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 ________.

Hint:

To find the real or imaginary part of a complex fraction, always start by making the denominator real by multiplying the numerator and denominator by the conjugate of the denominator.

Answer 1

Correct Answer: 1

Given \[ z=\frac{3+2i\cos\theta}{1-3i\cos\theta} \] Multiply numerator and denominator by the conjugate: \[ z=\frac{(3+2i\cos\theta)(1+3i\cos\theta)}{1+9\cos^2\theta} \] \[ z=\frac{(3-6\cos^2\theta)+i(11\cos\theta)}{1+9\cos^2\theta} \] Real part is zero: \[ 3-6\cos^2\theta=0 \Rightarrow \cos^2\theta=\frac12 \Rightarrow \theta=\frac{\pi}{4} \] \[ \sin^2 3\theta+\cos^2\theta =\sin^2\frac{3\pi}{4}+\cos^2\frac{\pi}{4} =\frac12+\frac12=1 \] \[ \boxed{1} \]