Equivalence of Definitions of Complex Inverse Hyperbolic Tangent

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

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

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

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

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

Definition 1 implies Definition 2
It is demonstrated that:


 * $\set {w \in \C: z = \tanh w} \subseteq \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$

Let $w \in \set {w \in \C: z = \tanh w}$.

Then:

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

Definition 2 implies Definition 1
It is demonstrated that:


 * $\set {w \in \C: z = \tanh w} \supseteq \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$

Let $w \in \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$.

Then:

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

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