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

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(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 $\sum 	heta^2 = \left(\frac{\pi}{4}ight)^2 + \left(\frac{3\pi}{4}ight)^2 + \left(\frac{5\pi}{4}ight)^2 + \left(\frac{7\pi}{4}ight)^2 = \frac{\pi^2}{16}(1 + 9 + 25 + 49) = \frac{84 \pi^2}{16} = \frac{21 \pi^2}{4}$.

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	$(4)$ $10$
	$(5)$ $7$

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

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