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Question Number 81966 by naka3546 last updated on 17/Feb/20

$${\left(\frac{1+i\sqrt{3}}{2}\right)}^{2020}+{\left(\frac{1-i\sqrt{3}}{2}\right)}^{2020}=A$$$$A^4=?$$

Answered by TANMAY PANACEA last updated on 17/Feb/20

$$\frac{1}{2}=cos\frac{\pi }{3}\frac{\sqrt{3}}{2}=sin\frac{\pi }{3}$$$${(cos\frac{\pi }{3}+isin\frac{\pi }{3})}^{2020}+{(cos\frac{\pi }{3}-isin\frac{\pi }{3})}^{2020}$$$${\left(e^{i\times \frac{\pi }{3}}\right)}^{2020}+{\left(e^{-i\times \frac{\pi }{3}}\right)}^{2020}$$$$e^{i\theta }+e^{-i\theta }=2cos\theta where[\theta =\frac{2020\pi }{3}]$$$$A=2cos\theta$$$$A^4=16{\left(cos\theta \right)}^4$$$$nowcos\left(\frac{2020\pi }{3}\right)$$$$cos\left(2020\times 60\right)$$$$=cos\left(121200\right)$$$$=cos(336\times 360+240)$$$$=-cos60$$$$=\frac{-1}{2}$$$$\mathit{s}\mathit{o}\mathit{r}\mathit{e}\mathit{q}\mathit{u}\mathit{i}\mathit{r}\mathit{e}\mathit{d}\mathit{a}\mathit{n}\mathit{s}\mathit{w}\mathit{e}\mathit{r}\mathit{i}\mathit{s}16\times {\left(\frac{-1}{2}\right)}^4=1$$

Commented by naka3546 last updated on 17/Feb/20

$$\theta =\frac{4\pi }{3}\Rightarrow A=2\cos \theta =-1$$

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