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JEE Mains · Maths · STD 11 - 4.1 complex nubers

Let \(\alpha, \beta\) be the roots of the equation \(x^2-\sqrt{2} x+2=0\). Then \(\alpha^{14}+\beta^{14}\) is equal to

  1. A \(-64 \sqrt{2}\)
  2. B \(-128 \sqrt{2}\)
  3. C \(-64\)
  4. D \(-128\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(-128\)

Step-by-step Solution

Detailed explanation

\(x^2-\sqrt{2} x+2=0\) \(x=\frac{\sqrt{2} \pm \sqrt{2-8}}{2}=\frac{\sqrt{2} \pm \sqrt{6} i}{2}\) \(\alpha=\frac{\sqrt{2}+\sqrt{6} i}{2}=\sqrt{2} e^{\frac{i \pi}{3}}\) \(\beta=\sqrt{2} e^{\frac{-i \pi}{3}}\)…
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