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Topic: Let Z be a complex number. How do you (elegantly) prove that |z - 1|
> 2 implies |z^3 - 1| > 1

Replies: 11   Last Post: Oct 29, 2013 2:49 AM

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dan.ms.chaos@gmail.com

Posts: 409
Registered: 3/1/08
Re: Let Z be a complex number.
Posted: Oct 28, 2013 2:48 AM
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On Monday, 28 October 2013 03:07:18 UTC+2, William Elliot wrote:
> On Sun, 27 Oct 2013, dan.ms.chaos@gmail.com wrote:
>
>
>

> > > Are you sure of "> 2"?
>
> >
>
> > sqr((x + 1)^2 + y^2) > 2 can be rewritten as
>
> > ((x + 1)^2 + y^2) > 4 , <=>
>
> > x^2 + 2x + 1 +y^2 > 4 <=>
>
> > x^2 +2x + y^2 > 3 which is what we've started with.
>
> > The same goes for the other relation.
>
>
>
> What happen to the problem statement?
>
> It disappeared and with it's disappearance, poof, no problem!


There are two equivalent statements of the same problem . The first is to prove that for complex z, |z - 1| > 2 implies |z^3 - 1| > 1 , the second is made by substituting z with |i*y - x| , and proving that one
x^2 + 2x + y^2 > 3 (which is equivalent to |z-1| > 2|
implies x^3 - 3xy^2 + 1)^2 + (3x^2 y - y^3)^2 > 1. (which is equivalent to
|z^3 - 1| > 1 ) .The second was actually the original statement of the problem, I've made the first by noticing the pattern that corresponds to complex multiplication .



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