
Re: Let Z be a complex number.
Posted:
Oct 28, 2013 2:48 AM


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 z1 > 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 .

