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MathGroup Archive 2007

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Re: Re-defining Log over it's branch cut

  • To: mathgroup at smc.vnet.net
  • Subject: [mg77498] Re: Re-defining Log over it's branch cut
  • From: dimitris <dimmechan at yahoo.com>
  • Date: Sun, 10 Jun 2007 07:25:51 -0400 (EDT)
  • References: <f4dsng$hhu$1@smc.vnet.net>

Hi.

In[11]:=
Log[z] /. z -> r*Exp[0*I]
Out[11]=
Log[r]

In[14]:=
PowerExpand[Log[z] /. z -> r*Exp[2*Pi*i]] /. i -> I
Out[14]=
2*I*Pi + Log[r]

In[15]:=
PowerExpand[Log[z] /. z -> r*Exp[(-Pi)*i]] /. i -> I
Out[15]=
(-I)*Pi + Log[r]

In[16]:=
PowerExpand[Log[z] /. z -> r*Exp[Pi*i]] /. i -> I
Out[16]=
I*Pi + Log[r]

Cheers
Dimitris

 /  chuck009       :
> Hello all,
>
> When integrating along a logarithmic branch cut, often have to define two values of z:
>
> z=r Exp[0 i] which is a path above the positive real axis
>
> z=r Exp[2 pi i] which is the path below the positive real axis
>
> However, when I make a transformation rule:
>
> Log[z] /.z->r Exp[0 i]  returns Log[r]
>
> but:
>
> Log[z] /.z->r Exp[2 pi i] should be Log[r]+2 pi i
>
> however Mathematica returns Log[r] in both cases.
>
>
> Is there a way to configure the Log function to return these two values?  I have the same question when the branch cut is along the negative real axis:
>
> Log[z] /. z->r Exp[-pi i] (needs to be Log[r]-pi i)
>
> Log[z] /. z->r Exp[pi i]  (needs to be Log[r]+pi i)



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