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 \( \boxed{281} \).
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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.