If e^{i theta} = operatorname{cis} theta, the | vedclass

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(C) Let $C = \sum_{n=0}^{\infty} \frac{\cos (n 	heta)}{2^n} = 1 + \frac{\cos 	heta}{2} + \frac{\cos (2 	heta)}{2^2} + \dots$ and $S = \sum_{n=0}^{\

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$\frac{4-2 \cos 	heta}{5-4 \cos 	heta}$

Vedclass · Answer

(C) Let $C = \sum_{n=0}^{\infty} \frac{\cos (n 	heta)}{2^n} = 1 + \frac{\cos 	heta}{2} + \frac{\cos (2 	heta)}{2^2} + \dots$ and $S = \sum_{n=0}^{\infty} \frac{\sin (n 	heta)}{2^n} = 0 + \frac{\sin 	heta}{2} + \frac{\sin (2 	heta)}{2^2} + \dots$ Consider $C + iS = \sum_{n=0}^{\infty} \frac{e^{in 	heta}}{2^n} = \sum_{n=0}^{\infty} \left(\frac{e^{i 	heta}}{2}ight)^n$. This is an infinite geometric series with first term $a = 1$ and common ratio $r = \frac{e^{i 	heta}}{2}$. Since $|r| = |\frac{e^{i 	heta}}{2}| = \frac{1}{2} Substituting $e^{i 	heta} = \cos 	heta + i \sin 	heta$, we get $C + iS = \frac{2}{2 - \cos 	heta - i \sin 	heta}$. Multiply the numerator and denominator by the conjugate $(2 - \cos 	heta + i \sin 	heta)$: $C + iS = \frac{2(2 - \cos 	heta + i \sin 	heta)}{(2 - \cos 	heta)^2 + \sin^2 	heta} = \frac{4 - 2 \cos 	heta + 2i \sin 	heta}{4 - 4 \cos 	heta + \cos^2 	heta + \sin^2 	heta}$. Since $\cos^2 	heta + \sin^2 	heta = 1$, the denominator is $4 - 4 \cos 	heta + 1 = 5 - 4 \cos 	heta$. Thus, $C + iS = \frac{4 - 2 \cos 	heta}{5 - 4 \cos 	heta} + i \frac{2 \sin 	heta}{5 - 4 \cos 	heta}$. Comparing the real parts, $C = \frac{4 - 2 \cos 	heta}{5 - 4 \cos 	heta}$.

If $e^{i \theta} = \operatorname{cis} \theta$, then find the value of $\sum_{n=0}^{\infty} \frac{\cos (n \theta)}{2^n}$.

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