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 = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
Let $a_1, a_2, a_3,....$ be a G.P. of increasing positive numbers. If $a_3a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to
All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_n$. Let the probability $P(W_n)$ of choosing the word $W_n$ satisfy $P(W_n) = 2P(W_{n-1})$, $n > 1$.
If $P(CDBEA) = \frac{2^\alpha}{2^\beta-1}$, $\alpha, \beta\in N$, then $\alpha + \beta$ is equal to :
- The largest $n \in N$ such that $3^n$ divides $50!$ is :
- The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to :
- Let A be the set of all functions $f: Z \to Z$ and R be a relation on A such that $R = \{(f, g) : f(0) = g(1) \text{ and } f(1) = g(0)\}$. Then R is :
Let $a \in \mathbf{R}$ and $A$ be a matrix of order $3 \times 3$ such that $\det (A) = -4$ and $A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$, where $I$ is the identity matrix of order $3 \times 3$. If $\det ((a+1)\text{adj}((a-1)A))$ is $2^m 3^n$, $m, n \in \{0, 1, 2, \dots, 20\}$, then $m+n$ is equal to :
Important Questions from Algebra
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
Let $a_1, a_2, a_3,....$ be a G.P. of increasing positive numbers. If $a_3a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to
