Key Concepts and Formulas

• Complex Number Conjugate: The conjugate of a complex number $a + bi$ is $a - bi$. Multiplying a complex number by its conjugate results in a real number: $(a + bi)(a - bi) = a^2 + b^2$.
• Real Part of a Complex Number: If $z = x + iy$, where $x$ and $y$ are real numbers, then the real part of $z$ is $x$, denoted as $\text{Re}(z) = x$.
• Trigonometric Identities: $\sin(\pi - x) = \sin(x)$ and $\sin^2(x) + \cos^2(x) = 1$.

Step-by-Step Solution

Step 1: Express the complex number in the form $a + bi$.

We are given $z = \frac{3 + 2i\cos \theta}{1 - 3i\cos \theta}$. To find the real part, we need to express $z$ in the standard form $a + bi$. This is done by multiplying the numerator and denominator by the conjugate of the denominator. The conjugate of $1 - 3i\cos \theta$ is $1 + 3i\cos \theta$.

$z = \frac{3 + 2i\cos \theta}{1 - 3i\cos \theta} \cdot \frac{1 + 3i\cos \theta}{1 + 3i\cos \theta}$

Step 2: Expand the numerator and denominator.

Expanding the numerator: $(3 + 2i\cos \theta)(1 + 3i\cos \theta) = 3 + 9i\cos \theta + 2i\cos \theta + 6i^2\cos^2 \theta = 3 + 11i\cos \theta - 6\cos^2 \theta = (3 - 6\cos^2 \theta) + i(11\cos \theta)$

Expanding the denominator: $(1 - 3i\cos \theta)(1 + 3i\cos \theta) = 1 - (3i\cos \theta)^2 = 1 - 9i^2\cos^2 \theta = 1 + 9\cos^2 \theta$

Therefore, $z = \frac{(3 - 6\cos^2 \theta) + i(11\cos \theta)}{1 + 9\cos^2 \theta} = \frac{3 - 6\cos^2 \theta}{1 + 9\cos^2 \theta} + i\frac{11\cos \theta}{1 + 9\cos^2 \theta}$

Step 3: Find the real part of $z$ and set it to zero.

The real part of $z$ is given by: $\text{Re}(z) = \frac{3 - 6\cos^2 \theta}{1 + 9\cos^2 \theta}$

We are given that the real part is zero, so: $\frac{3 - 6\cos^2 \theta}{1 + 9\cos^2 \theta} = 0$

Since the denominator is always positive ($1 + 9\cos^2 \theta \ge 1$), the numerator must be zero: $3 - 6\cos^2 \theta = 0$

Step 4: Solve for $\cos^2 \theta$.

From $3 - 6\cos^2 \theta = 0$, we have: $6\cos^2 \theta = 3$ $\cos^2 \theta = \frac{3}{6} = \frac{1}{2}$

Step 5: Solve for $\theta$.

Since $\theta \in (0, \frac{\pi}{2})$, $\cos \theta$ must be positive. Thus: $\cos \theta = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$ This means $\theta = \frac{\pi}{4}$.

Step 6: Calculate $3\theta$.

$3\theta = 3 \cdot \frac{\pi}{4} = \frac{3\pi}{4}$

Step 7: Evaluate $\sin^2(3\theta) + \cos^2(\theta)$.

We want to find the value of $\sin^2(3\theta) + \cos^2(\theta)$. Substituting $\theta = \frac{\pi}{4}$: $\sin^2\left(\frac{3\pi}{4}\right) + \cos^2\left(\frac{\pi}{4}\right) = \left(\sin\left(\frac{3\pi}{4}\right)\right)^2 + \left(\cos\left(\frac{\pi}{4}\right)\right)^2$

We know that $\sin(\frac{3\pi}{4}) = \sin(\pi - \frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. So, $\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} + \frac{1}{2} = 1$

Common Mistakes & Tips

• Remember that $i^2 = -1$. A sign error here will propagate through the entire solution.
• Pay close attention to the given interval for $\theta$. This restricts the possible values of $\cos \theta$.
• Always multiply by the conjugate of the denominator, not the denominator itself, to rationalize the complex number.

Summary

We found the real part of the complex number $z$ by rationalizing the denominator. Setting the real part to zero, we solved for $\cos^2 \theta$, found $\theta$, and then evaluated the expression $\sin^2(3\theta) + \cos^2(\theta)$. The final result is 1.

Final Answer

The final answer is \boxed{1}.