Question Number 62308 by aliesam last updated on 19/Jun/19
Commented by maxmathsup by imad last updated on 20/Jun/19
$${x}^{\mathrm{2}} −\mathrm{2}{x}\:\:+\mathrm{2}\:=\mathrm{0}\:\rightarrow\Delta^{'} \:=\mathrm{1}−\mathrm{2}\:=−\mathrm{1}\:={i}^{\mathrm{2}} \:\Rightarrow\alpha\:=\mathrm{1}+{i}\:\:{and}\:\beta\:=\mathrm{1}−{i}\:=\overset{−} {\alpha}\:\Rightarrow \\ $$$$\alpha^{{n}} \:+\beta^{{n}} \:=\alpha^{{n}} \:+\left(\overset{−} {\alpha}\right)^{{n}} \:=\mathrm{2}\:{Re}\left(\alpha^{{n}} \right)\:\:\:\:{we}\:{have}\:\:\alpha\:=\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow\alpha^{{n}} \:=\left(\sqrt{\mathrm{2}}\right)^{{n}} \:{e}^{\frac{{in}\pi}{\mathrm{4}}} \: \\ $$$$\Rightarrow{Re}\left(\alpha^{{n}} \right)\:=\mathrm{2}^{\frac{{n}}{\mathrm{2}}} \:{cos}\left(\frac{{n}\pi}{\mathrm{4}}\right)\:\Rightarrow\alpha^{{n}} \:+\beta^{{n}} \:=\mathrm{2}^{\mathrm{1}+\frac{{n}}{\mathrm{2}}} \:{cos}\left(\frac{{n}\pi}{\mathrm{4}}\right)\:=\:\mathrm{2}^{\frac{{n}+\mathrm{2}}{\mathrm{2}}} \:{cos}\left(\frac{{n}\pi}{\mathrm{4}}\right). \\ $$
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 α=1+i β=1−i α^n +β^n =(1+i)^n +(1−i)^n =[(√2) ((1/(√2))+i×(1/(√2)))]^n +[(√2) ((1/(√2))−i×(1/(√2)))]^n =[(√2) (cos(π/4)+isin(π/4))]^n +[(√2) (cos(π/4)−isin(π/4))]^n =[(√2) ×e^(i×(π/4)) ]^n +[(√2) ×e^(−i×(π/4)) ]^n =(2)^(n/2) [e^(i×((nπ)/4)) +e^(−i×((nπ)/4)) ] =2^(n/2) ×2cos(((nπ)/4)) =2^((n+2)/2) ×cos(((nπ)/4))](Q62309.png)
$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} −{i}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}−\mathrm{1}+{i}\right)\left({x}−\mathrm{1}−{i}\right)=\mathrm{0} \\ $$$$\alpha=\mathrm{1}+{i}\:\:\:\beta=\mathrm{1}−{i} \\ $$$$\alpha^{{n}} +\beta^{{n}} \\ $$$$=\left(\mathrm{1}+{i}\right)^{{n}} +\left(\mathrm{1}−{i}\right)^{{n}} \\ $$$$=\left[\sqrt{\mathrm{2}}\:\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+{i}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\right]^{{n}} +\left[\sqrt{\mathrm{2}}\:\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}−{i}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\right]^{{n}} \\ $$$$=\left[\sqrt{\mathrm{2}}\:\left({cos}\frac{\pi}{\mathrm{4}}+{isin}\frac{\pi}{\mathrm{4}}\right)\right]^{{n}} +\left[\sqrt{\mathrm{2}}\:\left({cos}\frac{\pi}{\mathrm{4}}−{isin}\frac{\pi}{\mathrm{4}}\right)\right]^{{n}} \\ $$$$=\left[\sqrt{\mathrm{2}}\:×{e}^{{i}×\frac{\pi}{\mathrm{4}}} \right]^{{n}} +\left[\sqrt{\mathrm{2}}\:×{e}^{−{i}×\frac{\pi}{\mathrm{4}}} \right]^{{n}} \\ $$$$=\left(\mathrm{2}\right)^{\frac{{n}}{\mathrm{2}}} \left[{e}^{{i}×\frac{{n}\pi}{\mathrm{4}}} +{e}^{−{i}×\frac{{n}\pi}{\mathrm{4}}} \right] \\ $$$$=\mathrm{2}^{\frac{{n}}{\mathrm{2}}} ×\mathrm{2}{cos}\left(\frac{{n}\pi}{\mathrm{4}}\right) \\ $$$$=\mathrm{2}^{\frac{{n}+\mathrm{2}}{\mathrm{2}}} ×{cos}\left(\frac{{n}\pi}{\mathrm{4}}\right) \\ $$
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
$${blessings}\:{shower}\:{to}\:{all}. \\ $$