JEE Advanced 2023 Paper 1, Mathematics Section 3 Q4: Introduction to Complex Numbers And Their Conjugates

Factor the numerator: $1967=281\cdot 7$ and $1686=281\cdot 6$, so

$A_{\theta }=\frac{281(7+6i\sin \theta )}{7-3i\cos \theta }$.

Rationalise by multiplying by $\frac{7+3i\cos \theta }{7+3i\cos \theta }$:

Numerator $=281(49-18\sin \theta \cos \theta +i(21\cos \theta +42\sin \theta ))$,

Denominator $=49+9{\cos }^2\theta$.

For $A_{\theta }$ to be a positive integer, the imaginary part must vanish:

$21\cos \theta +42\sin \theta =0\Rightarrow \tan \theta =-\frac{1}{2}$.

Then $\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }=-\frac{4}{5}$ and ${\cos }^2\theta =\frac{1}{1+{\tan }^2\theta }=\frac{4}{5}$.

Substitute:

$A_{\theta }=\frac{281(49-9\sin 2\theta )}{49+9{\cos }^2\theta }=\frac{281\left(49+\frac{36}{5}\right)}{49+\frac{36}{5}}=281$.

Hence the unique positive integer is $n=281$.