The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]

Posts: 409
Registered: 3/1/08
Re: Let Z be a complex number.
Posted: Oct 26, 2013 8:24 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Saturday, 26 October 2013 06:01:27 UTC+3, William Elliot wrote:
> On Fri, 25 Oct 2013, wrote:

> > Of course the proof boils down to elementary algebra, but I'm looking for a
> > way that avoids tedious calculations as much as possible (eg. some geometric
> > principle) ...
> What proof?

The proof I were to make substituting z as iy - x, and using Lagrange multipliers to solve the constraint. The original problem was not the one posted here. It 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 .

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
The question is, does this now shortened problem have a short solution?^3+-+3*x*y*y+%2B1%29^2+%2B+%283*x*x*y-y*y*y%29^2++%3D+1%29++and+%28x*x+%2B+2*x+%2B+y*y+%3D+3%29+++and+%28x*x+%2B+y*y+%3D+2^%282%2F3%29%29%29

The shape described by |z^3 - 1| > 1 looks like a sort of cycloid (hypotrochoid ?), other than that and the standard procedures for solving a function subject to inequality constraints, I'm stumped as to what a proper proof might look like.

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.