Equivalence of Definitions of Complex Inverse Secant Function

Proof
The proof strategy is to show that for all $z \in \C_{\ne 0}$:
 * $\left\{{w \in \C: \sec \left({w}\right) = z}\right\} = \left\{{\dfrac 1 i \ln \left({\dfrac {1 + \sqrt{\left|{1 - z^2}\right|} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}} z}\right) + 2 k \pi: k \in \Z}\right\}$

Thus let $z \in \C$.

Definition 1 implies Definition 2
It will be demonstrated that:


 * $\left\{{w \in \C: \sec \left({w}\right) = z}\right\} \subseteq \left\{{\dfrac 1 i \ln \left({\dfrac {1 + \sqrt{\left|{1 - z^2}\right|} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}} z}\right) + 2 k \pi: k \in \Z}\right\}$

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

From Secant Exponential Formulation:


 * $(1): \quad z = \dfrac 2 {e^{i w} + e^{-i w}}$

Let $v = e^{i w}$.

Then:

Let $s = 1 - z^2$.

Then:

We have that:

Thus from $(2)$ and $(3)$:

Thus by definition of subset:
 * $\left\{{w \in \C: \sec \left({w}\right) = z}\right\} \subseteq \left\{{\dfrac 1 i \ln \left({\dfrac {1 + \sqrt{\left|{1 - z^2}\right|} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}} z}\right) + 2 k \pi: k \in \Z}\right\}$

Definition 2 implies Definition 1
It will be demonstrated that:


 * $\left\{{w \in \C: \sec \left({w}\right) = z}\right\} \supseteq \left\{{\dfrac 1 i \ln \left({\dfrac {1 + \sqrt{\left|{1 - z^2}\right|} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} } z}\right) + 2 k \pi: k \in \Z}\right\}$

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

Then:

Thus by definition of superset:
 * $\left\{{w \in \C: \sec \left({w}\right) = z}\right\} \supseteq \left\{{\dfrac 1 i \ln \left({\dfrac {1 + \sqrt{\left|{1 - z^2}\right|} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}} z}\right) + 2 k \pi: k \in \Z}\right\}$

Thus by definition of set equality:
 * $\left\{{w \in \C: \sec \left({w}\right) = z}\right\} = \left\{{\dfrac 1 i \ln \left({\dfrac {1 + \sqrt{\left|{1 - z^2}\right|} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}} z}\right) + 2 k \pi: k \in \Z}\right\}$