Re: x^x = 1/4

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?

I know the answer is yes if f(z) is a polynomial with real
coefficients, and based on this thread, it's now clear that the answer
is also yes if f(z)=z^z.

How generally is this true?

As far as conditions on f, add them as needed.

quasi
.