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

**References**:**question: How to get Mathematica to show more than one value of a complex valued function?***From:*nma124@hotmail.com (steve_H)