If the real part of the complex number z = fr | vedclass

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(A) Given $z = \frac{3 + 2i \cos 	heta}{1 - 3i \cos 	heta}$.To find the real part, multiply the numerator and denominator by the conjugate of the de

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(A) Given $z = \frac{3 + 2i \cos 	heta}{1 - 3i \cos 	heta}$.To find the real part, multiply the numerator and denominator by the conjugate of the denominator $(1 + 3i \cos 	heta)$:$z = \frac{(3 + 2i \cos 	heta)(1 + 3i \cos 	heta)}{(1 - 3i \cos 	heta)(1 + 3i \cos 	heta)}$$z = \frac{3 + 9i \cos 	heta + 2i \cos 	heta + 6i^2 \cos^2 	heta}{1 + 9 \cos^2 	heta}$Since $i^2 = -1$, we have:$z = \frac{3 - 6 \cos^2 	heta + 11i \cos 	heta}{1 + 9 \cos^2 	heta}$The real part is $\operatorname{Re}(z) = \frac{3 - 6 \cos^2 	heta}{1 + 9 \cos^2 	heta}$.Given $\operatorname{Re}(z) = 0$, we have $3 - 6 \cos^2 	heta = 0$, which implies $\cos^2 	heta = \frac{1}{2}$.Since $	heta \in (0, \frac{\pi}{2})$, $\cos 	heta = \frac{1}{\sqrt{2}}$, so $	heta = \frac{\pi}{4}$.Now, calculate $\sin^2 3	heta + \cos^2 	heta$:$\sin^2 3(\frac{\pi}{4}) + \cos^2(\frac{\pi}{4}) = \sin^2(\frac{3\pi}{4}) + (\frac{1}{\sqrt{2}})^2$$= (\frac{1}{\sqrt{2}})^2 + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = 1$.

If the real part of the complex number $z = \frac{3 + 2i \cos \theta}{1 - 3i \cos \theta}$, where $\theta \in (0, \frac{\pi}{2})$, is zero, then the value of $\sin^2 3\theta + \cos^2 \theta$ is equal to:

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