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JEE Mains · Maths · STD 11 - 4.1 complex nubers

Let \(\alpha\) and \(\beta\) be the roots of the equation \(x^{2}+(2 i -\) \(1)=0\). Then, the value of \(\left|\alpha^{8}+\beta^{8}\right|\) is equal to

  1. A \(50\)
  2. B \(250\)
  3. C \(1250\)
  4. D \(1500\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(50\)

Step-by-step Solution

Detailed explanation

\(X^{2}=1-2 i \quad \Rightarrow \alpha^{2}=1-2 i , \quad \beta^{2}=1-2 i\) Hence \(\alpha^{8}=\beta^{8}\) \(\left|\alpha^{8}+\beta^{8}\right|=\left|2 \alpha^{8}\right|=2\left|\alpha^{2}\right|^{4}\) \(=2 \sqrt{5}^{4}=50\)
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