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mathman
#5
Oct15-09, 03:45 PM
Sci Advisor
P: 6,077
Algebraically if we have , then taking the cube root of both sides should yield -1, not a complex number.
There are three cube roots. -1 is one of them, the other two are complex conjugates
given by [tex]e^{i\frac{\pi}{3}}[/tex] and [tex]e^{-i\frac{\pi}{3}}[/tex].