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Re: Cube root of -1 and 1

  • To: mathgroup at smc.vnet.net
  • Subject: [mg90883] Re: [mg90875] Cube root of -1 and 1
  • From: Murray Eisenberg <murray at math.umass.edu>
  • Date: Mon, 28 Jul 2008 07:51:53 -0400 (EDT)
  • Organization: Mathematics & Statistics, Univ. of Mass./Amherst
  • References: <200807270632.CAA17232@smc.vnet.net>
  • Reply-to: murray at math.umass.edu

Because of the following?

   Arg[-1]
Pi

   Arg[1]
0

According to the reference page for Power, "For complex numbers x and y, 
Power gives the principal value of e^(y log (x))."  And an example there 
shows the result (-1)^(1/3) for (-1)^(1/3) and says, "The principal root 
is always used."

If you want an actual expansion for (-1)^(1/3) into real and complex 
parts, then use ComplexExpand:

   ComplexExpand[(-1)^(1/3)] // InputForm
1/2 + (I/2)*Sqrt[3]


Bob F wrote:
> Could someone explain why Mathematica evaluates these so differently?
> 
> In[53]:=
> 
> (Sqrt[36] - 7)^(1/3)
> (Sqrt[36] - 5)^(1/3)
> 
> Out[53]= (-1)^(1/3)
> 
> Out[54]= 1
> 
> In other words why isn't (-1)^1/3 expressed as -1 ??
> 
> Thanks...
> 
> -Bob
> 

-- 
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305


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