Question-77132

Question Number 77132 by peter frank last updated on 03/Jan/20

Commented by kaivan.ahmadi last updated on 03/Jan/20

z=4(√3)((1/2)+i((√3)/2))−4(−((√3)/2)+i(1/2))= 2(√3)+6i+2(√3)−2i=4(√3)+4i=4((√3)+i)=8(((√3)/2)+i(1/2))= 8e^(i(π/6)) ⇒(z/8)+i((z/8))^2 +((z/8))^3 =e^(i(π/6)) +ie^(i(π/3)) +e^(i(π/2)) =2e^(i(π/2)) since e^(i(π/6)) +ie^(i(π/3)) =((√3)/2)+i(1/2)+(1/2)i−((√3)/2)=i=e^(i(π/2))

$${z}=\mathrm{4}\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−\mathrm{4}\left(−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+{i}\frac{\mathrm{1}}{\mathrm{2}}\right)= \\ $$$$\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{6}{i}+\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{2}{i}=\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{4}{i}=\mathrm{4}\left(\sqrt{\mathrm{3}}+{i}\right)=\mathrm{8}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+{i}\frac{\mathrm{1}}{\mathrm{2}}\right)= \\ $$$$\mathrm{8}{e}^{{i}\frac{\pi}{\mathrm{6}}} \\ $$$$\Rightarrow\frac{{z}}{\mathrm{8}}+{i}\left(\frac{{z}}{\mathrm{8}}\right)^{\mathrm{2}} +\left(\frac{{z}}{\mathrm{8}}\right)^{\mathrm{3}} ={e}^{{i}\frac{\pi}{\mathrm{6}}} +{ie}^{{i}\frac{\pi}{\mathrm{3}}} +{e}^{{i}\frac{\pi}{\mathrm{2}}} =\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{2}}} \\ $$$${since} \\ $$$${e}^{{i}\frac{\pi}{\mathrm{6}}} +{ie}^{{i}\frac{\pi}{\mathrm{3}}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+{i}\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{i}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}={i}={e}^{{i}\frac{\pi}{\mathrm{2}}} \\ $$

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