Let A = {theta in [0, 2pi] : 1 + 10 operatorn | vedclass

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

Question

(A) Given the equation $1 + 10 \operatorname{Re}\left(\frac{2 \cos 	heta + i \sin 	heta}{\cos 	heta - 3i \sin 	heta}ight) = 0$. Multiply the num

Vedclass · Accepted Answer

$\frac{21}{4} \pi^2$

Vedclass · Answer

(A) Given the equation $1 + 10 \operatorname{Re}\left(\frac{2 \cos 	heta + i \sin 	heta}{\cos 	heta - 3i \sin 	heta}ight) = 0$. Multiply the numerator and denominator of the fraction by the conjugate of the denominator $(\cos 	heta + 3i \sin 	heta)$: $\frac{(2 \cos 	heta + i \sin 	heta)(\cos 	heta + 3i \sin 	heta)}{\cos^2 	heta + 9 \sin^2 	heta} = \frac{2 \cos^2 	heta + 6i \cos 	heta \sin 	heta + i \sin 	heta \cos 	heta - 3 \sin^2 	heta}{\cos^2 	heta + 9 \sin^2 	heta} = \frac{(2 \cos^2 	heta - 3 \sin^2 	heta) + i(7 \sin 	heta \cos 	heta)}{\cos^2 	heta + 9 \sin^2 	heta}$. The real part is $\frac{2 \cos^2 	heta - 3 \sin^2 	heta}{\cos^2 	heta + 9 \sin^2 	heta}$. Substituting this into the original equation: $1 + 10 \left(\frac{2 \cos^2 	heta - 3 \sin^2 	heta}{\cos^2 	heta + 9 \sin^2 	heta}ight) = 0$. $\frac{\cos^2 	heta + 9 \sin^2 	heta + 20 \cos^2 	heta - 30 \sin^2 	heta}{\cos^2 	heta + 9 \sin^2 	heta} = 0$. $21 \cos^2 	heta - 21 \sin^2 	heta = 0 \implies 21 \cos(2	heta) = 0$. Thus, $\cos(2	heta) = 0$, which implies $2	heta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$. So, $	heta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$. Then $\sum 	heta^2 = \left(\frac{\pi}{4}ight)^2 + \left(\frac{3\pi}{4}ight)^2 + \left(\frac{5\pi}{4}ight)^2 + \left(\frac{7\pi}{4}ight)^2 = \frac{\pi^2}{16}(1 + 9 + 25 + 49) = \frac{84 \pi^2}{16} = \frac{21 \pi^2}{4}$.

Let $A = \{\theta \in [0, 2\pi] : 1 + 10 \operatorname{Re}\left(\frac{2 \cos \theta + i \sin \theta}{\cos \theta - 3i \sin \theta}\right) = 0\}$. Then $\sum_{\theta \in A} \theta^2$ is equal to

Similar Questions

If $(3 + i)z = (3 - i)\bar{z}$,then the complex number $z$ is

Let $\alpha$ and $\beta$ be the roots of the equation $x^{2} + (2i - 1) = 0$. Then,the value of $|\alpha^{8} + \beta^{8}|$ is equal to

If $z_1$ and $z_2$ are complex numbers such that $\frac{2 z_1}{3 z_2}$ is a purely imaginary number, then the value of $\left|\frac{z_1-z_2}{z_1+z_2}\right|$ is

If $1+2i$ is a root of the equation $x^4-3x^3+8x^2-7x+5=0$,then the sum of the squares of the other roots is

Let $z$ be a complex number with $\operatorname{Im}(z)=10$ and satisfying $\frac{2z-n}{2z+n}=2i-1$, where $i=\sqrt{-1}$, for some natural number $n$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module