Re: x^x = 1/4

In article <4uk4s11veqoqhbr0rp5f89sqgs9m60t7h2@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:
>On 8 Jan 2006 23:54:17 -0500, "Keith F. Lynch" <kfl@xxxxxxxxxxxxxx>
>wrote:

>>If you're speaking of the change of sign of the imaginary, that's
>>meaningless. Any solution to x^x equals a real number has another
>>solution with the opposite sign on the imaginary, i.e. the complex
>>conjugate.
>
>Is this obvious?

Yes.

>Even if it is, can you explain it a little?

conjugate(x^y) = conjugate(exp(y ln(x)))
= exp(conjugate(y ln(x)))
= exp(conjugate(y) ln(conjugate(x)))
= conjugate(x)^conjugate(y)

Assuming you're taking the principal branch, this is valid except in the
case where x is a negative real: thus (-1)^y = exp(i pi y) and
(-1)^conjugate(y) = exp(i pi conjugate(y)), not exp(-i pi conjugate(y)).
If you allow all branches, the statement is that every value of
conjugate(x^y) is a value of conjugate(x)^conjugate(y) and vice versa.

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada


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Relevant Pages

  • Re: x^x = 1/4
    ... >>>If you're speaking of the change of sign of the imaginary, ... >>>meaningless. ... Any solution to x^x equals a real number has another ... >Let x be the general solution. ...
    (sci.math)
  • Re: x^x = 1/4
    ... >>If you're speaking of the change of sign of the imaginary, ... >>meaningless. ... Any solution to x^x equals a real number has another ... Let x be the general solution. ...
    (sci.math)
  • Re: x^x = 1/4
    ... >> If you're speaking of the change of sign of the imaginary, ... >> meaningless. ... Any solution to x^x equals a real number has another ...
    (sci.math)