Question Number 117188 by bemath last updated on 10/Oct/20
$${Given}\:{r}+\frac{\mathrm{1}}{{r}}\:=\:\sqrt{\mathrm{2}}\:,\:{then}\:{r}^{\mathrm{8}} +\frac{\mathrm{1}}{{r}^{\mathrm{8}} }\:=\:?\: \\ $$
Commented by bemath last updated on 10/Oct/20
$${thank}\:{you}\:{sirs} \\ $$
Answered by AbduraufKodiriy last updated on 10/Oct/20
$$\left(\boldsymbol{{r}}+\frac{\mathrm{1}}{\boldsymbol{{r}}}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:\Rightarrow\:\boldsymbol{{r}}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{\boldsymbol{{r}}^{\mathrm{2}} }=\mathrm{2}\:\Rightarrow\:\boldsymbol{{r}}^{\mathrm{2}} +\frac{\mathrm{1}}{\boldsymbol{{r}}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\boldsymbol{{r}}\:\boldsymbol{{is}}\:\boldsymbol{{Complex}} \\ $$$$\left(\boldsymbol{{r}}^{\mathrm{2}} +\frac{\mathrm{1}}{\boldsymbol{{r}}^{\mathrm{2}} }\right)=\mathrm{0}^{\mathrm{2}} \:\Rightarrow\:\boldsymbol{{r}}^{\mathrm{4}} +\mathrm{2}+\frac{\mathrm{1}}{\boldsymbol{{r}}^{\mathrm{4}} }=\mathrm{0}\:\Rightarrow\:\boldsymbol{{r}}^{\mathrm{4}} +\frac{\mathrm{1}}{\boldsymbol{{r}}^{\mathrm{4}} }=−\mathrm{2} \\ $$$$\left(\boldsymbol{{r}}^{\mathrm{4}} +\frac{\mathrm{1}}{\boldsymbol{{r}}^{\mathrm{4}} }\right)^{\mathrm{2}} =\left(−\mathrm{2}\right)^{\mathrm{2}} \:\Rightarrow\:\boldsymbol{{r}}^{\mathrm{8}} +\mathrm{2}+\frac{\mathrm{1}}{\boldsymbol{{r}}^{\mathrm{8}} }=\mathrm{4}\:\Rightarrow\:\boldsymbol{{r}}^{\mathrm{8}} +\frac{\mathrm{1}}{\boldsymbol{{r}}^{\mathrm{8}} }=\mathrm{2} \\ $$
Answered by TANMAY PANACEA last updated on 10/Oct/20
![r^2 −(√2) r+1=0 r=(((√2) ±(√(2−4)))/2)=(((√2) ±(√2) i)/2)→((1/( (√2)))±i(1/( (√2))))→(cos(π/4)±isin(π/4)) r=e^(i×(π/4)) or e^(−((iπ)/4)) r^8 +(1/r^8 ) =(e^(i×(π/4)) )^8 +(e^(−i×(π/4)) )^8 =e^(i2π) +e^(−i2π) [e^(iθ) =cosθ+isinθ] =2cos2π=2](https://www.tinkutara.com/question/Q117226.png)
$${r}^{\mathrm{2}} −\sqrt{\mathrm{2}}\:{r}+\mathrm{1}=\mathrm{0} \\ $$$${r}=\frac{\sqrt{\mathrm{2}}\:\pm\sqrt{\mathrm{2}−\mathrm{4}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{2}}\:\pm\sqrt{\mathrm{2}}\:{i}}{\mathrm{2}}\rightarrow\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\pm{i}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\rightarrow\left({cos}\frac{\pi}{\mathrm{4}}\pm{isin}\frac{\pi}{\mathrm{4}}\right) \\ $$$${r}={e}^{{i}×\frac{\pi}{\mathrm{4}}} \:{or}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$${r}^{\mathrm{8}} +\frac{\mathrm{1}}{{r}^{\mathrm{8}} } \\ $$$$=\left({e}^{{i}×\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{8}} +\left({e}^{−{i}×\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{8}} \\ $$$$={e}^{{i}\mathrm{2}\pi} +{e}^{−{i}\mathrm{2}\pi} \:\:\:\left[{e}^{{i}\theta} ={cos}\theta+{isin}\theta\right] \\ $$$$=\mathrm{2}{cos}\mathrm{2}\pi=\mathrm{2} \\ $$