ちょっとした式変形

$ \begin{align*} e^{2i\theta} &=\frac{e^{i\theta}}{e^{-i\theta}}\\ &=\frac{\cos\theta+i\sin\theta}{\cos\theta-i\sin\theta} \\ &=\frac{1+i\tan\theta}{1-i\tan\theta}\\ \end{align*} $
両辺の対数を取ると
$ \begin{align*} \theta =\displaystyle\frac{1}{2i}\ln\bigg(\frac{1+i\tan\theta}{1-i\tan\theta}\bigg)\\ \end{align*} $
$\tan\theta=x$と置くと
$ \begin{align*} \mathrm{arctan}x=\displaystyle\frac{1}{2i}\ln\bigg(\frac{1+ix}{1-ix}\bigg) \end{align*} $