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 $\frac{21}{4}\pi^2$.
Similar Questions
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
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- Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $\det(Q) = 2^{10}$. Then the value of $\det(\text{adj}(\text{adj } P))$ is:
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Important Questions from Algebra
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
- The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______
- Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $\det(Q) = 2^{10}$. Then the value of $\det(\text{adj}(\text{adj } P))$ is:
