Equivalence of Definitions of Complex Inverse Tangent Function

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

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

Note that when $z = 0 + i$:

Similarly, when $z = 0 - i$:

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

Definition 1 implies Definition 2
It is demonstrated that:


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

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

Then:

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

Definition 2 implies Definition 1
It is demonstrated that:


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

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

Then:

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

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