       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>

Because of the following?

Arg[-1]
Pi

Arg
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

Bob F wrote:
> Could someone explain why Mathematica evaluates these so differently?
>
> In:=
>
> (Sqrt - 7)^(1/3)
> (Sqrt - 5)^(1/3)
>
> Out= (-1)^(1/3)
>
> Out= 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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