JEE Mains · Maths · STD 11 - 4.1 complex nubers
If \(\alpha\) and \(\beta\) are the roots of the equation \(2 z^2-3 z-2 \mathrm{i}=0\), where \(\mathrm{i}=\sqrt{-1}\), then \(16 \cdot \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{lm}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)\) is equal to
- A \(441\)
- B \(398\)
- C \(312\)
- D \(409\)
Answer & Solution
Correct Answer
(A) \(441\)
Step-by-step Solution
Detailed explanation
\(2 z^2-3 z-2 i=0\) ...(i) \(2\left(z-\frac{i}{z}\right)=3\) As \(\alpha, \beta\) are roots of (i)…
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