It follows that $$ \begin{align} (1 + \frac{1}{3}\omega)(1 + \frac{1}{3}\overline{\omega}) &= 1 + \frac{1}{3}\omega + \frac{1}{3}\overline{\omega} + \frac{1}{9} \\ &= \frac{10}{9} + \frac{1}{3}\left(e^{5i\theta} + e^{-5i\theta}\right)\\ &=\frac{10}{9} + \frac{2}{3}\cos(5\theta)\end{align} $$

The sequence $C$ can be rewritten as $$ C = (e^{3i\theta} + e^{-3i\theta}) - \frac{1}{3}(e^{8i\theta} + e^{-8i\theta}) + \frac{1}{9}(e^{13i\theta} + e^{-13i\theta}) - \dots $$

This can now be split into two distinct geometric series,

$$C_+ = e^{3i\theta}\left(1 - \frac{1}{3}\omega + \frac{1}{3^2}\omega^2 - \dots \right) = \frac{e^{3i\theta}}{1 + \frac{1}{3}\omega} $$

$$C_- = e^{-3i\theta}\left(1 - \frac{1}{3}\overline{\omega} + \frac{1}{3^2}\overline{\omega^2} - \dots \right) = \frac{e^{-3i\theta}}{1 + \frac{1}{3}\overline{\omega}} $$

Their sum can now be manipulated to give the desired result.

$$\begin{align}C &= C_+ + C_- \\ &= \frac{e^{3i\theta}}{1 + \frac{1}{3}\omega}  + \frac{e^{-3i\theta}}{1 + \frac{1}{3}\overline{\omega}} \\ &= \frac{e^{3i\theta}(1 + \frac{1}{3}\overline{\omega}) + e^{-3i\theta}(1 + \frac{1}{3}\omega)}{(1 + \frac{1}{3}\omega)(1 + \frac{1}{3}\overline{\omega})}  \\ &= \frac{2\cos(3\theta) + \frac{2}{3}\cos(2\theta)}{\frac{10}{9} + \frac{2}{3}\cos(5\theta)} \\ &=  \frac{9\cos(3\theta) + 3\cos(2\theta)}{5 + 3\cos(5\theta)} \end{align}$$