View source for Log((1+z)/(1-z)) as continued fraction

==Theorem==
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]].

==Proof==

==References==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Log(1+z) as continued fraction|next=findme}}: $4.1.40$

[[Category:Theorem]]
[[Category:Unproven]]