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

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 ]
dan.ms.chaos@gmail.com

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
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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!

Now
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|.



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 1994-2017. All Rights Reserved.