If a complex number z = frac{4 + 3i sin theta | vedclass

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(D) For a complex number $z$ to be purely real,its imaginary part must be zero. Given $z = \frac{4 + 3i \sin 	heta}{1 - 2i \sin 	heta}$. Multiply th

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$n \pi, n \in Z$

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(D) For a complex number $z$ to be purely real,its imaginary part must be zero. Given $z = \frac{4 + 3i \sin 	heta}{1 - 2i \sin 	heta}$. Multiply the numerator and denominator by the conjugate of the denominator $(1 + 2i \sin 	heta)$: $z = \frac{(4 + 3i \sin 	heta)(1 + 2i \sin 	heta)}{(1 - 2i \sin 	heta)(1 + 2i \sin 	heta)}$ $z = \frac{4 + 8i \sin 	heta + 3i \sin 	heta + 6i^2 \sin^2 	heta}{1 + 4 \sin^2 	heta}$ Since $i^2 = -1$,we have: $z = \frac{4 - 6 \sin^2 	heta + i(11 \sin 	heta)}{1 + 4 \sin^2 	heta}$ For $z$ to be purely real,the imaginary part must be zero: $\frac{11 \sin 	heta}{1 + 4 \sin^2 	heta} = 0$ This implies $11 \sin 	heta = 0$,so $\sin 	heta = 0$. The general solution for $\sin 	heta = 0$ is $	heta = n \pi$,where $n \in Z$.

If a complex number $z = \frac{4 + 3i \sin \theta}{1 - 2i \sin \theta}$ (where $i = \sqrt{-1}$) is purely real,then the value of $\theta$ is

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