Equivalence of Definitions of Complex Inverse Tangent Function

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

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

Note that when $z = 0 + i$:

Similarly, when $z = 0 - i$:

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

Definition 1 implies Definition 2
It is demonstrated that:


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

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

Then:

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

Definition 2 implies Definition 1
It is demonstrated that:


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

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

Then:

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

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