Difference between revisions of "Log((1+z)/(1-z)) as continued fraction"

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(Created page with "==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-...")
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Revision as of 20:09, 25 June 2017

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