If e^{i θ} = {cis} θ then ₙ=0}^{∞} {(nθ)}{2ⁿ} =

Correct Answer is : 4−2cos⁡θ5−4cos⁡θ4-2θ/5-4θ5−4cosθ4−2cosθ

Correct Answer is : 4−2cos⁡θ5−4cos⁡θ4-2θ/5-4θ5−4cosθ4−2cosθ

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AP EAPCET 04-Jul-2022 Shift 2 Paper

Section: Mathematics

Question : 10 of 160

Marks: +1, -0

e^{i \theta} = \operatorname{cis} \theta

\sum\limits_{n=0}^{\infty} \frac{\cos(n\theta)}{2^{n}} =

Solution: 👈: Video Solution

e^{i \theta} = \operatorname{cis} \theta = \cos \theta + i \sin \theta

\sum\limits_{n=0}^{\infty} \frac{\cos(n\theta)}{2^{n}}

= \operatorname{Re} \left[ \sum\limits_{n=0}^{\infty} \frac{\cos n\theta}{2^{n}} \right]

= \operatorname{Re} \left[ \sum\limits_{n=0}^{\infty} \left( \frac{\cos \theta}{2} \right)^n \right] \text{ [Using De-moivre theorem] }

= \operatorname{Re} \left[ \sum\limits_{n=0}^{\infty} \left( \frac{e^{i \theta}}{2} \right)^n \right]

= \operatorname{Re} \left[ 1 + \frac{e^{i \theta}}{2} + \left( \frac{e^{i \theta}}{2} \right)^2 + \dots \infty \right]

1 + \frac{e^{i \theta}}{2} + \left( \frac{e^{i \theta}}{2} \right)^2 + \dots

is an GP

= \operatorname{Re} \left[ \frac{1}{1 - \frac{e^{i0}}{2}} \right] = \operatorname{Re} \left[ \frac{2}{2 - e^{i \theta}} \right]

= \operatorname{Re} \left[ \frac{2}{2 - \cos \theta - i \sin \theta} \right]

= \operatorname{Re} \left[ \frac{2(2 - \cos \theta - i \sin \theta)}{(2 - \cos \theta)^2 - i^2 \sin^2 \theta} \right]

= \operatorname{Re} \left[ \frac{4 - 2 \cos \theta}{4 + \cos^2 \theta - 4 \cos \theta + \sin^2 \theta} \right.

= \frac{i \sin \theta}{4 + \cos^2 \theta - 4 \cos \theta + \sin^2 \theta}

= \frac{4 - 2 \cos \theta}{5 - 4 \cos \theta}

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