Find (-64i)^1/3.

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- Dec 6th 2012, 06:18 PM #1

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- Dec 6th 2012, 06:40 PM #2

- Dec 19th 2012, 08:03 AM #3

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- Dec 19th 2012, 08:51 AM #4
## Re: how to (-64i)^1/3?

That is one of he cube roots of -64i. There are two others - do you know how to find them? You can use de Moivre's tehorem:

$\displaystyle ( R \cos \theta + i \sin \theta)^n = R^n (\cos (n \theta) + \sin (n \theta)) $

From this you can derive the fact that for $\displaystyle z = R(\cos \theta + i \sin \theta)$, $\displaystyle z^{1/n} = R^{1/n}[\cos(\frac {\theta + 2 \pi k}n) + i \sin (\frac {\theta + 2 k \pi} n)]$

Here you have $\displaystyle R = 64$, $\displaystyle \theta = \frac {3 \pi}2$, and $\displaystyle n = 3 $. Set k = 0, 1, and 2 to find the three cube roots.

- Dec 19th 2012, 09:09 AM #5

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- Dec 19th 2012, 10:35 AM #6
## Re: how to (-64i)^1/3?

- Dec 19th 2012, 11:07 AM #7
## Re: how to (-64i)^1/3?

I don't understand why all the replies assume that the question is about cube roots.

I would have thought that it is straightforward complex expatiation:

$\displaystyle z^w=\exp(w\log(z))$.

In this case**the principal value**:

$\displaystyle \left( { - 64i} \right)^{\frac{1}{3}} = \exp \left( {\frac{{\ln (64) - \frac{{i\pi }}{2}}}{3}} \right) = \exp \left( {\ln (4) - \frac{{i\pi }}{6}} \right)$

- Dec 19th 2012, 11:18 AM #8

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- Dec 19th 2012, 12:09 PM #9

- Dec 19th 2012, 05:39 PM #10

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- Dec 22nd 2012, 08:18 AM #11

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- Dec 22nd 2012, 12:49 PM #12

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## Re: how to (-64i)^1/3?

Did you consider putting k= 1 in the formula you are given?

$\displaystyle \theta= \frac{\pi}{2}+ \frac{2k\pi}{3}$

If k= 0, $\displaystyle \theta= \frac{\pi}{2}+ 0= \frac{\pi}{2}$ as you say.

If k= 1, $\displaystyle \theta= \frac{\pi}{2}+ \frac{2\pi}{3}= \frac{3\pi}{6}+ \frac{4\pi}{6}= \frac{7\pi}{6}$

If k= 2, $\displaystyle \theta= \frac{\pi}{2}+ \frac{4\pi}{3}= \frac{3\pi}{6}+ \frac{8\pi}{6}= \frac{11\pi}{6}$

If k= 3, $\displaystyle \theta= \frac{\pi}{2}+ \frac{6\pi}{3}= \frac{3\pi}{6}+ \frac{12\pi}{6}= \frac{15\pi}{6}$ which will give the same complex number as with k= 0 because sine and cosine (and so $\displaystyle e^{ix}$) have period $\displaystyle 2\pi$.