# SOLUTION: Find the cube roots of -i.

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 Click here to see ALL problems on test Question 340848: Find the cube roots of -i.Answer by Edwin McCravy(8924)   (Show Source): You can put this solution on YOUR website!There are three cube roots of every number [except 0, which only has 0 as all of its roots of any kind.] ``` i written as a complex number of the form a+bi is 0+1i. The complex number a+bi is the vector that goes from the origin to the point (a,b). Therefore the number 0+1i is the vector that goes from the origin to the point (0,1) We want to put that number in trigonometric form. First we draw the vector 0+1i which goes from the origin to the point (0,1) and we see that its angle with the right hand side of the x-axis is 90°: So we see that the vector's magnitude (length) is r=1, and its angle is Therefore the trigonometric form of i, or 0+1i, is or in this case Now the cosine and sine will not change if we add or subtract 360° any whole number of times. So we can write that as: Next we use DeMoivre's theorem, which states that with , , , since the cube root is the power. Now we substitute any three consecutive values of integer n We use the easiest ones, 0, 1 and 2 Substituting n=0 Substituting n=1 Substituting n=2 So the three cube roots of i are , , Edwin```