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 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
- If \( i = \sqrt{-1} \), then \[ \sum_{n=2}^{30} i^n + \sum_{n=30}^{65} i^{n+3} = \]
- AP EAPCET - 2025
- Mathematics
- Complex numbers
Let \( z \) satisfy \( |z| = 1, \ z = 1 - \overline{z} \text{ and } \operatorname{Im}(z)>0 \)
Then consider:
Statement-I: \( z \) is a real number
Statement-II: Principal argument of \( z \) is \( \dfrac{\pi}{3} \)
Then:
- AP EAPCET - 2025
- Mathematics
- Complex numbers
If \( z \) and \( \omega \) are two non-zero complex numbers such that \( |z\omega| = 1 \) and
\[ \arg(z) - \arg(\omega) = \frac{\pi}{2}, \]
Then the value of \( \overline{z\omega} \) is:
- AP EAPCET - 2025
- Mathematics
- Complex numbers
If \( x^a y^b = e^m, \)
and
\[ x^c y^d = e^n, \]
and
\[ \Delta_1 = \begin{vmatrix} m & b \\ n & d \\ \end{vmatrix}, \quad \Delta_2 = \begin{vmatrix} a & m \\ c & n \\ \end{vmatrix}, \quad \Delta_3 = \begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} \]
Then the values of \( x \) and \( y \) respectively (where \( e \) is the base of the natural logarithm) are:
- AP EAPCET - 2025
- Mathematics
- Complex numbers
Questions Asked in JEE Advanced exam
- Let $$ \alpha = \frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \cdots + \frac{1}{\sin 118^\circ \sin 119^\circ}. $$ Then the value of $$ \left( \frac{\csc 1^\circ}{\alpha} \right)^2 $$ is \rule{1cm}{0.15mm}.
- JEE Advanced - 2025
- Some Applications of Trigonometry
Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter $ D $ of a tube. The measured value of $ D $ is:
- JEE Advanced - 2025
- Units and measurement
- Let $ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $ and $ P = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} $. Let $ Q = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix} $ for some non-zero real numbers $ x, y, z $, for which there is a $ 2 \times 2 $ matrix $ R $ with all entries being non-zero real numbers, such that $$ QR = RP $$ Then which of the following statements is (are) TRUE?
- JEE Advanced - 2025
- Matrices
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
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
- 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.