Function Equivalence of ArcTan and Log

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

Hi, 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)] and 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 Log[(x^2+(1-y)^2)/(x^2+(1+y)^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