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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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Roland Franzius

Posts: 586
Registered: 12/7/04
Re: Let Z be a complex number.
Posted: Oct 28, 2013 5:57 AM
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Am 28.10.2013 07:48, schrieb dan.ms.chaos@gmail.com:
> 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 |z-1| > 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 .
>


Well, substituting z-1 <-> w you have

Implies[ |w > 2 |, |w| | w^2 + 3 w + 3 | >1 ]

which means that the polynomial map V = w^2 + 3 w + 3 maps the outer
of the circle |w| >2 onto the outer of the circle |V| >1/2

Since this is true for large w and since V absolutely squared is roughly
rotationally invariant and has no real local minima for
|w| > 2,
it suffices to show that for a unit w = e^(i?) on the bounding circle of
radius 2 the absolute square is larger than 1/4

Introducing w=q^2 or going to half angles ? = ?/2
(thats always better in order to avoid a cut on a single round on the
unit circle)

((4 w^2 + 6 w + 3 )(4/w^2 + 6/w +3 )) /. {w -> q^2}


(w^2 + 3 w + 3) (w^-2 + 3/w + 3)>1/4
-> 1 + 12 (q + q^-1)^4 - 6 (q + q^-1)^2 > 1/4
-> 3/4 + 6 x^2 (2 x^2 -1) > 0

The minimum 0 is at x = 2 cos (2?) = 1/4

Of course, this complex evaluation is nothing else but replacement of a
algebraic evaluation of a inequality by a trigonometric one. Since it
depends on the factorization of a (4,4)-rational expression into
positive squares there will probably exist no simple interpretation by
complex mappings.

--

Roland Franzius



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