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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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Posts: 409
Registered: 3/1/08
Re: Let Z be a complex number. How do you (elegantly) prove that |z -
1| > 2 implies |z^3 - 1| > 1

Posted: Oct 29, 2013 2:49 AM
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One of my colleagues found a proof:

First we show that:

( cosa+sqrt(3+cosa^2))^3>=2cos3a , for any real a. (1)

Let x:= cosa+sqrt(3+cosa^2). x > 1

We have: -cosa+sqrt(3+cosa^2)=3/ cosa+sqrt(3+cosa^2)=3/x, resulting that cosa=(1/2)(x-3/x) from where (using cos(3a) identity) 2cos3a=8((1/2)(x-3/x))^3-6(1/2)(x-3/x)=

=(x-3/x))^3-3x+9/x=x^3-3x^2*3/x+3x9/x^2-27/x^3-3x+9/x=x^3-9x+27/x-27/x^3-3x+9/x (1) is rewritten in terms of x as:

0>=-3x+9/x-9x^3-x+3/x=-4x+12/x-9x^3 meaning 4x-12/x+9/x^3>=0 | x^3 ;
We obtain:
0<=4x^4-12x^2+9=(2x^2-3)^2 which is true!

We rewrite |z-1|>2 as ( z=r(cosa +isina) ,r>0): r^2-2rcosa-3>0 resulting r>cosa+sqrt(3+cosa^2), and so, r^3>( cosa+sqrt(3+cosa^2))^3>=2cos3a (from (1))

We have 1<1+2r^3cos3a+r^6=|z^3-1|.

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