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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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}^{\

Vedclass · Accepted Answer

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

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, $C = \frac{4 - 2 \cos 	heta}{5 - 4 \cos 	heta}$.

List-$I$	List-$II$
$(P)$ $\|z\|^2$ is equal to	$(1)$ $12$
$(Q)$ $\|z-\bar{z}\|^2$ is equal to	$(2)$ $4$
$(R)$ $\|z\|^2+\|z+\bar{z}\|^2$ is equal to	$(3)$ $8$
$(S)$ $\|z+1\|^2$ is equal to	$(4)$ $10$
	$(5)$ $7$

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

Similar Questions

If $x = -5 + 2 \sqrt{-4}$,then the value of $x^4 + 9x^3 + 35x^2 - x + 4$ is

$\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right) \times \left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) \times \left(\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}\right) \times \ldots \infty =$

If $(x-iy)^{\frac{1}{3}} = a+ib$,then $\frac{ax-by}{a-b} = $

If $x=\frac{4}{5}+\frac{3}{5} i$ and $y=\frac{\sqrt{3}}{\sqrt{8}}-\frac{\sqrt{5}}{\sqrt{8}} i$,then $\left(x^2+\frac{1}{x^2}\right)\left(y^2-\frac{1}{y^2}\right)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module