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

[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.

[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.