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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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David C. Ullrich

Posts: 3,555
Registered: 12/13/04
Re: Let Z be a complex number. How do you (elegantly) prove that |z - 1| > 2 implies |z^3 - 1| > 1
Posted: Oct 26, 2013 2:02 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Fri, 25 Oct 2013 08:33:48 -0700 (PDT), dan.ms.chaos@gmail.com
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) ...

Don't put the question in the subject line!

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

I haven't checked whether this works. But I'd start by saying

z^3 - 1 = (z-1)(z-a)(z-b),

where a and b are the other two cube roots of 1.
Then I'd say

|z-a| >= |z-1| - |a-1| >= 2 - |a-1|

and see what developed...




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