Key Concepts and Formulas

• Real Part of a Complex Number: For a complex number $z = a + bi$, the real part is denoted as $\operatorname{Re}(z) = a$.
• Complex Conjugate: The conjugate of a complex number $z = a + bi$ is $\overline{z} = a - bi$. The product of a complex number and its conjugate is a real number: $z\overline{z} = (a+bi)(a-bi) = a^2 + b^2$.
• Trigonometric Identity: $\cos^2\theta - \sin^2\theta = \cos(2\theta)$

Step-by-Step Solution

Step 1: Isolate the Real Part Term

We begin by isolating the real part term in the given equation: $1 + 10\operatorname{Re}\left( \frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta} \right) = 0$ Subtract 1 from both sides: $10\operatorname{Re}\left( \frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta} \right) = -1$ Divide both sides by 10: $\operatorname{Re}\left( \frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta} \right) = -\frac{1}{10}$ This isolates the real part of the complex expression.

Step 2: Calculate the Real Part of the Complex Expression

To find the real part, we multiply the numerator and denominator of the complex fraction by the conjugate of the denominator: $\frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta} = \frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta} \cdot \frac{\cos\theta + 3i\sin\theta}{\cos\theta + 3i\sin\theta}$ We multiply by the conjugate to make the denominator a real number. Expanding the numerator: $(2\cos\theta + i\sin\theta)(\cos\theta + 3i\sin\theta) = 2\cos^2\theta + 6i\cos\theta\sin\theta + i\sin\theta\cos\theta + 3i^2\sin^2\theta$ $= 2\cos^2\theta + 7i\cos\theta\sin\theta - 3\sin^2\theta = (2\cos^2\theta - 3\sin^2\theta) + i(7\cos\theta\sin\theta)$ Expanding the denominator: $(\cos\theta - 3i\sin\theta)(\cos\theta + 3i\sin\theta) = \cos^2\theta - (3i\sin\theta)^2 = \cos^2\theta - 9i^2\sin^2\theta$ $= \cos^2\theta + 9\sin^2\theta$ Thus, the complex fraction becomes: $\frac{(2\cos^2\theta - 3\sin^2\theta) + i(7\cos\theta\sin\theta)}{\cos^2\theta + 9\sin^2\theta}$ The real part is: $\operatorname{Re}\left( \frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta} \right) = \frac{2\cos^2\theta - 3\sin^2\theta}{\cos^2\theta + 9\sin^2\theta}$

Step 3: Formulate and Solve the Equation for $\theta$

Now, equate the real part with $-\frac{1}{10}$: $\frac{2\cos^2\theta - 3\sin^2\theta}{\cos^2\theta + 9\sin^2\theta} = -\frac{1}{10}$ Cross-multiply: $10(2\cos^2\theta - 3\sin^2\theta) = -(\cos^2\theta + 9\sin^2\theta)$ $20\cos^2\theta - 30\sin^2\theta = -\cos^2\theta - 9\sin^2\theta$ $21\cos^2\theta - 21\sin^2\theta = 0$ $21(\cos^2\theta - \sin^2\theta) = 0$ $\cos^2\theta - \sin^2\theta = 0$ We have simplified the equation to a form suitable for using trigonometric identities.

Step 4: Apply Double Angle Identity for Cosine

Using the identity $\cos(2\theta) = \cos^2\theta - \sin^2\theta$, we have: $\cos(2\theta) = 0$ This simplifies the equation significantly.

Step 5: Find the Values of $\theta$

We need to find all $\theta \in [0, 2\pi]$ such that $\cos(2\theta) = 0$. Since $0 \le \theta \le 2\pi$, we have $0 \le 2\theta \le 4\pi$. The solutions for $\cos(2\theta) = 0$ in the interval $[0, 4\pi]$ are: $2\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$ Dividing by 2, we get: $\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ These are all the values of $\theta$ in the interval $[0, 2\pi]$ that satisfy the equation.

Step 6: Calculate the Sum of Squares of $\theta$

We need to calculate $\sum_{\theta \in A} \theta^2$, where $A = \left\{ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\}$. $\sum \theta^2 = \left(\frac{\pi}{4}\right)^2 + \left(\frac{3\pi}{4}\right)^2 + \left(\frac{5\pi}{4}\right)^2 + \left(\frac{7\pi}{4}\right)^2$ $= \frac{\pi^2}{16} + \frac{9\pi^2}{16} + \frac{25\pi^2}{16} + \frac{49\pi^2}{16} = \frac{(1+9+25+49)\pi^2}{16}$ $= \frac{84\pi^2}{16} = \frac{21\pi^2}{4}$

Common Mistakes & Tips

• Careless calculation of the complex conjugate and expansion can lead to errors in the real part.
• Forgetting to consider the expanded range for $2\theta$ when solving $\cos(2\theta) = 0$ will lead to missing solutions.
• Incorrect application of trigonometric identities will lead to an incorrect equation to solve.

Summary

We isolated the real part of the complex expression, multiplied by the conjugate to simplify, and then solved the resulting trigonometric equation using the double angle identity. We found four solutions for $\theta$ in the interval $[0, 2\pi]$ and calculated the sum of their squares. The final answer is $\frac{21\pi^2}{4}$.

Final Answer

The final answer is $\boxed{\frac{21}{4} \pi^2}$, which corresponds to option (A).