
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


On Friday, October 25, 2013 10:33:48 AM UTC5, 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.

