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

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  • Subject: [mg31301] Re: [mg31265] (-8)^(1/3)
  • From: Otto Linsuain <linsuain at>
  • Date: Sat, 27 Oct 2001 01:08:10 -0400 (EDT)
  • Sender: owner-wri-mathgroup at

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?

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