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 realvalued 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.

