A branch of $\tanh^{-1}z$?

Question

$\def\Log{\operatorname{Log}}$ How can I show that

$$\frac{1}{2}\Log\left(\frac{1+z}{1-z}\right)$$

defines a branch of $\tanh^{-1}(z)$ on $\mathbb{C}\backslash((-\infty,-1]\cup[1,\infty))$? (where $\Log$ is the main branch of the logarithmic function)

Actually I don't really see what I have to show to get that result...

Answer 1

If $\tanh w=z$, then $$ \frac{\mathrm{e}^w-\mathrm{e}^{-w}}{\mathrm{e}^w+\mathrm{e}^{-w}}=z, $$ or $$ \mathrm{e}^w-\mathrm{e}^{-w}=z(\mathrm{e}^w+\mathrm{e}^{-w}), $$ or $$ \mathrm{e}^w(1-z)-\mathrm{e}^{-w}(1+z)=0, $$ or $$ \mathrm{e}^{2w}=\frac{1+z}{1-z}. $$ The function $\,h(z)=\dfrac{1+z}{1-z}\,$ possesses a logarithm in $\mathbb C\smallsetminus\big((-\infty,-1]\cup[1,\infty)\big)$, since $(1-z)$ possesses a logarithm $g_1$ in $\mathbb C\smallsetminus[1,\infty)$, while $(1+z)$ possesses a logarithm $g_2$ in $\mathbb C\smallsetminus(-\infty,-1]$. Then $$ g(z)=g_1(z)-g_2(z), $$ is a logarithm of $\dfrac{1+z}{1-z}$, and thus $$ \tanh^{-1} z=w=\frac{1}{2}g(z)=\frac{1}{2}\log\left(\frac{1+z}{1-z}\right). $$