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

**mous99** · December 6th 2012, 06:18 PM

Find (-64i)^1/3.

**MarkFL** · December 6th 2012, 06:40 PM

$(-64i)^{\frac{1}{3}}=4(-i)^{\frac{1}{3}}=4(e^{\frac{\pi}{2}(4k+3)})^{\frac {1}{3}}=4e^{\frac{\pi}{6}(4k+3)}$

where $k\in\{0,1,2\}$

Now, use Euler's formula to find the 3 roots.

**mous99** · December 19th 2012, 08:03 AM

(-64 i) ^1/3 = - 2√(3) - 2i???

**ebaines** · December 19th 2012, 08:51 AM

Originally Posted by mous99

(-64 i) ^1/3 = - 2√(3) - 2i???

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:

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

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

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

**mous99** · December 19th 2012, 09:09 AM

(-64i)^1/3 = (-64)^1/3 *(i)^1/3
= -4 (i)^1/3
= -4(cos π/6 + i sin π/6)

why the 0= 3π/ 6, not π/6???

**ebaines** · December 19th 2012, 10:35 AM

Originally Posted by mous99

(-64i)^1/3 = (-64)^1/3 *(i)^1/3

Not quite right: $(-64 i)^{1/3} = 64^{1/3}(- i)^{1/3} = 4 [ \cos(\frac {\frac { 3\pi} 2 + 2k \pi} 3) + i \sin(\frac {\frac { 3\pi} 2 + 2k \pi} 3 ) ]$
$= -4i, \ 2\sqrt 3 - 2i,\ -2 \sqrt 3 - 2i$ for $k = 0, \ 1,\ 2$ respectively.

**Plato** · December 19th 2012, 11:07 AM

Originally Posted by mous99

Find (-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:
$z^w=\exp(w\log(z))$ .

In this case the principal value:
$\left( { - 64i} \right)^{\frac{1}{3}} = \exp \left( {\frac{{\ln (64) - \frac{{i\pi }}{2}}}{3}} \right) = \exp \left( {\ln (4) - \frac{{i\pi }}{6}} \right)$

**HallsofIvy** · December 19th 2012, 11:18 AM

Just possibly all replies assume that the question is about cube roots because the question specifically [b]asked[/b about a cube root.

**Plato** · December 19th 2012, 12:09 PM

Originally Posted by HallsofIvy

Just possibly all replies assume that the question is about cube roots because the question specifically asked about a cube root.

Where in

Originally Posted by mous99

Find (-64i)^1/3.

is there anything about cube roots?

Is that not just complex exponentiation?

**mous99** · December 19th 2012, 05:39 PM

how to find θ= 3π/2 +2kπ,

because my answer for θ = π/2.

its no same with u.

(-64i)^1/3 = 4(-i)^1/3

z = re^iθ

z^3 = (r^3)(e^3θ )

so, r^3 = 64
r = 4

how to find 3θ = 3π/2 +2kπ???

**mous99** · December 22nd 2012, 08:18 AM

z = (-64i)^1/3
z^3 = (-64i)1/3

r = √[(0)^2 + (-64)^2 ] =64

cos 3 π/2 = 0
sin 3π/2 = -1

z^3 = (r^3) ( e^(3π/2 + 2kπ)

theta = π / 2 + (2kπ/3)

when k = 0, theta = π/2
z0 = 4i

when k =1, theta = ??? (how to find this theta)

**HallsofIvy** · December 22nd 2012, 12:49 PM

Did you consider putting k= 1 in the formula you are given?
$\theta= \frac{\pi}{2}+ \frac{2k\pi}{3}$
If k= 0, $\theta= \frac{\pi}{2}+ 0= \frac{\pi}{2}$ as you say.

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

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

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

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