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Jan 22nd, 2005, 04:10 PM
#2
Thread Starter
Hyperactive Member
Re: 4,6,8... root of a negative
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??
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