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

## Homework Statement

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

## 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($$\Theta$$)

## The Attempt at a Solution

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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)

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

$$\theta = tan^{-1}(y/x) - \pi$$