Re: Roots of x^3 = 1;

From: Saju (saju.pillai_at_gmail.com)
Date: 10/28/04 

Date: 28 Oct 2004 01:26:55 -0700

Stan Brown <the_stan_brown@fastmail.fm> wrote in message news:<MPG.1be9e4e3fc18cfd298cbce@news.odyssey.net>...
> "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.

Thanks, will read up on this.

I managed to figure out deal on these roots on my own (After some
prodding on this group) using Euler's identity. I noticed that the
roots occur on 1, cos(2pi/3) + isin(2pi/3) and cos(4pi/3) +
isin(4pi/3) - the higher roots cycle back - i believe a poster here
called these roots cyclotomic.

It was also fun reaching Euler's identity by setting x = i*theta in
the power series expansion of e^x.

regards
srp

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