# Equivalence of Definitions of Complex Inverse Hyperbolic Tangent

Jump to navigation Jump to search

## Theorem

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

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

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

### Definition 1

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

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

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

### Definition 2

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

$\forall z \in S: \map {\tanh^{-1} } z := \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$

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

## Proof

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

$\set {w \in \C: z = \tanh w} = \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$

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

 $\ds 1 + z$ $=$ $\ds 0 + 0 i$ $\ds \leadsto \ \$ $\ds \frac {1 + z} {1 - z}$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \map \ln {\dfrac {1 + z} {1 - z} }$  $\ds \text {is undefined}$

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

 $\ds 1 - z$ $=$ $\ds 0 + 0 i$ $\ds \leadsto \ \$ $\ds \frac {1 + z} {1 - z}$  $\ds \text {is undefined}$

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

### Definition 1 implies Definition 2

It is demonstrated that:

$\set {w \in \C: z = \tanh w} \subseteq \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$

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

Then:

 $\ds z$ $=$ $\ds \frac {e^w - e^{- w} } {e^w + e^{- w} }$ Definition of Hyperbolic Tangent $\ds z$ $=$ $\ds \frac {e^{2 w} - 1} {e^{2 w} + 1}$ multiplying top and bottom by $e^w$ $\ds \leadsto \ \$ $\ds e^{2 w}$ $=$ $\ds \frac {1 + z} {1 - z}$ solving for $e^{2 w}$ $\ds \leadsto \ \$ $\ds \map \ln {e^{2 w} }$ $=$ $\ds \ln \frac {1 + z} {1 - z}$ $\ds \leadsto \ \$ $\ds \exists k' \in \Z: \,$ $\ds 2 w + 2 k' \pi i$ $=$ $\ds \ln \frac {1 + z} {1 - z}$ Definition of Complex Natural Logarithm $\ds \leadsto \ \$ $\ds \exists k \in \Z: \,$ $\ds w$ $=$ $\ds \frac 1 2 \ln \frac {1 + z} {1 - z} + k \pi i$ putting $k = -k'$

Thus by definition of subset:

$\set {w \in \C: z = \tanh w} \subseteq \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$

$\Box$

### Definition 2 implies Definition 1

It is demonstrated that:

$\set {w \in \C: z = \tanh w} \supseteq \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$

Let $w \in \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$.

Then:

 $\ds \exists k \in \Z: \,$ $\ds w$ $=$ $\ds \dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i$ $\ds \leadsto \ \$ $\ds \exists k \in \Z: \,$ $\ds 2 w + 2 \paren {-k} \pi i$ $=$ $\ds \map \ln {\dfrac {1 + z} {1 - z} }$ $\ds \leadsto \ \$ $\ds e^{2 w + 2 \paren {-k} \pi i}$ $=$ $\ds \dfrac {1 + z} {1 - z}$ Definition of Complex Natural Logarithm $\ds \leadsto \ \$ $\ds e^{2 w}$ $=$ $\ds \dfrac {1 + z} {1 - z}$ Complex Exponential Function has Imaginary Period $\ds \leadsto \ \$ $\ds z$ $=$ $\ds \dfrac {e^w - e^{- w} } {e^w + e^{- w} }$ $\ds \leadsto \ \$ $\ds z$ $=$ $\ds \tanh w$ Definition of Hyperbolic Tangent $\ds \leadsto \ \$ $\ds w$ $\in$ $\ds \set {w \in \C: z = \tanh w}$

Thus by definition of superset:

$\set {w \in \C: z = \tanh w} \supseteq \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$

$\Box$

Thus by definition of set equality:

$\set {w \in \C: z = \tanh w} = \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$

$\blacksquare$