Re: Roots of x^3 = 1;
From: Gib Bogle (bogle_at_ihug.too.much.spam.co.nz)
Date: 10/27/04
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Date: Wed, 27 Oct 2004 19:57:55 +1300
Saju wrote:
> Hello,
>
> I was teaching myself some algebra when I got stuck at the following
> ...
>
> I am trying to find roots of x^3 = 1
>
> x =1 is a solution. But keeping x = i^(4/3) also seems to work, so
> does i^(8/3), i^(16/3) and so on - I am able to generate 1 real root
> and infinite complex roots - this must be bogus.
>
> According to the theory, x^3 = 1 must have 3 roots (real+complex) - so
> what am I missing ?
>
> regards
> srp
1 = e^2nxi (where x = pi), for n = 0,1,2,3,...
1^(1/3) = e^(2nxi/3), for n = 0,1,2,3,...
This takes values that are points on the unit circle in the complex
plane with argument 0, 2x/3, 4x/3 for n = 0,1,2 (for higher values of n
these values are repeated). The first value is of course 1, the other
two are complex.
Gib
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