Re: (-8)^(1/3)

*To*: mathgroup at smc.vnet.net*Subject*: [mg31301] Re: [mg31265] (-8)^(1/3)*From*: Otto Linsuain <linsuain at andrew.cmu.edu>*Date*: Sat, 27 Oct 2001 01:08:10 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Dear LuisMa, the answer Mathematica give you is the correct one in the following sense. Every complex number has three complex cubic roots in the complex plane, one has to choose which root one wants to call z^(1/3). This freedom to choose is, however, limited, if you choose (1)^(1/3)=1, then when you move, say counterclockwise, to -1, you will find that (-1)^(1/3)=Exp[i Pi/3]. To have (1)^(1/3)=1 and (-1)^(1/3)=-1 would require a discontinuity at some point. This situation is refered to as to the function having 3 branches, and you just have to pick one branch. (1)^(1/3)=1 and (-1)^(1/3)=-1 are on different branches, so you can have either, but not both. Having said this, you can have (-8)^(1/3)=-2 (which is the same branch as (-1)^(1/3)=-1). I don't know that there is a way to tell the function Power (which is the underlying function here) which branch to choose, but you can define your one little function. All you would have to do would be to think of -8 as having phase 3 \Pi (rather than \Pi). Then (-8)^(1/3)=(8^(1/3)) Exp[i 3 \Pi /3]=2 Exp[i \Pi]=-2. Notice that you would then have to choose the phase of +8 to be 2 \Pi, or 4 \Pi, which will make 8^(1/3) != 2 Hope it helps, Otto Linsuain. On Fri, 26 Oct 2001, LuisMa wrote: > When I enter > > (-8)^(1/3) > > Mathematica answer > > 1 + 1.73205i, > > but I need > > (-8)^(1/3) = -2 > > How can I get it? > > >