### Related Formula Since alpha$\alpha$ and beta$\beta$ are roots of 2z^2 - 3z - 2i = 0$2z^2 - 3z - 2i = 0$, they satisfy the quadratic equation directly, meaning: 2alpha^2 - 3alpha - 2i = 0 implies 2left(alpha - fracialpharight) = 3 implies alpha - fracialpha = frac32$2\alpha^2 - 3\alpha - 2i = 0 \implies 2\left(\alpha - \frac{i}{\alpha}\right) = 3 \implies \alpha - \frac{i}{\alpha} = \frac{3}{2}$ Similarly for beta$\beta$: beta - fracibeta = frac32$\beta - \frac{i}{\beta} = \frac{3}{2}$ ### Core Logic Square the baseline relation to transition to higher exponential powers: left(alpha - fracialpharight)^2 = left(frac32right)^2 implies alpha^2 - frac1alpha^2 - 2i = frac94$\left(\alpha - \frac{i}{\alpha}\right)^2 = \left(\frac{3}{2}\right)^2 \implies \alpha^2 - \frac{1}{\alpha^2} - 2i = \frac{9}{4}$ alpha^2 - frac1alpha^2 = frac94 + 2i$\alpha^2 - \frac{1}{\alpha^2} = \frac{9}{4} + 2i$ Squaring once more to isolate the fourth powers: left(alpha^2 - frac1alpha^2right)^2 = left(frac94 + 2iright)^2$\left(\alpha^2 - \frac{1}{\alpha^2}\right)^2 = \left(\frac{9}{4} + 2i\right)^2$ alpha^4 + frac1alpha^4 - 2 = frac8116 - 4 + 9i$\alpha^4 + \frac{1}{\alpha^4} - 2 = \frac{81}{16} - 4 + 9i$ alpha^4 + frac1alpha^4 = frac4916 + 9i$\alpha^4 + \frac{1}{\alpha^4} = \frac{49}{16} + 9i$ ### Step 1: Simplify the Target Expression Fraction Rearrange the given complex algebraic fraction by factoring out powers: fracalpha^19 + alpha^11 + beta^19 + beta^11alpha^15 + beta^15 = fracalpha^15left(alpha^4 + frac1alpha^4right) + beta^15left(beta^4 + frac1beta^4right)alpha^15 + beta^15$\frac{\alpha^{19} + \alpha^{11} + \beta^{19} + \beta^{11}}{\alpha^{15} + \beta^{15}} = \frac{\alpha^{15}\left(\alpha^4 + \frac{1}{\alpha^4}\right) + \beta^{15}\left(\beta^4 + \frac{1}{\beta^4}\right)}{\alpha^{15} + \beta^{15}}$ Since both alpha$\alpha$ and beta$\beta$ satisfy the exact same symmetric relational identity: alpha^4 + frac1alpha^4 = beta^4 + frac1beta^4 = frac4916 + 9i$\alpha^4 + \frac{1}{\alpha^4} = \beta^4 + \frac{1}{\beta^4} = \frac{49}{16} + 9i$ Substitute this uniform value back into the algebraic expression: = frac(alpha^15 + beta^15)left(frac4916 + 9iright)alpha^15 + beta^15 = frac4916 + 9i$= \frac{(\alpha^{15} + \beta^{15})\left(\frac{49}{16} + 9i\right)}{\alpha^{15} + \beta^{15}} = \frac{49}{16} + 9i$ ### Step 2: Extract Real and Imaginary Components From our simplified expression: mathrmRe = frac4916$\mathrm{Re} = \frac{49}{16}$ operatornameIm = 9$\operatorname{Im} = 9$ Now, substitute these into the evaluation formula: textResult = 16 cdot left(frac4916right) cdot 9 = 49 cdot 9 = 441$\text{Result} = 16 \cdot \left(\frac{49}{16}\right) \cdot 9 = 49 \cdot 9 = 441$ ### Pattern Recognition Symmetric rational polynomials in roots alpha, beta$\alpha, \beta$ that can be split into identical numeric multipliers for alpha^n$\alpha^n$ and beta^n$\beta^n$ allow direct cancellation of the polynomial bases without evaluating the individual roots explicitly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers