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Question Number 62308 by aliesam last updated on 19/Jun/19

Commented by maxmathsup by imad last updated on 20/Jun/19

$$x^2-2x+2=0\to {\mathrm{\Delta }}^{{}^{\prime }}=1-2=-1=i^2\Rightarrow \alpha =1+iand\beta =1-i=\stackrel{-}{\alpha }\Rightarrow$$$${\alpha }^n+{\beta }^n={\alpha }^n+{\left(\stackrel{-}{\alpha }\right)}^n=2Re\left({\alpha }^n\right)wehave\alpha =\sqrt{2}{e}^{\frac{i\pi }{4}}\Rightarrow {\alpha }^n={\left(\sqrt{2}\right)}^n{e}^{\frac{in\pi }{4}}$$$$\Rightarrow Re\left({\alpha }^n\right)=2^{\frac{n}{2}}cos\left(\frac{n\pi }{4}\right)\Rightarrow {\alpha }^n+{\beta }^n=2^{1+\frac{n}{2}}cos\left(\frac{n\pi }{4}\right)=2^{\frac{n+2}{2}}cos\left(\frac{n\pi }{4}\right).$$

Commented by aliesam last updated on 20/Jun/19

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by tanmay last updated on 19/Jun/19

$${(x-1)}^2+1=0$$$${(x-1)}^2-i^2=0$$$$(x-1+i)(x-1-i)=0$$$$\alpha =1+i\beta =1-i$$$${\alpha }^n+{\beta }^n$$$$={(1+i)}^n+{(1-i)}^n$$$$={\left[\sqrt{2}(\frac{1}{\sqrt{2}}+i\times \frac{1}{\sqrt{2}})\right]}^n+{\left[\sqrt{2}(\frac{1}{\sqrt{2}}-i\times \frac{1}{\sqrt{2}})\right]}^n$$$$={\left[\sqrt{2}(cos\frac{\pi }{4}+isin\frac{\pi }{4})\right]}^n+{\left[\sqrt{2}(cos\frac{\pi }{4}-isin\frac{\pi }{4})\right]}^n$$$$={\left[\sqrt{2}\times e^{i\times \frac{\pi }{4}}\right]}^n+{\left[\sqrt{2}\times e^{-i\times \frac{\pi }{4}}\right]}^n$$$$={\left(2\right)}^{\frac{n}{2}}[e^{i\times \frac{n\pi }{4}}+e^{-i\times \frac{n\pi }{4}}]$$$$=2^{\frac{n}{2}}\times 2cos\left(\frac{n\pi }{4}\right)$$$$=2^{\frac{n+2}{2}}\times cos\left(\frac{n\pi }{4}\right)$$

Commented by aliesam last updated on 19/Jun/19

$$\mathrm{god}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by tanmay last updated on 19/Jun/19

$$blessingsshowertoall.$$

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