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Function Equivalence of ArcTan and Log

  • To: mathgroup at
  • Subject: [mg114062] Function Equivalence of ArcTan and Log
  • From: "Dr. Robert Kragler" <kragler at>
  • Date: Mon, 22 Nov 2010 07:38:53 -0500 (EST)
  • Reply-to: kragler at


I have a problem to show with Mathematica (using already TrigToExp and 
ComplexExpand) that

        In[1]:= s1=(ArcTan[x-I y]-ArcTan[x+I y])//TrigToExp
        Out[1]:= 1/2 I Log[1-I (x-I y)]-1/2 I Log[1+I (x-I y)]-1/2 I Log[1-I 
(x+I y)]+1/2 I Log[1+I (x+I y)]

        In[2]:= s2=I/2 Log[(x^2+(1-y)^2)/(x^2+(1+y)^2)]//ComplexExpand
Out[2]:= I (1/2 Log[x^2+(1-y)^2]-1/2 Log[x^2+(1+y)^2])

are identical, i.e. s1===s2 should give "True". But FullSimplify applied to lhs 
or rhs does not help.
Of course, the identity wanted can be derived by hand making use of the inverse 
function Tan[s1]==I z and Sin[s1], Cos[s1] etc.
which gives rise to the following replacement rule :

ArcTanToLog = {(ArcTan[x_-I y_]-ArcTan[x_+I y_])-> I/2 

It is, however, non-trivial to prove this identity purely with term rewriting by 
means of Mathematica.

Any suggestion is appreciated. Thanks in advance,
Robert Kragler

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