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Topic: Square roots of complex numbers
Replies: 25   Last Post: Jul 22, 2011 10:50 PM

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 W^3 Posts: 29 Registered: 4/19/11
Re: Square roots of complex numbers
Posted: Jul 21, 2011 3:25 PM

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.