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Question Number 180653 by Shrinava last updated on 14/Nov/22

Answered by aleks041103 last updated on 17/Nov/22

$$A=\left(\begin{array}{cc}a& b\\ -b& a\end{array}\right)=a\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)+b\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)=aE+bS$$$$AB=(aE+bS)(cE+dS)=$$$$=acE+{bdS}^2+(ad+bc)S$$$$S^2=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)=-E$$$$\Rightarrow AB=(ac-bd)E+(ad+bc)S$$$$\Rightarrow E\equiv 1andS\equiv i\left(complexnums\right)$$$$\Rightarrow A^n\equiv x_n+{iy}_n={(x_1+{iy}_1)}^n$$$$...$$

Answered by aleks041103 last updated on 17/Nov/22

$$...$$$$z_1=\sqrt{\frac{x}{x^2+1}}(\frac{1}{\sqrt{3}}+\frac{i}{2})=\sqrt{\frac{7}{12}\frac{x}{x^2+1}}{e}^{i\theta }={re}^{i\theta }$$$$\theta =arctan\left(\frac{\sqrt{3}}{2}\right)$$$$\Rightarrow z_n=r^n{e}^{in\theta }=r^{n}cos\left(n\theta \right)+{ir}^{n}sin\left(n\theta \right)=x_n+{iy}_n$$$$\Rightarrow s_n=x_n+y_n=r^n(cos\left(n\theta \right)+sin\left(n\theta \right))=$$$$=\sqrt{2}{r}^{n}sin(\frac{\pi }{4}+n\theta )$$$$r\left(x\right)=\sqrt{\frac{7}{12}\frac{x}{x^2+1}}$$$$forx>0,\frac{x}{x^2+1}>0$$$${\left(\frac{x}{x^2+1}\right)}^{\prime }=\frac{x^2+1-2x^2}{{(x^2+1)}^2}=\frac{1-x^2}{{(1+x^2)}^2}$$$$\Rightarrow x=1\to max\Rightarrow x>0,0<\frac{x}{1+x^2}⩽\frac{1}{2}$$$$\Rightarrow x>0\Rightarrow 0<r\left(x\right)⩽\sqrt{\frac{7}{24}}<1$$$$\Rightarrow {\underset{n\to \mathrm{\infty }}{limr}}^n=0$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{lim}(x_n+y_n)=0$$

Commented by Shrinava last updated on 17/Nov/22

$$\mathrm{Thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{dear}\mathrm{professor},\mathrm{perfect}$$

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