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 A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- 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 ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
- The number of elements in the relation $R = \{(x, y) : 4x^2 + y^2 < 52, x, y \in \mathbb{Z}\}$ is
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Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
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 ________.
- Let $S = \frac{1}{25!} + \frac{1}{3!23!} + \frac{1}{5!21!} + \dots$ up to 13 terms. If $13S = \frac{2^k}{n!}, k \in \mathbb{N}$, then $n + k$ is equal to
Important Questions from Algebra
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- 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 ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
