W^3
Posts:
28
Registered:
4/19/11
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Re: Square roots of complex numbers
Posted:
Jul 21, 2011 3:25 PM
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In article <4E2796A3.2080700@netscape.net>,
"Stephen J. Herschkorn" <sjherschko@netscape.net> wrote:
> W^3 wrote:
>
> >In article <4E2676DB.80109@netscape.net>,
> > "Stephen J. Herschkorn" <sjherschko@netscape.net> wrote:
> >
> >
> >
> >>So using principal values correctly, if complex z is nonnegative and
> >>r is real, then conj(z^r) = (conj(z))^r. Makes much more sense.
> >>
> >>
> >
> >The equation conj(f(z)) = f(conj(z)) holds for lots of analytic
> >functions f.
> >
> >Thm: Suppose f is analytic on a connected open subset U of the complex
> >plane that intersects R. Assume further that U = conj(U), i.e., U is
> >symmetric about R. If f is real-valued on a nonempty open subinterval
> >of U intersect R, then conj(f(z)) = f(conj(z)) for all z in U (in
> >particular, f is real on all of U intersect R).
> >
> >
> Does that follow from my observation above and the fact that an analytic
> funciton can be represented by a power series?
In simple situations, yes. But in general you there won't be a power
series that equals f in the whole domain, so you need a different
argument.
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