Am 28.10.2013 07:48, schrieb firstname.lastname@example.org: > On Monday, 28 October 2013 03:07:18 UTC+2, William Elliot wrote: >> On Sun, 27 Oct 2013, email@example.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 . >
Well, substituting z-1 <-> w you have
Implies[ |w > 2 |, |w| | w^2 + 3 w + 3 | >1 ]
which means that the polynomial map V = w^2 + 3 w + 3 maps the outer of the circle |w| >2 onto the outer of the circle |V| >1/2
Since this is true for large w and since V absolutely squared is roughly rotationally invariant and has no real local minima for |w| > 2, it suffices to show that for a unit w = e^(i?) on the bounding circle of radius 2 the absolute square is larger than 1/4
Introducing w=q^2 or going to half angles ? = ?/2 (thats always better in order to avoid a cut on a single round on the unit circle)
Of course, this complex evaluation is nothing else but replacement of a algebraic evaluation of a inequality by a trigonometric one. Since it depends on the factorization of a (4,4)-rational expression into positive squares there will probably exist no simple interpretation by complex mappings.