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

[dogfish227]

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

[dogfish227]

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??

[Disiance]

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

[dogfish227]

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

[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

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


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.

[Halsafar]

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.

[wossname]

{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

It's suddenly SOOOO obvious!

[Halsafar]

It's suddenly SOOOO obvious!

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

[dogfish227]

thanks for that formula LHK. ill try it out.

[dogfish227]

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

[dogfish227]

nm found what i was looking for

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