
Re: Let Z be a complex number.
Posted:
Oct 28, 2013 4:12 AM


> 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 z1 > 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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