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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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Rock Brentwood

Posts: 129
Registered: 6/18/10
Re: Let Z be a complex number. How do you (elegantly) prove that |z -
1| > 2 implies |z^3 - 1| > 1

Posted: Oct 28, 2013 8:24 PM
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On Friday, October 25, 2013 10:33:48 AM UTC-5, dan.ms...@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) ...

For convenience, define 1^{x} = exp(2 pi i x). Then
|z^3 - 1| = |z - 1| |z - 1^{+1/3}| |z - 1^{-1/3}|
> 2 |z - 1^{+1/3}| |z - 1^{-1/3}|.

Since |z - 1| > 2 and 1^{+/- 1/3} = -1/2 +/- i root(3/4) and
1 - 1^{+/- 1/3} = 3/2 -/+ i root(3/4)
then
|1 - 1^{+/- 1/3}| = root(9/4 + 9/16) = 3/4 root(5)
and
|z - 1^{+/- 1/3}|
>= | |z - 1| - |1 - 1^{+/- 1/3}| |
= | |z - 1| - 3/4 root(5)|
> 2 - 3/4 root(5).
Therefore
|z^3 - 1| > 2 (2 - 3/4 root(5))^2 = 109/8 - 6 root(5).

That's the best I can do. The right hand side is slightly less than 1:
109/8 - 6 root(5) > 1
iff 101/8 > 6 root(5)
iff 101/48 > root(5)
iff 10201/2304 > 5
iff 10201 > 11520.



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