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Re: question: How to get Mathematica to show more than one value of a complex valued function?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg46341] Re: question: How to get Mathematica to show more than one value of a complex valued function?
*From*: "David W. Cantrell" <DWCantrell at sigmaxi.org>
*Date*: Sat, 14 Feb 2004 22:19:51 -0500 (EST)
*References*: <200402140256.VAA08476@smc.vnet.net> <c0kr2e$fob$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
> On 14 Feb 2004, at 03:56, steve_H wrote:
>
> > hello;
> >
> > this is Mathematica 5.0
> >
> > given a multi-valued complex function, how can I get Mathematica to
> > show
> > more than one value? or to show the general formula for all solutions?
> >
> > for example, (-i)^i is multivalued. The result returned by
> > Mathematica is
> > Exp[Pi/2], which is the first possible value. But there are infinitly
> > many
> > others (if my math is correct), such as Exp[-3/2 Pi], Exp[5/2 Pi]
> > etc...
> > where the general function value is Exp[Pi/2 +- 2 n Pi]
> > n=0,1,2,3,....
> >
> Mathematica returns in such cases the so called principal value.
> However, in version 5 you can get multivalued answer by using:
>
> Reduce[x^I == -I, x]
I'd thought of mentioning the same sort of thing myself. But anyway, FWIW,
if you were trying use Steve's own example, then you made a small error.
You probably intended something like what I give below.
> C[1] $B":(B Integers &&
> x == E^((-I)*(2*I*Pi*C[1] -
> (I*Pi)/2))
>
> etc
In[1]:= Simplify[Reduce[z^(1/I) == -I, z]]
Out[1]= E^(Pi/2 + 2*Pi*C[1]) == z && C[1] ∈ Integers
Regards,
David Cantrell
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