# Log((1+z)/(1-z)) as continued fraction

The following formula holds for $z \in \mathbb{C} \setminus \left[ (-\infty,-1) \bigcup (1,\infty) \right]$: $$\log \left( \dfrac{1+z}{1-z} \right) = \cfrac{2z}{1-\cfrac{z^2}{3-\cfrac{4z^2}{5-\cfrac{9z^2}{7-\cfrac{16z^2}{9-\cfrac{25z^2}{11-\cfrac{36z^2}{13-\ddots}}}}}}},$$ where $\log$ denotes the logarithm.