Re: x^x = 1/4

In article <q3e7s1l7fkorsbs4s4plrs96ocut6905lr@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:
>On 9 Jan 2006 16:26:53 GMT, israel@xxxxxxxxxxx (Robert Israel) wrote:
>
>>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.
>>
>
>This makes sense. Thanks.
>
>Let me ask a related question.
>
>Suppose f(z) is a function from C to C which can be represented as "an
>expression in closed form with real coefficients" (whatever that
>means). Is it necessarily true that the nonreal zeros of f come in
>complex conjugate pairs?

If what it means is that f is an entire function whose Taylor series
about some real point has real coefficients, then the answer is yes.
Or more generally if U is a connected open set that is symmetric under
complex conjugation and intersects the reals R, f is analytic in U, and
f is real on some interval [a,b] with a < b in R intersect U, then
conjugate(f(z)) = f(conjugate(z)) for all z in U.
In particular, then, f(z) = 0 implies f(conjugate(z)) = 0.

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

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