# Equivalence of Definitions of Complex Inverse Hyperbolic Secant

## Theorem

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

### Definition 1

The inverse hyperbolic secant is a multifunction defined as:

$\forall z \in \C_{\ne 0}: \operatorname{sech}^{-1} \left({z}\right) := \left\{{w \in \C: z = \operatorname{sech} \left({w}\right)}\right\}$

where $\operatorname{sech} \left({w}\right)$ is the hyperbolic secant function.

### Definition 2

The inverse hyperbolic secant is a multifunction defined as:

$\forall z \in \C_{\ne 0}: \map {\sech^{-1} } z := \set {\map \ln {\dfrac {1 + \sqrt {\size {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi i: k \in \Z}$

where:

$\sqrt {\size {1 - z^2} }$ denotes the positive square root of the complex modulus of $1 - z^2$
$\map \arg {1 - z^2}$ denotes the argument of $1 - z^2$
$\ln$ denotes the complex natural logarithm as a multifunction.

## Proof

The proof strategy is to show that for all $z \in \C_{\ne 0}$:

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

Thus let $z \in \C_{\ne 0}$.

### Definition 1 implies Definition 2

It will be demonstrated that:

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

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

From the definition of hyperbolic secant:

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

Let $v = e^w$.

Then:

 $\displaystyle z \left({v + \frac 1 v}\right)$ $=$ $\displaystyle 2$ multiplying $(1)$ by $v + \dfrac 1 v$ $\displaystyle \implies \ \$ $\displaystyle z v^2 - 2 v + z$ $=$ $\displaystyle 0$ multiplying by $v$ and rearranging $\displaystyle \implies \ \$ $\displaystyle v$ $=$ $\displaystyle \frac {1 + \left({1 - z^2}\right)^{1/2} } z$ Quadratic Formula

Let $s = 1 - z^2$.

Then:

 $\displaystyle v$ $=$ $\displaystyle \frac {1 + s^{1/2} } z$ $\displaystyle$ $=$ $\displaystyle \frac {1 + \sqrt{\left\vert{s}\right\vert} \left({\cos \left({\dfrac {\arg \left({s}\right)} 2}\right) + i \sin \left({\dfrac {\arg \left({s}\right)} 2}\right)}\right) } z$ Definition of Complex Square Root $(2):\quad$ $\displaystyle \implies \ \$ $\displaystyle \ln v$ $=$ $\displaystyle \ln \left({\frac {1 + \sqrt{\left\vert{s}\right\vert} \left({\cos \left({\dfrac {\arg \left({s}\right)} 2}\right) + i \sin \left({\dfrac {\arg \left({s}\right)} 2}\right)}\right) } z}\right)$ where $\ln$ denotes the Complex Natural Logarithm

We have that:

 $\displaystyle v$ $=$ $\displaystyle e^w$ $\displaystyle \implies \ \$ $\displaystyle \ln v$ $=$ $\displaystyle \ln \left({e^w}\right)$ $(3):\quad$ $\displaystyle$ $=$ $\displaystyle w + 2 k' \pi i: k' \in \Z$ Definition of Complex Natural Logarithm

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

 $\displaystyle w + 2 k' \pi i$ $=$ $\displaystyle \ln \left({\frac {1 + \sqrt{\left\vert{s}\right\vert} \left({\cos \left({\dfrac {\arg \left({s}\right)} 2}\right) + i \sin \left({\dfrac {\arg \left({s}\right)} 2}\right)}\right) } z}\right)$ $\displaystyle \implies \ \$ $\displaystyle w$ $=$ $\displaystyle \ln \left({\frac {1 + \sqrt{\left\vert{s}\right\vert} \left({\cos \left({\dfrac {\arg \left({s}\right)} 2}\right) + i \sin \left({\dfrac {\arg \left({s}\right)} 2}\right)}\right) } z}\right) + 2 k \pi i$ putting $k = -k'$ $\displaystyle \implies \ \$ $\displaystyle w$ $=$ $\displaystyle \ln \left({\frac {1 + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} } z}\right) + 2 k \pi i$ Definition of Exponential Form of Complex Number

Thus by definition of subset:

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

$\Box$

### Definition 2 implies Definition 1

It will be demonstrated that:

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

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

Then:

 $\displaystyle \exists k \in \Z: \ \$ $\displaystyle w + 2 \left({-k}\right) \pi i$ $=$ $\displaystyle \ln \left({\dfrac {1 + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} } z}\right)$ $\displaystyle \implies \ \$ $\displaystyle e^{w + 2 \left({-k}\right) \pi i}$ $=$ $\displaystyle \dfrac {1 + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} } z$ Definition of Complex Natural Logarithm $\displaystyle \implies \ \$ $\displaystyle e^w$ $=$ $\displaystyle \dfrac {1 + \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)} } z$ Complex Exponential Function has Imaginary Period $\displaystyle \implies \ \$ $\displaystyle z e^w - 1$ $=$ $\displaystyle \sqrt{\left\vert{1 - z^2}\right\vert} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}$ $\displaystyle \implies \ \$ $\displaystyle \left({z e^w - 1}\right)^2$ $=$ $\displaystyle \left\vert{1 - z^2}\right\vert e^{i \arg \left({1 - z^2}\right)}$ Roots of Complex Number $\displaystyle \implies \ \$ $\displaystyle \left({z e^w - 1}\right)^2$ $=$ $\displaystyle 1 - z^2$ Definition of Exponential Form of Complex Number $\displaystyle \implies \ \$ $\displaystyle z^2 e^{2 w} - 2 z e^w + 1$ $=$ $\displaystyle 1 - z^2$ Square of Difference $\displaystyle \implies \ \$ $\displaystyle z^2 e^{2 w} - 2 z e^w$ $=$ $\displaystyle - z^2$ $\displaystyle \implies \ \$ $\displaystyle z e^{2 w} + z$ $=$ $\displaystyle 2 e^w$ $\displaystyle \implies \ \$ $\displaystyle z \left({e^w + \frac 1 {e^w} }\right)$ $=$ $\displaystyle 2$ $\displaystyle \implies \ \$ $\displaystyle z$ $=$ $\displaystyle \frac 2 {e^w + e^{- w} }$ $\displaystyle \implies \ \$ $\displaystyle z$ $=$ $\displaystyle \operatorname{sech} \left({w}\right)$ Definition of Hyperbolic Secant $\displaystyle \implies \ \$ $\displaystyle w$ $\in$ $\displaystyle \left\{ {w \in \C: z = \operatorname{sech} \left({w}\right)}\right\}$

Thus by definition of superset:

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

$\Box$

Thus by definition of set equality:

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

$\blacksquare$