Equivalence of Definitions of Complex Inverse Cosine Function

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

Thus let $z \in \C$.

Definition 1 implies Definition 2
It will be demonstrated that:


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

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

From Cosine Exponential Formulation:


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

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

Then:

Let $s = z^2 - 1$.

Then:

We have that:

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

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

Definition 2 implies Definition 1
It will be demonstrated that:


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

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

Then:

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

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