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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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William Elliot

Posts: 2,634
Registered: 1/8/12
Re: Let Z be a complex number.
Posted: Oct 28, 2013 4:12 AM
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> 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 ) .


What's 2|?

Let z = -x + yi. Then |z - 1| = |-1 - x + yi| = sqr((-1 - x)^2 + y^2)
. . = sqr(x^2 + 2x + y^2)
|z^3 - 1| = |-x^3 + 3x^2 yi - 3xy^2 - y^3 i|
. . = sqr((x^3 + 3xy^2)^2 + (3x^2 y - y^3)^2)

Thus |z - 1| > 2 iff sqr(x^2 + 2x + y^2) > 2 iff x^2 + 2x + y^2 > 4.
Are sure about "> 2"? Also
|z^3 - 1| > 1 iff sqr((x^3 + 3xy^2)^2 + (3x^2 y - y^3)^2) > 1.
. . iff (x^3 + 3xy^2)^2 + (3x^2 y - y^3)^2 > 1.

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