De Moivre's Theorem (-12-5i)^-3

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Homework Statement



Using De Moivres Theorem, solve (-12-5i)^-3

Homework Equations





The Attempt at a Solution



The solution i get for this problem is different from the one given in the exercise text. This is 1/2197cis(8.241)

Note: cis is equivalent to cos(\Theta)+isin(\Theta)
 
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well you haven't actually written an equation down so there's nothing to solve.
i'm assuming you meant to rewrite it in polar form and then use de moivre

write -12-5i in polar form, that is z=r cis(theta)
where r will be 13 if my mental pythagoras is correct and theta is arctan(y/x)

then use de moivre (-12-5i)^(-3)=z^(-3)=r^(-3) cis(-3 theta)
 
latentcorpse said:
well you haven't actually written an equation down so there's nothing to solve.
i'm assuming you meant to rewrite it in polar form and then use de moivre

write -12-5i in polar form, that is z=r cis(theta)
where r will be 13 if my mental pythagoras is correct and theta is arctan(y/x)

then use de moivre (-12-5i)^(-3)=z^(-3)=r^(-3) cis(-3 theta)

Remember that in the Argand diagram,-12-5i lies in the 3rd quadrant. Thus

\theta = tan^{-1}(y/x) - \pi
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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