If alpha and beta are the roots of the equati | vedclass

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(D) Given the equation $2z^2 - 3z - 2i = 0$. Dividing by $z$,we get $2z - 3 - \frac{2i}{z} = 0$,which implies $2(z - \frac{i}{z}) = 3$,or $z - \frac{i

Vedclass · Accepted Answer

$441$

Vedclass · Answer

(D) Given the equation $2z^2 - 3z - 2i = 0$. Dividing by $z$,we get $2z - 3 - \frac{2i}{z} = 0$,which implies $2(z - \frac{i}{z}) = 3$,or $z - \frac{i}{z} = \frac{3}{2}$. Since $\alpha$ and $\beta$ are roots,$\alpha - \frac{i}{\alpha} = \frac{3}{2}$ and $\beta - \frac{i}{\beta} = \frac{3}{2}$. Squaring both sides: $(\alpha - \frac{i}{\alpha})^2 = \alpha^2 + \frac{i^2}{\alpha^2} - 2i = \alpha^2 - \frac{1}{\alpha^2} - 2i = \frac{9}{4}$. Thus,$\alpha^2 - \frac{1}{\alpha^2} = \frac{9}{4} + 2i$. Similarly,$\beta^2 - \frac{1}{\beta^2} = \frac{9}{4} + 2i$. Consider the expression $E = \frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}} = \frac{\alpha^{15}(\alpha^4 + \alpha^{-4}) + \beta^{15}(\beta^4 + \beta^{-4})}{\alpha^{15} + \beta^{15}}$. Note that $(\alpha^2 - \alpha^{-2})^2 = \alpha^4 + \alpha^{-4} - 2 = (\frac{9}{4} + 2i)^2 = \frac{81}{16} - 4 + 9i = \frac{17}{16} + 9i$. So,$\alpha^4 + \alpha^{-4} = \frac{17}{16} + 9i + 2 = \frac{49}{16} + 9i$. Substituting this into $E$: $E = \frac{(\alpha^{15} + \beta^{15})(\frac{49}{16} + 9i)}{\alpha^{15} + \beta^{15}} = \frac{49}{16} + 9i$. Then $\operatorname{Re}(E) = \frac{49}{16}$ and $\operatorname{Im}(E) = 9$. The required value is $16 \cdot \operatorname{Re}(E) \cdot \operatorname{Im}(E) = 16 \cdot \frac{49}{16} \cdot 9 = 49 \cdot 9 = 441$.

List-$I$	List-$II$
$(i)$ $a \bar{a}$	$(A)$ $-\frac{\pi}{3}$
$(ii)$ $\arg \left(\frac{1}{\bar{a}}\right)$	$(B)$ $-i \sqrt{3}$
$(iii)$ $a - \bar{a}$	$(C)$ $2i / \sqrt{3}$
$(iv)$ $\operatorname{Im}\left(\frac{4}{3a}\right)$	$(D)$ $1$
	$(E)$ $\pi / 3$
	$(F)$ $\frac{2}{\sqrt{3}}$

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