If $\alpha$ and $\beta$ are the roots of the equation $2 z^2-3 z-2 \mathrm{i}=0$, where…

If $\alpha$ and $\beta$ are the roots of the equation $2 z^2-3 z-2 \mathrm{i}=0$, where $\mathrm{i}=\sqrt{-1}$, then $16 \cdot \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{lm}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)$ is equal to
  1. $441$
  2. $398$
  3. $312$
  4. $409$

Solution

$2 z^2-3 z-2 i=0$ ...(i)
$2\left(z-\frac{i}{z}\right)=3$
As $\alpha, \beta$ are roots of (i)
$\begin{aligned}
& \alpha-\frac{i}{\alpha}=\frac{3}{2} \\ & \Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4} \\ & \Rightarrow \alpha^2-\frac{1}{\alpha^2}=\frac{9}{4}+2 i
\end{aligned}$
Squaring both sides
$\begin{aligned}
& \Rightarrow \alpha^4+\frac{1}{\alpha^4}-2=\frac{81}{16}-4+9 i \\ & \Rightarrow \alpha^4+\frac{1}{\alpha^4}=\frac{49}{16}+9 i
\end{aligned}$
Similarly, $\beta^4+\frac{1}{\beta^4}=\frac{49}{16}+9 i$
$\begin{aligned}
& \frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}} \\ & =\frac{\alpha^{15}\left(\alpha^4+\frac{1}{\alpha^4}\right)+\beta^{15}\left(\beta^4+\frac{1}{\beta^4}\right)}{\alpha^{15}+\beta^{15}} \\ & =\frac{49}{16}+9 i \\ & \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)=\frac{49}{16} \\ & \operatorname{Im}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)=9 \\ & \Rightarrow 16 \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{lm}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \\ & =16 \times \frac{49}{16} \times 9 \\ & =441
\end{aligned}$

Asked in: JEE Main 2025 (24 Jan Shift 1)

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