Equivalence of Definitions of Complex Inverse Hyperbolic Secant

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

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

Definition 1 implies Definition 2
It will be demonstrated that:


 * $\set {w \in \C: z = \sech w} \subseteq \set {\map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi i: k \in \Z}$

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

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:
 * $\set {w \in \C: z = \sech w} \subseteq \set {\map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi i: k \in \Z}$

Definition 2 implies Definition 1
It will be demonstrated that:


 * $\set {w \in \C: z = \sech w} \supseteq \set {\map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi i: k \in \Z}$

Let $w \in \set {\map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi i: k \in \Z}$.

Then:

Thus by definition of superset:
 * $\set {w \in \C: z = \sech w} \supseteq \set {\map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi i: k \in \Z}$

Thus by definition of set equality:
 * $\set {w \in \C: z = \sech w} = \set {\map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi i: k \in \Z}$