Test Index

Complex Numbers Part 2

Section: Mathematics

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Question : 34 of 106

Marks: +1, -0

If

\alpha

\beta

are the roots of the equation

2z^2-3z-2i=

\; \;\text{ 0, where }\; \;\; i=\sqrt{-1}, \;\; \;\text{ then }\;

\; \;\text{ 16 }\; \cdot \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{lm}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)

[24 Jan 2025 Shift 1]

441
398
312
409

Solution: 👈: Video Solution

\;2z^2-3z-2i=0

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

As

\alpha, \beta

are roots of (i)

\;\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

Squaring both sides

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

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

Similarly,

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

\;\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

\;\operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)=\frac{49}{16}

\;\operatorname{lm}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)=9

\;\Rightarrow 16 \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{lm}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)

\;=16 \times \frac{49}{16} \times 9

\;=441

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