TS EAMCET · Maths · Complex Number
If \(\alpha\) is a root of \(z^2-z+1=0\), then \(\left(\alpha^{2014}+\frac{1}{\alpha^{2014}}\right)+\left(\alpha^{2015}+\frac{1}{\alpha^{2015}}\right)^2\) \[ \begin{aligned} & +\left(\alpha^{2016}+\frac{1}{\alpha^{2016}}\right)^3+\left(\alpha^{2017}+\frac{1}{\alpha^{2017}}\right)^4+ \ & \left(\alpha^{2018}+\frac{1}{\alpha^{2018}}\right)^5= \end{aligned} \]
- A 8
- B 5
- C 3
- D -5
Answer & Solution
Correct Answer
(A) 8
Step-by-step Solution
Detailed explanation
\(z^2-z+1=0\), then \(z=-\omega\)…
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