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4-1.Complex numbers
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If $\alpha$ and $\beta$ are the roots of the equation $2 z^2-3 z-2 i=0$, where $i=\sqrt{-1}$, then $\text { 16. } \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot$ $\operatorname{Im}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}} {\alpha^{15}+\beta^{15}}\right)$ is equal to
A$398$
B$312$
C$409$
D$441$
(JEE MAIN-2025)
Solution
$2 z^2-32-2 i=0$
$2\left(z-\frac{i}{z}\right)=3$
$\alpha-\frac{i}{\alpha}=\frac{3}{2}$
$\Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4}$
$\Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4}$
$\Rightarrow \frac{9}{4}+2 i=\alpha^2-\frac{1}{\alpha^2}$
$\Rightarrow \frac{81}{16}-4+9 i=\alpha^4+\frac{1}{\alpha^4}-2$
$\Rightarrow \frac{49}{16}+9 i=\alpha^4+\frac{1}{\alpha^4}$
Similarly
$\Rightarrow \frac{49}{16}+9 i=\beta^4+\frac{1}{\beta^4}$
$\Rightarrow \frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}=\frac{\alpha^{15}\left(\alpha^4+\frac{1}{\alpha^4}\right)+\beta^{15}\left(\beta^4+\frac{1}{\beta^4}\right)}{\alpha^{15}+\beta^{15}}$
$=\frac{\left(\alpha^{15}+\beta^{15}\right)\left(\frac{49}{16}+9 i\right)}{\left(\alpha^{15}+\beta^{15}\right)}$
$\text { Real }=\frac{49}{16}$
$\text { Im }=9$
$\text { Ans. } 441$
$2\left(z-\frac{i}{z}\right)=3$
$\alpha-\frac{i}{\alpha}=\frac{3}{2}$
$\Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4}$
$\Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4}$
$\Rightarrow \frac{9}{4}+2 i=\alpha^2-\frac{1}{\alpha^2}$
$\Rightarrow \frac{81}{16}-4+9 i=\alpha^4+\frac{1}{\alpha^4}-2$
$\Rightarrow \frac{49}{16}+9 i=\alpha^4+\frac{1}{\alpha^4}$
Similarly
$\Rightarrow \frac{49}{16}+9 i=\beta^4+\frac{1}{\beta^4}$
$\Rightarrow \frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}=\frac{\alpha^{15}\left(\alpha^4+\frac{1}{\alpha^4}\right)+\beta^{15}\left(\beta^4+\frac{1}{\beta^4}\right)}{\alpha^{15}+\beta^{15}}$
$=\frac{\left(\alpha^{15}+\beta^{15}\right)\left(\frac{49}{16}+9 i\right)}{\left(\alpha^{15}+\beta^{15}\right)}$
$\text { Real }=\frac{49}{16}$
$\text { Im }=9$
$\text { Ans. } 441$
Standard 11
Mathematics
