Insight on why this equality may be true

Question

I was messing around and noticed that it seems like $$\text{arctanh}(1/x) = \Re(\text{arctanh}(x))$$ Does anyone have any insight on why this is true, or how to prove it?

Answer 1

$$\begin{align}\text{arctanh}(x)&=\frac12\log\left(\frac{1+x}{1-x}\right)\\ \implies\text{arctanh}(1/x)&=\frac12\log\left(\frac{1+\frac1x}{1-\frac1x}\right)\\ &=\frac12\log\left(-\frac{1+x}{1-x}\right)\\&=\text{arctanh}(x)+\frac12\log(-1)\\&=\text{arctanh}(x)+\frac12\log(1)+\frac {i\pi}2\\&=\text{arctanh}(x)+\frac{i\pi}2\end{align}$$ So the real parts are equal.

Answer 2

Use $$\text{arctanh}(x)=\frac{1}{2}\log(1+x)-\frac{1}{2}\log(1-x)$$ for real $x$ observe $\text{arctanh}(1/x)$: $$\frac{1}{2}\log\left(1+\frac{1}{x}\right)-\frac{1}{2}\log\left(1-\frac{1}{x}\right)=\frac{1}{2}\log\left(\pm\frac{1+x}{|x|}\right)-\frac{1}{2}\log\left(\mp\frac{1-x}{|x|}\right)$$

use $\log(a b)=\log(a)+\log(b)$ to get rid of denominator. Then use $\log(-1)=\pi i$ to get $$=\frac{1}{2}\log\left(1+x\right)-\frac{1}{2}\log\left(1-x\right)-\text{sign}(x)\frac{\pi i}{2}=\text{arctanh}(x)-\text{sign}(x)\frac{\pi i}{2}$$ which means the function $\text{arctanh}(x)$ is not completely invariant under inversion.

EDIT:

We can plot the imaginary part of the difference of the functions arctanh$(x)-$arctanh$(1/x)$ for example in the computer algebra program Mathematica:

This confirms that the sign$(x)$ is indeed necessary in front of the $-\frac{\pi i}{2}$ imaginary part of the function difference.