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
So, let's write the numbers i.e (1967, 1686) as factors of 7 and 3
1967 = 7 × 281 and 1686 = 6 × 562
So, the expression will be as follows :
\(\frac{281(7)+281(6i\sin(x))}{7-3i(\cos(x))}\) \(⇒281(\frac{7+6i\sin(x)}{7-3i\cos(x)})\) …. (i)
Now, by multiplying both numerator and denominator by \(7 + 3i\sin(x)\), we get :
\(=281(\frac{49+21i(\cos(x)+2\sin(x))+18\sin(x)\cos(x)}{49+9\cos^2(x)})\)
\(⇒281(\frac{49+18\sin(x)\cos(x)}{49+9\cos^2}+i\frac{21(\cos(x)+2(x)\sin)}{49+9\cos^2})\)
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
\((\frac{21(\cos(x)+2\sin(x))}{49+9\cos^2})=0\)
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
\(⇒281(\frac{7+6i\sin(x)}{7+6i\sin(x)})=281\)
Therefore, the value of n is 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.