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Problem: find three solutions to z^3=-1.
DeMoivre's theorem is that (cos b + i sin b)^n = cos bn + i sin bn
So we can set
z= (cos b + i sin b),
n = 3
cos bn + i sin bn = -1.
From the last equation, we know that cos bn = -1, and sin bn = 0.
Three possible solutions are bn=pi, bn=3pi, bn=5pi. This gives three possible values of b:
b=pi/3
b=pi
b = 5pi/3.
Now using z= (cos b + i sin b), we can get three possible cube roots of -1:
z= (cos pi/3 + i sin pi/3),
z= (cos pi + i sin pi),
z= (cos 5pi/3 + i sin 5pi/3).
Working these out gives
-1/2+i*sqrt(3)/2
-1
-1/2-i*sqrt(3)/2
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