JEE Advanced 2023 · Paper 1, Mathematics question 11

Multiply the numerator and denominator by $7+3i\cos\theta$:
$$z=\frac{(1967+1686i\sin\theta)(7+3i\cos\theta)}{49+9\cos^2\theta}.$$

Imaginary part zero.
$$1967\cdot3\cos\theta+1686\cdot7\sin\theta=0\implies 5901\cos\theta=-11802\sin\theta\implies\tan\theta=-\tfrac12.$$
Then $\cos^2\theta=\frac45$ and $\sin\theta\cos\theta=\frac{\tan\theta}{1+\tan^2\theta}=-\frac25$.

Real part.
$$z=\frac{13769-5058\sin\theta\cos\theta}{49+9\cos^2\theta}=\frac{13769+2023.2}{56.2}=281.$$

Only real values can be positive integers, and every real value of $z$ equals 281. So $n=281$.