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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

 Messages: [ Previous | Next ]
 William Elliot Posts: 2,637 Registered: 1/8/12
Re: Let Z be a complex number.
Posted: Oct 27, 2013 4:09 AM

On Sat, 26 Oct 2013, dan.ms.chaos@gmail.com wrote:

> The original problem ... was to prove that
>
> (x*x + 2*x + y*y > 3) implies that
> ((x^3 - 3*x*y*y +1)^2 + (3*x*x*y-y*y*y)^2 > 1)
> for x ,y real numbers .
>

25 - 10 > 3
-125 + 1

Egads, terrible notation. For all x,y in R
. . x^2 + 2x + y^2 > 3
implies
. . (x^3 - 3xy^2 + 1)^2 + (3x^2 y - y^3)^2 > 1.

> I've figured that if you write z = iy - x , you can rewrite it in a simple
> way, namely:

> |z - 1| > 2 implies |z^3 - 1| > 1

Are you sure of "> 2"?

|z - 1| = sqr((x + 1)^2 + y^2)
|z^3 - 1| = sqr((-x^3 + 3xy^2 - 1)^2 + (3x^2 y - y^3)^2)

|z^3 - 1| = |z - 1|.|z^2 + z + 1|

If |z - 1| > sqr3, is |z^2 + z + 1| >= 1/sqr 3?

> The question is, does this now shortened problem have a short solution?

I don't see any.

Date Subject Author
10/25/13 dan.ms.chaos@gmail.com
10/25/13 William Elliot
10/26/13 dan.ms.chaos@gmail.com
10/27/13 William Elliot
10/27/13 dan.ms.chaos@gmail.com
10/27/13 William Elliot
10/28/13 dan.ms.chaos@gmail.com
10/28/13 William Elliot
10/28/13 Roland Franzius
10/26/13 David C. Ullrich
10/28/13 Rock Brentwood
10/29/13 dan.ms.chaos@gmail.com