What is #(-\frac{1}{2}+\frac{i\sqrt{3}}{2})^{10}#?

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Answer 1 · 2014-10-29T17:53:22

By rewriting in trigonometric form,

#(-1/2+sqrt{3}/2i)^{10}=[cos({2pi}/3)+i sin({2pi}/3)]^{10}#

by De Moivre's Theorem,

#=cos(10cdot{2pi}/3)+i sin(10cdot{2pi}/3)#

#=cos({20pi}/3)+i sin({20pi}/3)#

#=cos({2pi}/3)+i sin({2pi}/3)#

#=-1/2+sqrt{3}/2i#

I hope that this was helpful.