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

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$z = \frac{3 + 2 i \cos θ}{1 - 3 i \cos θ} = \frac{(3 + 2 i \cos θ) (1 + 3 i \cos θ)}{(1 - 3 i \cos θ) (1 + 3 i \cos θ)} Re (z) = \frac{3 - 6 \cos^{2} θ}{1 + 9 \cos^{2} θ} = 0 \Rightarrow θ = \frac{π}{4} Hence, \sin^{2} 3 θ + \cos^{2} θ = 1$

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