How do i write this in polar form? (cos(-2π/9)+isin(-2π/9))^3

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Answer 1 · 2018-02-12T17:28:29

#Cis((4pi)/3)#

#z=((cos((-2pi)/9)+isin((-2pi)/9))^3#

=#(cos((-2pi)/9*3)+isin((-2pi)/9*3)#

=#cos((-2pi)/3)+isin((-2pi)/3)#

=#cos((-2pi)/3+2pi)+isin((-2pi)/3+2pi)#

=#cos((4pi)/3)+isin((4pi)/3)#

=#1*Cis((4pi)/3)#

=#Cis((4pi)/3)#