Is this valid when using arctanh ln identity?

[LAHLH]

Hi,

I start with $$arctanh\left(\frac{A}{\sqrt{A^2-1}}\right)=\frac{1}{2}ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{1-\frac{A}{\sqrt{A^2-1}}}\right)$$

The function $$\frac{A}{\sqrt{A^2-1}}$$ is real, and since A>1, it too is always greater than 1.

Is it true that it should really be the modulus around the argument of ln? therefore I can manipulate it as follows:

$$ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{1-\frac{A}{\sqrt{A^2-1}}}\right)=ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{-(1-\frac{A}{\sqrt{A^2-1}})}\right)=ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{(1-\frac{A}{\sqrt{A^2-1}})}\right)$$
[/tex]

thanks

 

[mathman]

In general |tanh(x)| < 1 for x real. Therefore it is to be expected that you will have a complex x for arctanh(u) when u > 1.