Equivalence of Definitions of Complex Inverse Secant Function

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Theorem

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

Definition 1

Let $z \in \C_{\ne 0}$ be a non-zero complex number.

The inverse secant of $z$ is the multifunction defined as:

$\sec^{-1} \left({z}\right) := \left\{{w \in \C: \sec \left({w}\right) = z}\right\}$

where $\sec \left({w}\right)$ is the secant of $w$.

Definition 2

Let $z \in \C_{\ne 0}$ be a non-zero complex number.

The inverse secant of $z$ is the multifunction defined as:

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

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 $z \in \C_{\ne 0}$:

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


Thus let $z \in \C$.


Definition 1 implies Definition 2

It will be demonstrated that:

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


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

From Secant Exponential Formulation:

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


Let $v = e^{i w}$.

Then:

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


Let $s = 1 - z^2$.

Then:

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


We have that:

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


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

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


Thus by definition of subset:

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

$\Box$


Definition 2 implies Definition 1

It will be demonstrated that:

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

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

Then:

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


Thus by definition of superset:

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

$\Box$


Thus by definition of set equality:

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

$\blacksquare$