JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Complex Numbers

Divide once: $\frac{z^2+8iz-15}{z^2-3iz-2}=1+\frac{11iz-13}{z^2-3iz-2}$, so the expression is real exactly when $\frac{11iz-13}{z^2-3iz-2}$ is real. Put $z=\alpha -\frac{13}{11}i$ with $\alpha \in \mathbb{R}$. The numerator becomes $11i\left(\alpha -\frac{13}{11}i\right)-13=11i\alpha +13-13=11i\alpha$, which is purely imaginary and non-zero because $\alpha \ne 0$. A non-zero purely imaginary number divided by a complex number is real only when that complex number is itself purely imaginary, so the real part of $z^2-3iz-2$ must vanish. With $z=x+iy$, $z^2-3iz-2=(x^2-y^2+3y-2)+i(2xy-3x)$, so $x^2-y^2+3y-2=0$. Substituting $x=\alpha$, $y=-\frac{13}{11}$ gives ${\alpha }^2=\frac{169}{121}+\frac{39}{11}+2=\frac{840}{121}$. Therefore $242{\alpha }^2=242\cdot \frac{840}{121}=1680$.