Equivalence of Definitions of Complex Inverse Hyperbolic Secant

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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:

$\forall z \in \C_{\ne 0}: \operatorname{sech}^{-1} \left({z}\right) := \left\{{\ln \left({\dfrac {1 + \sqrt{\left|{1 - z^2}\right|} e^{\left({i / 2}\right) \arg \left({1 - z^2}\right)}} z}\right) + 2 k \pi i: k \in \Z}\right\}$

where:

$\sqrt{\left|{1 - z^2}\right|}$ denotes the positive square root of the complex modulus of $1 - z^2$
$\arg \left({1 - z^2}\right)$ 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:

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

$\blacksquare$