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Re: branch of (-1)^(1/3)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg94543] Re: branch of (-1)^(1/3)
  • From: dh <dh at metrohm.com>
  • Date: Tue, 16 Dec 2008 02:32:18 -0500 (EST)
  • References: <ghtjcj$r7f$1@smc.vnet.net>


Hi Slawek,

back to the roots. After having chosen a branch cut (the most common 

one: 0..-Infinity), Log will be unique and always return the principle 

value. With this we may construct the brach we want "by hand":

Exp[Log[-1]/3+ n 2 Pi I/3]

with n==1 we get the branch you want.

If you know what you are doing (side effects!) you may define:

Unprotect[Power]

a_^(1/3):=Exp[Log[-1]/3+1 2 Pi I/3]

Now, if you type (-1)^(1/3) you will get: -1

Daniel



slawek wrote:

> Is a simple way to choose the branch of (-1)^(1/3) ?

> 

> Mathematica gives a (correct) non-real answer. It is ok, but I need the (-1) 

> as the output when I input (1)^(1/3)  because I know that it is a solution 

> of real-valued problem.

> 

> Is any "standard way" to pick up a correct (i.e. arbitrary) root of 

> (-1)^(1/n) instead the default?

> 

> slawek

> 

> 




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