Equivalence of Definitions of Complex Inverse Hyperbolic Tangent

Theorem
Let $S$ be the subset of the complex plane:
 * $S = \C \setminus \left\{{-1 + 0 i, 1 + 0 i}\right\}$

Proof
The proof strategy is to how that for all $z \in S$:
 * $\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} = \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$

Note that when $z = -1 + 0 i$:

Similarly, when $z = 1 + 0 i$:

Thus let $z \in \C \setminus \left\{{-1 + 0 i, 1 + 0 i}\right\}$.

Definition 1 implies Definition 2
It is demonstrated that:


 * $\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} \subseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$

Let $w \in \left\{{w \in \C: z = \tanh \left({w}\right)}\right\}$.

Then:

Thus by definition of subset:
 * $\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} \subseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$

Definition 2 implies Definition 1
It is demonstrated that:


 * $\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} \supseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$

Let $w \in \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$.

Then:

Thus by definition of superset:
 * $\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} \supseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$

Thus by definition of set equality:
 * $\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} = \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$