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 > >