# Equivalence of Definitions of Complex Inverse Hyperbolic Tangent

## Theorem

The following definitions of the concept of Complex Inverse Hyperbolic Tangent are equivalent:

Let $S$ be the subset of the complex plane:

$S = \C \setminus \left\{{-1 + 0 i, 1 + 0 i}\right\}$

### Definition 1

The inverse hyperbolic tangent is a multifunction defined on $S$ as:

$\forall z \in S: \tanh^{-1} \paren z := \set {w \in \C: z = \tanh \paren w}$

where $\tanh \paren w$ is the hyperbolic tangent function.

### Definition 2

The inverse hyperbolic tangent is a multifunction defined on $S$ as:

$\forall z \in S: \tanh^{-1} \left({z}\right) := \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$

where $\ln$ denotes the complex natural logarithm considered as a multifunction.

## Proof

The proof strategy is to how that for all $z \in S$:

$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} = \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$

Note that when $z = -1 + 0 i$:

 $\displaystyle 1 + z$ $=$ $\displaystyle 0 + 0 i$ $\displaystyle \implies \ \$ $\displaystyle \frac {1 + z} {1 - z}$ $=$ $\displaystyle 0$ $\displaystyle \implies \ \$ $\displaystyle \ln \left({\dfrac {1 + z} {1 - z} }\right)$  $\displaystyle \text {is undefined}$

Similarly, when $z = 1 + 0 i$:

 $\displaystyle 1 - z$ $=$ $\displaystyle 0 + 0 i$ $\displaystyle \implies \ \$ $\displaystyle \frac {1 + z} {1 - z}$  $\displaystyle \text {is undefined}$

Thus let $z \in \C \setminus \left\{{-1 + 0 i, 1 + 0 i}\right\}$.

### Definition 1 implies Definition 2

It is demonstrated that:

$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} \subseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$

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

Then:

 $\displaystyle z$ $=$ $\displaystyle \frac {e^w - e^{- w} } {e^w + e^{- w} }$ Definition of Hyperbolic Tangent $\displaystyle z$ $=$ $\displaystyle \frac {e^{2 w} - 1} {e^{2 w} + 1}$ multiplying top and bottom by $e^w$ $\displaystyle \implies \ \$ $\displaystyle e^{2 w}$ $=$ $\displaystyle \frac {1 + z} {1 - z}$ solving for $e^{2 w}$ $\displaystyle \implies \ \$ $\displaystyle \ln \left({e^{2 w} }\right)$ $=$ $\displaystyle \ln \frac {1 + z} {1 - z}$ $\displaystyle \implies \ \$ $\displaystyle 2 w + 2 k' \pi i: k' \in \Z$ $=$ $\displaystyle \ln \frac {1 + z} {1 - z}$ Definition of Complex Natural Logarithm $\displaystyle \implies \ \$ $\displaystyle w$ $=$ $\displaystyle \frac 1 2 \ln \frac {1 + z} {1 - z} + k \pi i: k \in \Z$ putting $k = -k'$

Thus by definition of subset:

$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} \subseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$

$\Box$

### Definition 2 implies Definition 1

It is demonstrated that:

$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} \supseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$

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

Then:

 $\displaystyle \exists k \in \Z: \ \$ $\displaystyle w$ $=$ $\displaystyle \dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i$ $\displaystyle \implies \ \$ $\displaystyle \exists k \in \Z: \ \$ $\displaystyle 2 w + 2 \left({-k}\right) \pi i$ $=$ $\displaystyle \ln \left({\dfrac {1 + z} {1 - z} }\right)$ $\displaystyle \implies \ \$ $\displaystyle e^{2 w + 2 \left({-k}\right) \pi i}$ $=$ $\displaystyle \dfrac {1 + z} {1 - z}$ Definition of Complex Natural Logarithm $\displaystyle \implies \ \$ $\displaystyle e^{2 w}$ $=$ $\displaystyle \dfrac {1 + z} {1 - z}$ Complex Exponential Function has Imaginary Period $\displaystyle \implies \ \$ $\displaystyle z$ $=$ $\displaystyle \dfrac {e^w - e^{- w} } {e^w + e^{- w} }$ $\displaystyle \implies \ \$ $\displaystyle z$ $=$ $\displaystyle \tanh w$ Definition of Hyperbolic Tangent $\displaystyle \implies \ \$ $\displaystyle w$ $\in$ $\displaystyle \left\{ {w \in \C: \tanh \left({w}\right) = z}\right\}$

Thus by definition of superset:

$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} \supseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$

$\Box$

Thus by definition of set equality:

$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} = \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$

$\blacksquare$