If and are the roots of the equation , where , then is equal to [2025]

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Solution

Q.
If $α$ and $β$ are the roots of the equation $2 z^{2} - 3 z - 2 i = 0$ , where $i = \sqrt{- 1}$ , then $16 \cdot Re (\frac{α^{19} + β^{19} + α^{11} + β^{11}}{α^{15} + β^{15}}) \cdot Im (\frac{α^{19} + β^{19} + α^{11} + β^{11}}{α^{15} + β^{15}})$ is equal to [2025]

1 441

2 409

3 312

4 398

Ans.
(1)

We have, $2 z^{2} - 3 z - 2 i = 0$ ... (i)

$\Rightarrow 2 (z - \frac{i}{z}) = 3$

$\Rightarrow α - \frac{i}{α} = \frac{3}{2}$ [ $∵ α$ is a root of equation (i)]

$\Rightarrow α^{2} - \frac{1}{α^{2}} - 2 i = \frac{9}{4} \Rightarrow α^{2} - \frac{1}{α^{2}} = \frac{9}{4} + 2 i$

$\Rightarrow α^{4} + \frac{1}{α^{4}} - 2 = \frac{81}{16} - 4 + 9 i \Rightarrow α^{4} + \frac{1}{α^{4}} = \frac{49}{16} + 9 i$

Similarly, $β^{4} + \frac{1}{β^{4}} = \frac{49}{16} + 9 i$

$\frac{α^{19} + β^{19} + α^{11} + β^{11}}{α^{15} + β^{15}} = \frac{α^{15} (α^{4} + \frac{1}{α^{4}}) + β^{15} (β^{4} + \frac{1}{β^{4}})}{α^{15} + β^{15}}$

$= \frac{(\frac{49}{16} + 9 i) (α^{15} + β^{15})}{α^{15} + β^{15}} = \frac{49}{16} + 9 i$

$∴ 16 \cdot Re (\frac{α^{19} + β^{19} + α^{11} + β^{11}}{α^{15} + β^{15}}) \cdot Im (\frac{α^{19} + β^{19} + α^{11} + β^{11}}{α^{15} + β^{15}})$

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

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