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Nov26-06, 04:26 PM
P: 134
then we have from trigonometry that:

[tex] Artanh(ix)=iartan(x) [/tex]

On the other hand...

[tex] 2artanh(x)=log(1-x)-log(1+x) [/tex]

[tex] 2artanh(ix)=log(1-ix)-log(1+ix) [/tex]

[tex] 2artanh(i/x)=log(1-i/x)-log(1+i/x) [/tex]

then [tex] artanh(ix)+artanh(i/x)=i(artan(x)+artan(1/x)) [/tex]

taking the sum of all the logs you have..

[tex] -log(1+ix)-log(1+i/x)+log(1-ix)+log(1-i/x) [/tex]

[tex] -log(2i/x)+log(-2i/x)=log(-1)=i\pi [/tex]

- i have taken the first "branch" of log ..(the angle goes from 0 to 360 degrees), the factor "2" comes from the definition of artanh(x) in the form of log (log in basis e)..