JEE Main 2025 Apr 8 Shift 2, Mathematics Q14: Introduction to Complex Numbers And Their Conjugates

For any complex $z$, $2\mathrm{Re}(z)=z+\bar{z}$; applying this to $z=\frac{2\cos \theta +i\sin \theta }{\cos \theta -3i\sin \theta }$ gives $2\mathrm{Re}(z)=\frac{2\cos \theta +i\sin \theta }{\cos \theta -3i\sin \theta }+\frac{2\cos \theta -i\sin \theta }{\cos \theta +3i\sin \theta }$.

Combining fractions gives $2\mathrm{Re}(z)=\frac{(2{\cos }^2\theta -3{\sin }^2\theta )+(2{\cos }^2\theta -3{\sin }^2\theta )}{{\cos }^2\theta +9{\sin }^2\theta }=\frac{4{\cos }^2\theta -6{\sin }^2\theta }{{\cos }^2\theta +9{\sin }^2\theta }$; so $\mathrm{Re}(z)=\frac{2{\cos }^2\theta -3{\sin }^2\theta }{{\cos }^2\theta +9{\sin }^2\theta }$.

Setting $1+10\mathrm{Re}(z)=0$: $\frac{2{\cos }^2\theta -3{\sin }^2\theta }{{\cos }^2\theta +9{\sin }^2\theta }=-\frac{1}{10}$.

Cross-multiplying: $20{\cos }^2\theta -30{\sin }^2\theta =-{\cos }^2\theta -9{\sin }^2\theta$, giving $21{\cos }^2\theta -21{\sin }^2\theta =0$, i.e. $\cos 2\theta =0$.

Solutions in $[0,2\pi ]$: $2\theta =\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{7\pi }{2}$, so $\theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$.

Sum of squares: $\frac{{\pi }^2}{16}(1+9+25+49)=\frac{84{\pi }^2}{16}=\frac{21{\pi }^2}{4}$, matching option (1).