Question

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

Answer 1

$z=\frac{3+2i\cos \theta }{1-3i\cos \theta }$
$\Rightarrow z=\frac{(3+2i\cos \theta )(1+3i\cos \theta )}{1+9{\cos }^2\theta }$
$\because Re\left(z\right)=0\Rightarrow \frac{3-6{\cos }^2\theta }{1+9{\cos }^2\theta }=0$
$\Rightarrow {\cos }^2\theta =\frac{1}{2}$
$\Rightarrow \theta =\frac{\pi }{4}$
Now, ${\sin }^{2}3\theta +{\cos }^2\theta =\frac{1}{2}+\frac{1}{2}=1$