# Is this valid when using arctanh ln identity?

1. Jun 1, 2010

### 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 fucntion $$\frac{A}{\sqrt{A^2-1}}$$ 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:

$$ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{1-\frac{A}{\sqrt{A^2-1}}}\right)=ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{-(1-\frac{A}{\sqrt{A^2-1}})}\right)=ln\left( \frac{1+\frac{A}{\sqrt{A^2-1}}}{(1-\frac{A}{\sqrt{A^2-1}})}\right)$$
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thanks

2. Jun 1, 2010

### 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.