JEE Advanced 2023 Paper 1 · Q11 · Complex Numbers
Let $A = \left\{\dfrac{1967 + 1686 i \sin\theta}{7 - 3i\cos\theta} : \theta \in \mathbb{R}\right\}$. If $A$ contains exactly one positive integer $n$, then the value of $n$ is
Reveal answer + step-by-step solution
Correct answer:281
Solution
For $z = \dfrac{1967+1686i\sin\theta}{7-3i\cos\theta}$ to be a positive integer, imaginary part = 0. Multiply by conjugate: numerator becomes $(1967+1686i\sin\theta)(7+3i\cos\theta)$; imaginary part = $1967 \cdot 3\cos\theta + 1686\sin\theta\cdot 7 = 0 \Rightarrow 5901\cos\theta + 11802\sin\theta = 0 \Rightarrow \tan\theta = -1/2$. Then $\sin^2\theta = 1/5$. Real part of numerator over denominator-modulus: $(1967\cdot 7 - 1686\sin\theta\cdot 3\cos\theta)/(49 + 9\cos^2\theta)$. Substituting $\sin^2\theta = 1/5$, $\cos^2\theta = 4/5$, $\sin\theta\cos\theta = -2/5$: value $= (13769 + 1686 \cdot 6/5)/(49 + 36/5) = $ simplifies to $281$.
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