Question

Let $ A = \left\{ \theta \in [0, 2\pi] : \Re\left( \frac{2 \cos \theta + i \sin \theta}{\cos \theta - 3i \sin \theta} \right) = 0 \right\} $. Then $ \sum_{\theta \in A} \theta^2 $ is equal to:

• A. \( \frac{27}{4} \pi^2 \)
• B. \( \frac{21}{4} \pi^2 \)
• C. \( 6\pi^2 \)
• D. \( 8\pi^2 \)

Hint:

N/A

Answer 1

The Correct Option is B

1. Simplify the Complex Fraction

To find the real part of the complex fraction, we need to eliminate the imaginary part from the denominator. Multiply the numerator and denominator by the conjugate of the denominator:

\(\frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta} \cdot \frac{\cos\theta + 3i\sin\theta}{\cos\theta + 3i\sin\theta}\)

\(= \frac{(2\cos\theta + i\sin\theta)(\cos\theta + 3i\sin\theta)}{(\cos\theta - 3i\sin\theta)(\cos\theta + 3i\sin\theta)}\)

\( = \frac{2\cos^2\theta + 6i\cos\theta\sin\theta + i\cos\theta\sin\theta - 3\sin^2\theta}{\cos^2\theta + 9\sin^2\theta}\)

\( = \frac{2\cos^2\theta - 3\sin^2\theta + 7i\cos\theta\sin\theta}{\cos^2\theta + 9\sin^2\theta}\)

2. Extract the Real Part

The real part of the complex fraction is:

$\text{Re}\left(\frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta}\right) = \frac{2\cos^2\theta - 3\sin^2\theta}{\cos^2\theta + 9\sin^2\theta}$

3. Set Up the Equation

We are given that $1 + 10 \cdot \text{Re}\left(\frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta}\right) = 0$. Substitute the real part we found:

$1 + 10 \cdot \frac{2\cos^2\theta - 3\sin^2\theta}{\cos^2\theta + 9\sin^2\theta} = 0$

$\implies 1 + \frac{20\cos^2\theta - 30\sin^2\theta}{\cos^2\theta + 9\sin^2\theta} = 0$

$\implies \frac{\cos^2\theta + 9\sin^2\theta + 20\cos^2\theta - 30\sin^2\theta}{\cos^2\theta + 9\sin^2\theta} = 0$

$\implies 21\cos^2\theta - 21\sin^2\theta = 0 \implies \cos^2\theta - \sin^2\theta = 0 $

$\implies \cos^2\theta = \sin^2\theta$

4. Solve for θ

$\cos^2\theta = \sin^2\theta$ implies $\tan^2\theta = 1$, so $\tan\theta = \pm 1$.

In the interval $[0, 2\pi]$:

$\tan\theta = 1 \implies \theta = \frac{\pi}{4}, \frac{5\pi}{4}$

$\tan\theta = -1 \implies \theta = \frac{3\pi}{4}, \frac{7\pi}{4}$

Therefore, the set $A = \left\{ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\}$

5. Calculate the Sum of Squares

$\sum \theta^2 = \left(\frac{\pi}{4}\right)^2 + \left(\frac{3\pi}{4}\right)^2 + \left(\frac{5\pi}{4}\right)^2 + \left(\frac{7\pi}{4}\right)^2 = \frac{\pi^2}{16} (1 + 9 + 25 + 49) = \frac{\pi^2}{16} (84) = \frac{21}{4}\pi^2$

Answer: The sum of the squares of the values of θ in set A is $\frac{21}{4}\pi^2$. So the answer is option 2.