Find (-64i)^1/3.

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- 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?

where

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

If k= 0, as you say.

If k= 1,

If k= 2,

If k= 3, which will give the same complex number as with k= 0 because sine and cosine (and so ) have period .