How to show $\tanh^{-1}(3i)=i\tan^{-1}(3)$? I want to find all the complex values associated with $\tanh^{-1}(3i)$. On WolframAlpha, it computes $\tanh^{-1}(3i)=i\tan^{-1}(3)$. I know that $\tanh^{-1}(z)=\frac{1}{2} \log{\frac{1+z}{1-z}}$ and so
$$\begin{align}
\tanh^{-1}(3i)&=\frac{1}{2} \log{\frac{1+3i}{1-3i}}\\
&=\frac{1}{2}\log \frac{-8+6i}{10}\\
&=\frac{1}{2}\log \frac{-4+3i}{5}\\
&=\frac{1}{2}\log \frac{-4+3i}{5}\\
&=\frac{1}{2}\Big[\ln(5)+\arg(-4+3i)-[\ln(5)+\arg(5)]\Big]\\
&=\frac{1}{2}\Big[\arg(-4+3i)+2\pi n\Big]\\
\end{align}$$
Where do I go from here to simplify it to $i\tan^{-1}(3)$?

put $\arg(-4+3i)=2\theta$
$\displaystyle\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}=-\frac{3}{4} $
$\displaystyle\tan\theta=3$
$\displaystyle\theta=i\tan^{-1}3$
