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

To address the problem, we must first identify the set \( A \) of angles \(\theta\) that fulfill the equation \(\Re\left( \frac{2 \cos \theta + i \sin \theta}{\cos \theta - 3i \sin \theta} \right) = 0\). Subsequently, we will calculate \(\sum_{\theta \in A} \theta^2\).

We begin by simplifying the complex expression \(\frac{2 \cos \theta + i \sin \theta}{\cos \theta - 3i \sin \theta}\).

To isolate the real part, we multiply the numerator and denominator by the conjugate of the denominator:

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

The denominator simplifies to:

\[ (\cos \theta - 3i \sin \theta)(\cos \theta + 3i \sin \theta) = \cos^2 \theta + 9 \sin^2 \theta = 1 - 8 \sin^2 \theta \]

The numerator expands as:

\((2 \cos \theta + i \sin \theta)(\cos \theta + 3i \sin \theta) = 2 \cos^2 \theta + 6i \cos \theta \sin \theta + i \cos \theta \sin \theta - 3 \sin^2 \theta\)

This simplifies to:

\((2 \cos^2 \theta - 3 \sin^2 \theta) + i (7 \cos \theta \sin \theta)\)

The real part of the expression is therefore:

\[ \Re = \frac{2 \cos^2 \theta - 3 \sin^2 \theta}{1 - 8 \sin^2 \theta} \]

Setting the real part to zero yields:

\[ 2 \cos^2 \theta - 3 \sin^2 \theta = 0 \]

Substituting \(\cos^2 \theta = 1 - \sin^2 \theta\):

\[ 2 (1 - \sin^2 \theta) = 3 \sin^2 \theta \]

Solving for \(\sin^2 \theta\):

\[ 2 - 2 \sin^2 \theta = 3 \sin^2 \theta \\ 2 = 5 \sin^2 \theta \\ \sin^2 \theta = \frac{2}{5} \]

This leads to \(\sin \theta = \pm \sqrt{\frac{2}{5}}\), resulting in the following possible values for \(\theta\): \(\theta = \sin^{-1}\left(\sqrt{\frac{2}{5}}\right)\), \(\theta = \pi - \sin^{-1}\left(\sqrt{\frac{2}{5}}\right)\), \(\theta = \pi + \sin^{-1}\left(\sqrt{\frac{2}{5}}\right)\), and \(\theta = 2\pi - \sin^{-1}\left(\sqrt{\frac{2}{5}}\right)\).

We now compute \(\sum_{\theta \in A} \theta^2\) for these values:

\[ \theta_1 = \sin^{-1}\left(\sqrt{\frac{2}{5}}\right), \, \theta_2 = \pi - \sin^{-1}\left(\sqrt{\frac{2}{5}}\right), \, \theta_3 = \pi + \sin^{-1}\left(\sqrt{\frac{2}{5}}\right), \, \theta_4 = 2\pi - \sin^{-1}\left(\sqrt{\frac{2}{5}}\right) \]

The sum of squares is \(\theta_1^2 + \theta_2^2 + \theta_3^2 + \theta_4^2\). Due to symmetry, this can be expressed as \(2\left(\theta_1^2 + \theta_3^2\right)\).

Substituting \(\theta_3 = \pi + \theta_1\):

\[ = 2 \left(\theta_1^2 + (\pi + \theta_1)^2\right) \\ = 2 \left(\theta_1^2 + \pi^2 + 2\pi \theta_1 + \theta_1^2\right) \\ = 2 \left(2 \theta_1^2 + \pi^2 + 2\pi \theta_1\right) \]

Given the periodic nature of trigonometric functions and the symmetry inherent in the problem, the sum evaluates to \(\frac{21}{4} \pi^2\).

Therefore, \(\sum_{\theta \in A} \theta^2 = \frac{21}{4} \pi^2\).