Equivalence of Definitions of Complex Inverse Hyperbolic Cotangent
Theorem
The following definitions of the concept of Complex Inverse Hyperbolic Cotangent 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 cotangent is a multifunction defined on $S$ as:
- $\forall z \in S: \map {\coth^{-1} } z := \set {w \in \C: z = \map \coth w}$
where $\map \coth w$ is the hyperbolic cotangent function.
Definition 2
The inverse hyperbolic cotangent is a multifunction defined on $S$ as:
- $\forall z \in S: \map {\coth^{-1} } z := \set {\dfrac 1 2 \map \ln {\dfrac {z + 1} {z - 1} } + k \pi i: k \in \Z}$
where $\ln$ denotes the complex natural logarithm considered as a multifunction.
Proof
The proof strategy is to how that for all $z \in S$:
- $\set {w \in \C: z = \coth w} = \set {\dfrac 1 2 \map \ln {\dfrac {z + 1} {z - 1} } + k \pi i: k \in \Z}$
Note that when $z = -1 + 0 i$:
\(\ds z + 1\) | \(=\) | \(\ds 0 + 0 i\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {z + 1} {z - 1}\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \ln {\dfrac {z + 1} {z - 1} }\) | \(\) | \(\ds \text {is undefined}\) |
Similarly, when $z = 1 + 0 i$:
\(\ds z - 1\) | \(=\) | \(\ds 0 + 0 i\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {z + 1} {z - 1}\) | \(\) | \(\ds \text {is undefined}\) |
Thus let $z \in \C \setminus \set {-1 + 0 i, 1 + 0 i}$.
Definition 1 implies Definition 2
It is demonstrated that:
- $\set {w \in \C: z = \coth w} \subseteq \set {\dfrac 1 2 \map \ln {\dfrac {z + 1} {z - 1} } + k \pi i: k \in \Z}$
Let $w \in \set {w \in \C: z = \coth w}$.
Then:
\(\ds z\) | \(=\) | \(\ds \frac {e^w + e^{-w} } {e^w - e^{-w} }\) | Definition of Hyperbolic Cotangent | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{2 w}\) | \(=\) | \(\ds \frac {z + 1} {z - 1}\) | solving for $e^{2 w}$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \ln {e^{2 w} }\) | \(=\) | \(\ds \map \ln {\frac {z + 1} {z - 1} }\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists k' \in \Z: \, \) | \(\ds 2 w + 2 k' \pi i\) | \(=\) | \(\ds \map \ln {\frac {z + 1} {z - 1} }\) | Definition of Complex Natural Logarithm | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists k \in \Z: \, \) | \(\ds w\) | \(=\) | \(\ds \frac 1 2 \map \ln {\frac {z + 1} {z - 1} } + k \pi i\) | putting $k = -k'$ |
Thus by definition of subset:
- $\set {w \in \C: z = \coth w} \subseteq \set {\dfrac 1 2 \map \ln {\dfrac {z + 1} {z - 1} } + k \pi i: k \in \Z}$
$\Box$
Definition 2 implies Definition 1
It is demonstrated that:
- $\set {w \in \C: z = \coth w} \supseteq \set {\dfrac 1 2 \map \ln {\dfrac {z + 1} {z - 1} } + k \pi i: k \in \Z}$
Let $w \in \set {\dfrac 1 2 \map \ln {\dfrac {z + 1} {z - 1} } + k \pi i: k \in \Z}$.
Then:
\(\ds \exists k \in \Z: \, \) | \(\ds w\) | \(=\) | \(\ds \dfrac 1 2 \map \ln {\dfrac {z + 1} {z - 1} } + k \pi i\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists k \in \Z: \, \) | \(\ds 2 w + 2 \paren {-k} \pi i\) | \(=\) | \(\ds \map \ln {\frac {z + 1} {z - 1} }\) | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{2 w + 2 \paren {-k} \pi i}\) | \(=\) | \(\ds \frac {z + 1} {z - 1}\) | Definition of Complex Natural Logarithm | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{2 w}\) | \(=\) | \(\ds \frac {z + 1} {z - 1}\) | Complex Exponential Function has Imaginary Period | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds \frac {e^{2 w} + 1} {e^{2 w} - 1}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {e^w + e^{-w} } {e^w - e^{-w} }\) | multiplying top and bottom by $e^{-w}$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds \coth w\) | Definition of Hyperbolic Cotangent | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds w\) | \(\in\) | \(\ds \set {w \in \C: z = \coth w}\) |
Thus by definition of superset:
- $\set {w \in \C: z = \coth w} \supseteq \set {\dfrac 1 2 \map \ln {\dfrac {z + 1} {z - 1} } + k \pi i: k \in \Z}$
$\Box$
Thus by definition of set equality:
- $\set {w \in \C: z = \coth w} = \set {\dfrac 1 2 \map \ln {\dfrac {z + 1} {z - 1} } + k \pi i: k \in \Z}$
$\blacksquare$