where

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

Results 1 to 12 of 12

- December 6th 2012, 07:18 PM #1

- Joined
- Oct 2012
- From
- malaysia
- Posts
- 32

- December 6th 2012, 07:40 PM #2

- December 19th 2012, 09:03 AM #3

- Joined
- Oct 2012
- From
- malaysia
- Posts
- 32

- December 19th 2012, 09:51 AM #4
## Re: how to (-64i)^1/3?

- December 19th 2012, 10:09 AM #5

- Joined
- Oct 2012
- From
- malaysia
- Posts
- 32

- December 19th 2012, 11:35 AM #6

- December 19th 2012, 12:07 PM #7

- December 19th 2012, 12:18 PM #8

- Joined
- Apr 2005
- Posts
- 17,730
- Thanks
- 2271

- December 19th 2012, 01:09 PM #9

- December 19th 2012, 06:39 PM #10

- Joined
- Oct 2012
- From
- malaysia
- Posts
- 32

- December 22nd 2012, 09:18 AM #11

- Joined
- Oct 2012
- From
- malaysia
- Posts
- 32

- December 22nd 2012, 01:49 PM #12

- Joined
- Apr 2005
- Posts
- 17,730
- Thanks
- 2271

## Re: 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 .