Given that 1+Z6=0
Z6=−1Z=(−1)1/6Z=(−1+0∗i)1/6Zk=(cosπ+i∗sin(2∗k∗π+π))1/6k=0,1,2,3,4,5Zk=cos(2∗k+1)∗π/6
Z0=cosπ/6+isinπ/6=3/2+i∗1/2
Z1=cos(3∗π/6)+i∗sin(3∗π/6)=cosπ/2+i∗sinπ/2=0+i∗1
Z2=cos(5∗π/6)+i∗sin(5∗π/6)=−3/2+i∗1/2
Z3=cos(7∗π/6)+i∗sin(7∗π/6)=−3/2+i∗(−1/2)
Z4=cos(9∗π/6)+i∗sin(9∗π/6)=cos(3∗π/2)+i∗sin3∗π/2=0+i∗(−1)=0−i
Z5=cos(11∗π/6)+i∗sin(11∗π/6)=cos(360−30)+i∗sin(360−30)=3/2−i∗1/2
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