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Question Number 169533 by cortano1 last updated on 02/May/22

$$\{\begin{array}{l}x+\frac{1}{y}=1\\ \frac{1}{x}+y=2\end{array}\Rightarrow x^{2022}+\frac{1}{y^{2022}}=?$$

Commented by greougoury555 last updated on 02/May/22

$$x=\frac{y-1}{y}\wedge \frac{y}{y-1}+y=2$$$$\Rightarrow y+y^2-y=2y-2$$$$\Rightarrow y^2-2y+2=0$$$$\Rightarrow {(y-1)}^2+1=0$$$$\Rightarrow y=1+i;x=\frac{i}{1+i}.\frac{1-i}{1-i}=\frac{1+i}{2}=\frac{1}{\sqrt{2}}{e}^{i\frac{\pi }{4}}$$$$\Rightarrow \frac{1}{y}=\frac{1}{1+i}.\frac{1-i}{1-i}=\frac{1-i}{2}=\frac{1}{\sqrt{2}}{e}^{-i\frac{\pi }{4}}$$$$\Rightarrow x^{2022}+\frac{1}{y^{2022}}=2^{-1011}{e}^{i\left(\frac{1011\pi }{2}\right)}+2^{-1011}{e}^{-i\left(\frac{1011\pi }{2}\right)}$$$$=2^{-1011}\left(\cos \frac{1011\pi }{2}\right)$$$$=2^{-1011}\times 0=0$$

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