[resolved] 4,6,8... root of a negative [/resolved]

4,6,8... root of a negative

ok i know how to take the sqaure root of a negative but ive been searching the web for an hour know and havent found any thing about even roots higher than that (well techincly lower 1/2 > 1/4)

i know from typing it into a calc thats its possible and it will return an number like xi + c but i cant figure out how it comes about getting that answer.

so please post any links or advise on how to solve thanks.

-Nate

well ive figured it out for 4th root just treat it like (-4)^(1/4) as ((-4)^(1/2))^(1/2) then you get

2i^(1/2)

so you must find the number A + Bi such that it ^2 = 2i so just say

(A +Bi)^2 = 2i
(A + Bi)(A + Bi) = 2i
A^2 + ABi - B^2 = 2i

A^2 - B^2 must = 0 as there is no constant
so A^2 - B^2 = 0 and solve
A^2 = B^2 and so
A = B

so A^2 + A^2 i - A^2 = 2i
A^2 i = 2i
A^2 = 2
A = 2^(1/2)

and you get your answer but i cant seem to get it to work out for 6,8,10 ect.

any ideas??

First of all you can't have negative numbers from an even power...
Second to get a root just do number^(1/power).

well you can if you use the imaginary number system. which it what im asking about

{r*(cos x + i sin x)}^(1/N) = r^(1/N) [cos(x/N+2PIk/N) + i sin(x/N+2PIk/N)] for k = 0,1,2,...,N

From:
http://oakroadsystems.com/twt/twtnotes.htm#eq82
:wave:

So, the trick is to determine, from some number A+iB which you want to take the root of, the values r and x.

Well, lets see.
If we say r = (A² + B²)^(1/2)
Then
A+iB = r*(A/r + i(B/r))
So cos(x) = A/r, and sin(x) = B/r.

:)

The root of a negetive is an imiginary number.
There are no two alike numbers when multiplied that will equal a negetive.

I hope you understand this much:
Sqrt(-1)...You can't take -1 * -1 = 1, nor can you 1 * 1 = 1.
So we call the sqrt(-1) an imiginary number i, or k.

Now i * i or i^2 = -1

So the sqrt(-4)
= (i^2 * 2)
= -1 * 2
= -2

I believe that is how it is done.
Just remeber, imaginary number * imaginary number always equals -1.
Follow that rule and you can "imagine" the sqrt of any negetive number.

It this idea that will lead into Quaternions.

Quote:

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Originally Posted by NotLKH

{r*(cos x + i sin x)}^(1/N) = r^(1/N) [cos(x/N+2PIk/N) + i sin(x/N+2PIk/N)] for k = 0,1,2,...,N

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It's suddenly SOOOO obvious! :D:D:D

Quote:

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Originally Posted by wossname

It's suddenly SOOOO obvious! :D:D:D

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Is that suppose to be sarcastic.
That equation went right over me head.

thanks for that formula LHK. ill try it out.

um LKH could you show me an example of how to use the equation you've given me.

like how could i solve (3i+5)^(1/3)

thanks
-nate

nm found what i was looking for

here:
http://www.vibrationdata.com/arbit_root.htm