The first thing to do is to put the number in the form of rhoe^(thetai)ρeθi
rho=sqrt((1/2)^2+(sqrt(3)/2)^2)=sqrt(1/4+3/4)=1ρ=
⎷(12)2+(√32)2=√14+34=1
theta=arctan((sqrt(3)/2)/(-1/2))=arctan(-sqrt(3))=-pi/3+kpiθ=arctan⎛⎜⎝√32−12⎞⎟⎠=arctan(−√3)=−π3+kπ. Let's choose (2pi)/32π3since we are in the second quadrant. Pay attention that -pi/3−π3 is in the fourth quadrant, and this is wrong.
Your number is now:
1e^((2pii)/3)1e2πi3
Now the roots are:
root(3)(1)e^(((2kpi+(2pi)/3)i)/3), k in ZZ
=e^((((6kpi+2pi)i)/9), k in ZZ
so you can choose k= 0, 1, 2 and obtain:
e^((2pii)/9, e^((8kpii)/9 and e^((14kpii)/9
or cos((2pi)/9)+ isin((2pi)/9), cos((8pi)/9)+ isin((8pi)/9) and
cos((14pi)/9)+ isin((14pi)/9).
For me this is a dead end, because I cannot compute trigonometric functions of multiples of pi/9. We must rely on a calculator:
0.7660+0.6428i
-0.9397+0.3420i
0.1736-0.9848i
