Equivalence of Definitions of Complex Inverse Hyperbolic Secant

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

Thus let $z \in \C_{\ne 0}$.

Definition 1 implies Definition 2
It will be demonstrated that:


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

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

From the definition of hyperbolic secant:


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

Let $v = e^w$.

Then:

Let $s = 1 - z^2$.

Then:

We have that:

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

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

Definition 2 implies Definition 1
It will be demonstrated that:


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

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

Then:

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

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