Re: Roots of x^3 = 1;
From: Stan Brown (the_stan_brown_at_fastmail.fm)
Date: 10/27/04
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Date: Wed, 27 Oct 2004 18:01:55 -0400
"Saju" <saju.pillai@gmail.com> wrote in sci.math:
>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 ?
My "Powers and Roots of a Complex Number" at
http://oakroadsystems.com/twt/twtnotes.htm#ComplexRoot might be
helpful. If you need to review polar form, that's the immediately
preceding section on the same page.
I'll give you the three cube roots of 1:
1 itself
-(1/2) + i*sqrt(3)/2
-(1/2) - i*sqrt(3)/2
You can easily verify them, but the above page tells you how to
derive Nth roots of any complex number (which of course includes any
real number).
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
Fortunately, I live in the United States of America, where we are
gradually coming to understand that nothing we do is ever our
fault, especially if it is really stupid. --Dave Barry
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