If the real part of the complex number z = (3 + 2icosθ)/(1 -...

We have, $z = \dfrac{3 + 2icos\theta}{1 - 3icos\theta}, \theta \in \left(0, \dfrac{\pi}{2}\right)$

$⇒ z = \dfrac{3 + 2icos\theta}{1 - 3icos\theta} \times \dfrac{1 + 3icos\theta}{1 + 3icos\theta}$

$⇒ z = \dfrac{\left(3 + 2icos\theta\right)\left(1 + 3icos\theta\right)}{1 + 9\cos^{2}\theta}$

$⇒ z = \dfrac{\left(3 - 6\cos^{2}\theta + 11icos\theta\right)}{1 + 9\cos^{2}\theta}$

Now, $Re \left(z\right) = \dfrac{3 - 6\cos^{2}\theta}{1 + 9\cos^{2}\theta} = 0$

$⇒ 3 - 6\cos^{2}\theta = 0$

$⇒ \cos^{2}\theta = \dfrac{3}{6}$

$⇒ \cos^{2}\theta = \dfrac{1}{2}$

$⇒ cos\theta = \pm \dfrac{1}{\sqrt{2}}$

$⇒ \theta = \dfrac{\pi}{4}, ∵ \theta \in \left(0, \dfrac{\pi}{2}\right)$

Hence, $\sin^{2}3\theta + \cos^{2}\theta$

$= \sin^{2}\left(\dfrac{3\pi}{4}\right) + \cos^{2}\left(\dfrac{\pi}{4}\right)$

$= {\left(\sin \left(\pi - \dfrac{\pi}{4}\right)\right)}^{2} + {\left(\cos \left(\dfrac{\pi}{4}\right)\right)}^{2}$

$= {\left(\sin \left(\dfrac{\pi}{4}\right)\right)}^{2} + {\left(\cos \left(\dfrac{\pi}{4}\right)\right)}^{2} = 1$

Since, $\sin^{2}\theta + \cos^{2}\theta = 1$