# Relationship between hyperbolic arctan and logarithm

## Homework Statement

The relationship between arctanh and log is:

arctanh(x)=$\frac{1}{2}log(\frac{1+x}{1-x})$
but if i take x=1.5,
I have:
arctanh(1.5)=0.8047 + 1.5708i
and
$\frac{1}{2}log(\frac{1+1.5}{1-1.5})$=0.8047 + 1.5708i
as expected, but using the laws of logarithm, why does this not equal to:
$(\frac{1}{2})\frac{1}{2}log((\frac{1+1.5}{1-1.5})^2)$=0.8047
?

## The Attempt at a Solution

I like Serena
Homework Helper
Hi sara_87!

You have applied the log to a negative number yielding a complex number.
So implicitly you have used the complex logarithm instead of the regular logarithm.

The complex logarithm happens to be a multivalued function:
$$\log re^{i\phi} = \log r + i (\phi + 2k\pi)$$

This means you have to consider the other solutions.
By taking the square you introduce a new solution.

In your example there is another solution, which is the right one:
$$(\frac{1}{2})\frac{1}{2}log((\frac{1+1.5}{1-1.5})^2) = (\frac{1}{2})\frac{1}{2}(\log 25 + i 2\pi)=\frac{1}{2}\log 5 + i \frac \pi 2$$

thank you :)

so, in general, can i say that if x is less than 1,

$arctanh(x)=\frac{1}{2}*log(\frac{1+x}{1-x})=\frac{1}{2}*\frac{1}{2}log((\frac{1+x}{1-x})^2)$

if x is more than 1, then

$arctanh(x)=\frac{1}{2}*log(\frac{1+x}{1-x})=\frac{1}{2}*\frac{1}{2}log((\frac{1+x}{1-x})^2)+i*pi/2$

I like Serena
Homework Helper
Yes.

Thanks again :)