Complex numbers – Quadratic Equations – JEE Main 24 January 2025 Shift 1

Solution:

Given the equation $2z^2 – 3z – 2i = 0$.
Rearrange terms: $2z^2 – 2i = 3z \Rightarrow 2(z^2 – i) = 3z$.
Divide by $z$ (since $z \ne 0$): $2(z – \frac{i}{z}) = 3 \Rightarrow z – \frac{i}{z} = \frac{3}{2}$.

Squaring both sides:
$$ \left(z – \frac{i}{z}\right)^2 = \frac{9}{4} $$
$$ z^2 + \frac{i^2}{z^2} – 2(z)\left(\frac{i}{z}\right) = \frac{9}{4} $$
$$ z^2 – \frac{1}{z^2} – 2i = \frac{9}{4} \Rightarrow z^2 – \frac{1}{z^2} = \frac{9}{4} + 2i $$

Squaring again:
$$ \left(z^2 – \frac{1}{z^2}\right)^2 = \left(\frac{9}{4} + 2i\right)^2 $$
$$ z^4 + \frac{1}{z^4} – 2 = \frac{81}{16} – 4 + 9i $$
$$ z^4 + \frac{1}{z^4} = \frac{81}{16} – 2 + 9i = \frac{49}{16} + 9i $$
Since both $\alpha$ and $\beta$ are roots, this holds for both. Let $K = z^4 + \frac{1}{z^4} = \frac{49}{16} + 9i$.

Now simplify the required expression $E$:
$$ E = \frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}} = \frac{\alpha^{15}(\alpha^4 + \alpha^{-4}) + \beta^{15}(\beta^4 + \beta^{-4})}{\alpha^{15} + \beta^{15}} $$
Note: $\alpha^{-4} = 1/\alpha^4$.
$$ E = \frac{\alpha^{15}(K) + \beta^{15}(K)}{\alpha^{15} + \beta^{15}} = \frac{K(\alpha^{15} + \beta^{15})}{\alpha^{15} + \beta^{15}} = K $$

So, $E = \frac{49}{16} + 9i$.
The real part $Re(E) = \frac{49}{16}$ and imaginary part $Im(E) = 9$.

We need to calculate:
$$ 16 \cdot Re(E) \cdot Im(E) = 16 \cdot \frac{49}{16} \cdot 9 $$
$$ = 49 \cdot 9 = 441 $$

Ans. (4)