If frac{3+i sin theta}{4-i cos theta}, theta | vedclass

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(C) Let $z = \frac{3 + i \sin 	heta}{4 - i \cos 	heta}$. For $z$ to be a real number,its imaginary part must be zero.Multiply the numerator and deno

Vedclass · Accepted Answer

$\pi - 	an^{-1}\left(\frac{4}{3}ight)$

Vedclass · Answer

(C) Let $z = \frac{3 + i \sin 	heta}{4 - i \cos 	heta}$. For $z$ to be a real number,its imaginary part must be zero.Multiply the numerator and denominator by the conjugate of the denominator $(4 + i \cos 	heta)$:$z = \frac{(3 + i \sin 	heta)(4 + i \cos 	heta)}{(4 - i \cos 	heta)(4 + i \cos 	heta)} = \frac{12 + 3i \cos 	heta + 4i \sin 	heta - \sin 	heta \cos 	heta}{16 + \cos^2 	heta}$.The imaginary part is $\frac{3 \cos 	heta + 4 \sin 	heta}{16 + \cos^2 	heta} = 0$.This implies $3 \cos 	heta + 4 \sin 	heta = 0$,so $	an 	heta = -\frac{3}{4}$.Since $	an 	heta = -\frac{3}{4}$,$\sin 	heta$ and $\cos 	heta$ have opposite signs.We want the argument of $w = \sin 	heta + i \cos 	heta$.Since $	an 	heta = -\frac{3}{4}$,let $\alpha = 	an^{-1}(\frac{3}{4})$. Then $	heta$ is in the second or fourth quadrant.If $	heta$ is in the second quadrant,$\sin 	heta = \frac{3}{5}, \cos 	heta = -\frac{4}{5}$. Then $w = \frac{3}{5} - i \frac{4}{5}$. The argument is $-	an^{-1}(\frac{4}{3})$.If $	heta$ is in the fourth quadrant,$\sin 	heta = -\frac{3}{5}, \cos 	heta = \frac{4}{5}$. Then $w = -\frac{3}{5} + i \frac{4}{5}$. The argument is $\pi - 	an^{-1}(\frac{4}{3})$.

List-$I$ (Complex number)	List-$II$ (Polar form)
$(i) \sqrt{3}-i$	$(a) 2 \operatorname{cis} \frac{\pi}{6}$
$(ii) \sqrt{3}+i$	$(b) 2 \operatorname{cis} \frac{5 \pi}{6}$
$(iii) -\sqrt{3}+i$	$(c) 2 \operatorname{cis}\left(-\frac{5 \pi}{6}\right)$
$(iv) -\sqrt{3}-i$	$(d) 2 \operatorname{cis}\left(-\frac{\pi}{6}\right)$

If $\frac{3+i \sin \theta}{4-i \cos \theta}, \theta \in [0, 2 \pi],$ is a real number,then an argument of $\sin \theta + i \cos \theta$ is

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