JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(z_0\) be a root of the quadratic equation, \(x^2 + x + 1= 0.\) If \(z = 3 + \,6iz_0^{81}\, - 3iz_0^{93}, \) then arg \(z\) is equal to
- A \(\frac {\pi }{4}\)
- B \(\frac {\pi }{3}\)
- C \(0\)
- D \(\frac {\pi }{6}\)
Answer & Solution
Correct Answer
(A) \(\frac {\pi }{4}\)
Step-by-step Solution
Detailed explanation
\(z_{0}=\omega\) or \(\omega^{2}\) (where \(\omega\) is a non - real cube root of unity) \(z=3+6 i(\omega)^{81}-3 i(\omega)^{93}\) \(z=3+3 i\) \(\Rightarrow \arg z=\frac{\pi}{4}\)
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