Let A = {\(\frac{1967+1686 i\,sin\theta}{7-3i\,cos\theta}:\theta\in R\) }. If A contains exactly one positive integer n, then the value of n is ______
Solution and Explanation
Let's write the numbers \( (1967, 1686) \) as factors of 7 and 3:
\( 1967 = 7 \times 281 \) and \( 1686 = 6 \times 562 \)
So, the expression will be as follows:
\[ \frac{281(7) + 281(6i\sin(x))}{7 - 3i(\cos(x))} \]
Thus, we have:
\[ 281\left(\frac{7 + 6i\sin(x)}{7 - 3i\cos(x)}\right) \quad \text{(i)} \]
Now, by multiplying both the numerator and denominator by \( 7 + 3i\sin(x) \), we get:
\[ = 281\left(\frac{49 + 21i(\cos(x) + 2\sin(x)) + 18\sin(x)\cos(x)}{49 + 9\cos^2(x)}\right) \]
Which simplifies to:
\[ \Rightarrow 281\left(\frac{49 + 18\sin(x)\cos(x)}{49 + 9\cos^2(x)} + i \frac{21(\cos(x) + 2\sin(x))}{49 + 9\cos^2(x)}\right) \]
Since we know the expression will be an integer after simplification, we can conclude that the imaginary part of the expression is zero.
Therefore, we can say that:
\[ \left(\frac{21(\cos(x) + 2\sin(x))}{49 + 9\cos^2(x)}\right) = 0 \]
After further simplification, it turns out to be:
\[ \cos(x) = -2\sin(x) \]
Now, substitute the value of \( \cos(x) \) back into the equation, and the equation becomes:
\[ \Rightarrow 281\left(\frac{7 + 6i\sin(x)}{7 + 6i\sin(x)}\right) = 281 \]
Final Answer:
The value of \( n \) is \( \boxed{281} \).
Top Questions on Complex numbers
- Let the curve \(z(1 + i) +\) \(\overline{z(1 - i) = 4}\), z \(\in\)\( \mathbb{C}\), divide the region \(|z - 3| \leq 1\) into two parts of areas \( \alpha \) and \( \beta \). Then \( |\alpha - \beta| \) equals:}
- JEE Main - 2025
- Mathematics
- Complex numbers
If \( z \) is a complex number and \( k \in \mathbb{R} \), such that \( |z| = 1 \), \[ \frac{2 + k^2 z}{k + \overline{z}} = kz, \] then the maximum distance from \( k + i k^2 \) to the circle \( |z - (1 + 2i)| = 1 \) is:
- JEE Main - 2025
- Mathematics
- Complex numbers
- Let \( z_1, z_2, z_3 \) be three complex numbers on the circle \( |z| = 1 \) with \( \arg(z_1) = -\frac{\pi}{4}, \arg(z_2) = 0 \) and \( \arg(z_3) = \frac{\pi}{4} \). If \( |z_1 \overline{z_2} + z_2 \overline{z_3} + z_3 \overline{z_1}|^2 = \alpha + \beta \sqrt{2} \), where \( \alpha, \beta \in \mathbb{Z} \), then the value of \( \alpha^2 + \beta^2 \) is :
- JEE Main - 2025
- Mathematics
- Complex numbers
- Let integers \( a, b \in [-3, 3] \) be such that \( a + b \neq 0 \). \(\text{Then the number of all possible ordered pairs}\) \( (a, b) \), \(\text{for which}\) \[ \left| \frac{z - a}{z + b} \right| = 1 \quad \text{and} \quad \left| \begin{matrix} z + 1 & \omega & \omega^2 \\ \omega^2 & 1 & z + \omega \\ \omega^2 & 1 & z + \omega \end{matrix} \right| = 1, \] \(\text{is equal to:}\)
- JEE Main - 2025
- Mathematics
- Complex numbers
- Among the statements: (S1): The set $ \{ z \in \mathbb{C} - \{-i\} : |z| = 1 \text{ and } \frac{z - i}{z + i} \text{ is purely real} \} $ contains exactly two elements.
(S2): The set $ \{ z \in \mathbb{C} - \{-1\} : |z| = 1 \text{ and } \frac{z - 1}{z + 1} \text{ is purely imaginary} \} $ contains infinitely many elements. Then, which of the following is correct?- JEE Main - 2025
- Mathematics
- Complex numbers
Questions Asked in JEE Advanced exam
The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is ____.
Use: Atomic mass (in amu): H = 1, C = 12, O = 16, Br = 80- JEE Advanced - 2025
- Organic Chemistry
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
- JEE Advanced - 2025
- binomial expansion formula
- The total number of real solutions of the equation $$ \theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2} \sin^{-1} \left( \frac{6 \tan \theta}{9 + \tan^2 \theta} \right) $$ is
(Here, the inverse trigonometric functions $ \sin^{-1} x $ and $ \tan^{-1} x $ assume values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $(-\frac{\pi}{2}, \frac{\pi}{2})$, respectively.)- JEE Advanced - 2025
- Inverse Trigonometric Functions
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
- JEE Advanced - 2025
- Coordinate Geometry
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity):
- JEE Advanced - 2025
- Waves and Oscillations
Concepts Used:
Complex Number
A Complex Number is written in the form
a + ib
where,
- “a” is a real number
- “b” is an imaginary number
The Complex Number consists of a symbol “i” which satisfies the condition i^2 = −1. Complex Numbers are mentioned as the extension of one-dimensional number lines. In a complex plane, a Complex Number indicated as a + bi is usually represented in the form of the point (a, b). We have to pay attention that a Complex Number with absolutely no real part, such as – i, -5i, etc, is called purely imaginary. Also, a Complex Number with perfectly no imaginary part is known as a real number.