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Question Number 82843 by M±th+et£s last updated on 24/Feb/20

show that (((1+(√3) i)^4 (1+i)^8 )/((cos100°−i sin100)^3 ))=−256

$${show}\:{that} \\ $$$$\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\:{i}\right)^{\mathrm{4}} \left(\mathrm{1}+{i}\right)^{\mathrm{8}} }{\left({cos}\mathrm{100}°−{i}\:{sin}\mathrm{100}\right)^{\mathrm{3}} }=−\mathrm{256} \\ $$

Answered by mind is power last updated on 24/Feb/20

cos(100)−isin(100)=cos(((5π)/9))−isin(((5π)/9)) =e^(−((i5π)/9)) (((1+i(√3))^4 (1+i)^8 )/((e^(−((i5π)/9)) )^3 ))=(((2((1/2)+((i(√3))/2)))^4 ((√2)((1/(√2))+(i/(√2))))^8 )/e^(−((5iπ)/3)) ) =((2^8 (e^(i(π/3)) )^4 (e^((iπ)/4) )^8 )/e^(−5i(π/3)) )=((2^8 e^(i(((4π)/3)+2π+((5π)/3))) )/)=2^8 e^(i(5π)) =2^8 (−1)=−256

$${cos}\left(\mathrm{100}\right)−{isin}\left(\mathrm{100}\right)={cos}\left(\frac{\mathrm{5}\pi}{\mathrm{9}}\right)−{isin}\left(\frac{\mathrm{5}\pi}{\mathrm{9}}\right) \\ $$$$={e}^{−\frac{{i}\mathrm{5}\pi}{\mathrm{9}}} \\ $$$$\frac{\left(\mathrm{1}+{i}\sqrt{\mathrm{3}}\right)^{\mathrm{4}} \left(\mathrm{1}+{i}\right)^{\mathrm{8}} }{\left({e}^{−\frac{{i}\mathrm{5}\pi}{\mathrm{9}}} \right)^{\mathrm{3}} }=\frac{\left(\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right)^{\mathrm{4}} \left(\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+\frac{{i}}{\sqrt{\mathrm{2}}}\right)\right)^{\mathrm{8}} }{{e}^{−\frac{\mathrm{5}{i}\pi}{\mathrm{3}}} } \\ $$$$=\frac{\mathrm{2}^{\mathrm{8}} \left({e}^{{i}\frac{\pi}{\mathrm{3}}} \right)^{\mathrm{4}} \left({e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{8}} }{{e}^{−\mathrm{5}{i}\frac{\pi}{\mathrm{3}}} }=\frac{\mathrm{2}^{\mathrm{8}} {e}^{{i}\left(\frac{\mathrm{4}\pi}{\mathrm{3}}+\mathrm{2}\pi+\frac{\mathrm{5}\pi}{\mathrm{3}}\right)} }{}=\mathrm{2}^{\mathrm{8}} {e}^{{i}\left(\mathrm{5}\pi\right)} =\mathrm{2}^{\mathrm{8}} \left(−\mathrm{1}\right)=−\mathrm{256} \\ $$

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