JEE Main 2025 Jan 24 Shift 1, Mathematics Q6: Introduction to Complex Numbers And Their Conjugates

The two roots $\alpha ,\beta$ satisfy $2z^2-3z-2i=0$, i.e. each root $z$ obeys $z-\frac{i}{z}=\frac{3}{2}$ (divide by $2z$).

Square both sides:

$z^2-2i-\frac{1}{z^2}\cdot (-1)=\frac{9}{4}\Rightarrow z^2-\frac{1}{z^2}=\frac{9}{4}+2i$

Square again:

$z^4+\frac{1}{z^4}-2={\left(\frac{9}{4}+2i\right)}^2=\frac{81}{16}-4+9i=\frac{17}{16}+9i$

$z^4+\frac{1}{z^4}=\frac{49}{16}+9i$

This identity holds for both roots, so for $z\in \{\alpha ,\beta \}$:

${\alpha }^4+\frac{1}{{\alpha }^4}={\beta }^4+\frac{1}{{\beta }^4}=\frac{49}{16}+9i$

Factor the target expression by pulling out ${\alpha }^{15}$ and ${\beta }^{15}$:

${\alpha }^{19}+{\alpha }^{11}={\alpha }^{15}\left({\alpha }^4+\frac{1}{{\alpha }^4}\right)$, similarly for $\beta$. So

$\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}=\frac{{\alpha }^{15}\left(\frac{49}{16}+9i\right)+{\beta }^{15}\left(\frac{49}{16}+9i\right)}{{\alpha }^{15}+{\beta }^{15}}=\frac{49}{16}+9i$

Therefore $\mathrm{Re}=\frac{49}{16}$, $\mathrm{Im}=9$, and

$16\cdot \frac{49}{16}\cdot 9=49\cdot 9=441$.

$\therefore$ Option (4).