Mathematics - Complex Number Question with Solution | TestHub

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