How do you find three cube roots of #-i#?

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Answer 1 · 2017-10-23T21:14:43

#i#, #" "-sqrt(3)/2i-1/2i" "# and #" "sqrt(3)/2i-1/2i#

The primitive complex cube root of #1# is:

#omega = cos((2pi)/3)+isin((2pi)/3) = -1/2+sqrt(3)/2i#

Note that:

#i^3 = i^2 * i = -i#

So one of the cube roots is #i#. The other two cube roots can be found by multiplying by powers of #omega#.

#iomega = i(-1/2+sqrt(3)/2i) = -sqrt(3)/2i-1/2i#

#iomega^2 = i(-1/2-sqrt(3)/2i) = sqrt(3)/2i-1/2i#

Here are the three cube roots of #-i# plotted in the complex plane, together with the unit circle on which they lie...

graph{(x^2+(y-1)^2-0.002)((x-sqrt(3)/2)^2+(y+1/2)^2-0.002)((x+sqrt(3)/2)^2+(y+1/2)^2-0.002)(x^2+y^2-1) = 0 [-2.5, 2.5, -1.25, 1.25]}