Find (-64i)^1/3.

Printable View

- Dec 6th 2012, 06:18 PMmous99how to (-64i)^1/3?
Find (-64i)^1/3.

- Dec 6th 2012, 06:40 PMMarkFLRe: how to (-64i)^1/3?
$\displaystyle (-64i)^{\frac{1}{3}}=4(-i)^{\frac{1}{3}}=4(e^{\frac{\pi}{2}(4k+3)})^{\frac {1}{3}}=4e^{\frac{\pi}{6}(4k+3)}$

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

Now, use Euler's formula to find the 3 roots. - Dec 19th 2012, 08:03 AMmous99Re: how to (-64i)^1/3?
(-64 i) ^1/3 = - 2√(3) - 2i???

- Dec 19th 2012, 08:51 AMebainesRe: 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 AMmous99Re: how to (-64i)^1/3?
(-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??? - Dec 19th 2012, 10:35 AMebainesRe: how to (-64i)^1/3?
Not quite right: $\displaystyle (-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 ) ] $

$\displaystyle = -4i, \ 2\sqrt 3 - 2i,\ -2 \sqrt 3 - 2i$ for $\displaystyle k = 0, \ 1,\ 2$ respectively. - Dec 19th 2012, 11:07 AMPlatoRe: 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 AMHallsofIvyRe: how to (-64i)^1/3?
Just possibly all replies assume that the question is about cube roots because the question specifically [b]asked[/b about a cube root.

- Dec 19th 2012, 12:09 PMPlatoRe: how to (-64i)^1/3?
- Dec 19th 2012, 05:39 PMmous99Re: how to (-64i)^1/3?
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??? - Dec 22nd 2012, 08:18 AMmous99Re: how to (-64i)^1/3?
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)

- Dec 22nd 2012, 12:49 PMHallsofIvyRe: 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 $\displaystyle 2\pi$.