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ikoreanx3
 one year ago
Best ResponseYou've already chosen the best response.0\[\sqrt[3]{{125}}\]

felavin
 one year ago
Best ResponseYou've already chosen the best response.1r= 5 , Theta= pi , n=3 , K=0,1,2 then \[\sqrt[3]{r} (\cos \frac{ 2k \pi+\theta }{ n ) +i \sin \frac{ 2k \pi+\theta }{ n } ) \] so, \[\[\sqrt[3]{r} (\cos \frac{ 2k \pi+\pi }{ n } ) +i \sin \frac{ 2k \pi+\pi }{ n} ) \] k=0 : w0=1.71 k=1 : \[\sqrt[3]{5} (\cos \frac{ 2*1 \pi+\pi }{ 3 } ) +i \sin \frac{ 2*1 \pi+\pi }{ 3} ) \]w1= 1.71+0.09i k=2 : w2: 1.70+0.16
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