Equivalence of Definitions of Complex Inverse Hyperbolic Cosine

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

Thus let $z \in \C$.

Definition 1 implies Definition 2
It is demonstrated that:


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

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

By definition of hyperbolic cosine:


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

Let $v = e^w$.

Then:

Let $s = z^2 - 1$.

Then:

We have that:

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

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

Definition 2 implies Definition 1
It is demonstrated that:


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

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

Then:

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

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