If α and β are the roots of the equation 2z^2-3z-2i=0, where i=√(-1), then 16.

Question

If \(\alpha\) and \(\beta\) are the roots of the equation \(2 z^{2}-3 z-2 i=0,\) where \(i=\sqrt{-1},\) then 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

(1) 398

(2) 312

(3) 409

(4) 441

Answer 1

Correct option is (4) 441 

\(2 z^{2}-32-2 i=0\)

\( 2\left(z-\frac{i}{z}\right)=3 \)

\(\alpha-\frac{i}{\alpha}=\frac{3}{2} \)

\(\Rightarrow \alpha^{2}-\frac{1}{\alpha^{2}}-2 i=\frac{9}{4} \)

\(\Rightarrow \alpha^{2}-\frac{1}{\alpha^{2}}-2 i=\frac{9}{4} \)

\(\Rightarrow \frac{9}{4}+2 i=\alpha^{2}-\frac{1}{\alpha^{2}} \)

\(\Rightarrow \frac{81}{16}-4+9 i=\alpha^{4}+\frac{1}{\alpha^{4}}-2 \)

\(\Rightarrow \frac{49}{16}+9 i=\alpha^{4}+\frac{1}{\alpha^{4}}\)

Similarly

\(\Rightarrow \frac{49}{16}+9 i=\beta^{4}+\frac{1}{\beta^{4}}\)

\(\Rightarrow \frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}=\frac{\alpha^{15}\left(\alpha^{4}+\frac{1}{\alpha^{4}}\right)+\beta^{15}\left(\beta^{4}+\frac{1}{\beta^{4}}\right)}{\alpha^{15}+\beta^{15}} \)

\(=\frac{\left(\alpha^{15}+\beta^{15}\right)\left(\frac{49}{16}+9 i\right)}{\left(\alpha^{15}+\beta^{15}\right)}\)

Real = \(\frac{49}{16}\)

\(\operatorname{Im}=9\)