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Re: Cube root of -1 and 1

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  • Subject: [mg90893] Re: [mg90875] Cube root of -1 and 1
  • From: Andrzej Kozlowski <akoz at>
  • Date: Mon, 28 Jul 2008 07:53:46 -0400 (EDT)
  • References: <>

On 27 Jul 2008, at 08:32, 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

The function z->z^n is a complex multivalued function, and, of course,  
there are three values that could serve as (-1)^(1/3). The most  
convenient choice is the so called "principal value", which in case is  
not -1. To see this clearly, note that Mathematica uses the following  
definition of Power:

Power[x,y] = Exp[y Log[x]]

So the choice of the principal value of Power[-1,1/3] amounts to the  
choice principal value of Log[-1].  What should be that?
Well, we would like the following relationship between Log and Arg to  

Log[z] = log[Abs[z]]+I Arg[z]


Log[-1]= I Arg[-1]

Since clearly one would not want Arg to be anything else but:


we have to choose Log[-1] as Pi I and hence Power[-1,1/3] as Exp[Pi I/ 
3]. Any other choice would make many formulas and computations more  

Andrzej Kozlowski

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