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Question

Let $A = \{\theta\in [0,2\pi]:1+10\text{Re}\left(\frac{2\cos\theta+i\sin\theta}{\cos\theta-3i\sin\theta}\right) = 0\}$. Then $\sum_{\theta\in A}\theta^2$ is equal to

The correct answer is
$\frac{21}{4}\pi^2$

Solving the Complex Number Equation

We need to find the set $A = \{\theta\in [0,2\pi]:1+10\text{Re}\left(\frac{2\cos\theta+i\sin\theta}{\cos\theta-3i\sin\theta}\right) = 0\}$ and then calculate $\sum_{\theta\in A}\theta^2$.

Complex Number Simplification

Let the complex number be $z = \frac{2\cos\theta+i\sin\theta}{\cos\theta-3i\sin\theta}$. To find its real part, we multiply the numerator and denominator by the conjugate of the denominator, which is $\cos\theta+3i\sin\theta$.

Numerator: $(2\cos\theta+i\sin\theta)(\cos\theta+3i\sin\theta) = 2\cos^2\theta + 6i\cos\theta\sin\theta + i\sin\theta\cos\theta + 3i^2\sin^2\theta$ $= (2\cos^2\theta - 3\sin^2\theta) + i(7\cos\theta\sin\theta)$ Denominator: $(\cos\theta-3i\sin\theta)(\cos\theta+3i\sin\theta) = \cos^2\theta - (3i\sin\theta)^2 = \cos^2\theta - 9i^2\sin^2\theta$ $= \cos^2\theta + 9\sin^2\theta$ So, $z = \frac{(2\cos^2\theta - 3\sin^2\theta) + i(7\cos\theta\sin\theta)}{\cos^2\theta + 9\sin^2\theta}$

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

To simplify further, divide the numerator and denominator by $\cos^2\theta$ (assuming $\cos\theta \neq 0$, which holds as shown below): $\text{Re}(z) = \frac{2 - 3\frac{\sin^2\theta}{\cos^2\theta}}{1 + 9\frac{\sin^2\theta}{\cos^2\theta}} = \frac{2 - 3\tan^2\theta}{1 + 9\tan^2\theta}$ If $\cos\theta = 0$, then $\theta = \pi/2$ or $3\pi/2$. The original expression simplifies to $\frac{i\sin\theta}{-3i\sin\theta} = -1/3$. Then $1+10(-1/3) = 1-10/3 = -7/3 \neq 0$. Thus, $\cos\theta \neq 0$.

Equation Solving for $\theta$

Substitute the real part into the given equation $1+10\text{Re}(z) = 0$: $1 + 10 \left( \frac{2 - 3\tan^2\theta}{1 + 9\tan^2\theta} \right) = 0$ $1 = -10 \left( \frac{2 - 3\tan^2\theta}{1 + 9\tan^2\theta} \right)$ $1 + 9\tan^2\theta = -10(2 - 3\tan^2\theta)$ $1 + 9\tan^2\theta = -20 + 30\tan^2\theta$ $21 = 21\tan^2\theta$ $\tan^2\theta = 1$

This means $\tan\theta = 1$ or $\tan\theta = -1$. For $\theta \in [0, 2\pi]$, the solutions are:

  • If $\tan\theta = 1$, then $\theta = \frac{\pi}{4}$ or $\theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$.
  • If $\tan\theta = -1$, then $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ or $\theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$.

Therefore, the set of angles is $A = \{ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \}$.

Sum of Squares Calculation

Now, we calculate the sum of the squares of these angles:

$ \sum_{\theta\in A}\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{\pi^2}{16} (1 + 9 + 25 + 49) $ $ = \frac{\pi^2}{16} (84) $ $ = \frac{84}{16} \pi^2 $

Simplifying the fraction $\frac{84}{16}$ by dividing both numerator and denominator by 4:

$ \frac{84 \div 4}{16 \div 4} \pi^2 = \frac{21}{4} \pi^2 $

The sum is $\frac{21}{4}\pi^2$.

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