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:

$\map {\sec^{-1} } z := \set {w \in \C: \map \sec w = z}$

where $\map \sec w$ 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 $z \in \C_{\ne 0}$:

$\set {w \in \C: \map \sec w = z} = \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$


Thus let $z \in \C$.


Definition 1 implies Definition 2

It will be demonstrated that:

$\set {w \in \C: \map \sec w = z} \subseteq \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$


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

From Secant Exponential Formulation:

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


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

Then:

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


Let $s = 1 - z^2$.

Then:

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


We have that:

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


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

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


Thus by definition of subset:

$\set {w \in \C: \map \sec w = z} \subseteq \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$

$\Box$


Definition 2 implies Definition 1

It will be demonstrated that:

$\set {w \in \C: \map \sec w = z} \supseteq \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$

Let $w \in \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$.

Then:

\(\, \displaystyle \exists k \in \Z: \, \) \(\displaystyle i w + 2 \paren {-k} \pi i\) \(=\) \(\displaystyle \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle e^{i w + 2 \paren {-k} \pi i}\) \(=\) \(\displaystyle \dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z\) Definition of Complex Natural Logarithm
\(\displaystyle \leadsto \ \ \) \(\displaystyle e^{i w}\) \(=\) \(\displaystyle \dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z\) Complex Exponential Function has Imaginary Period
\(\displaystyle \leadsto \ \ \) \(\displaystyle z e^{i w} - 1\) \(=\) \(\displaystyle \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {z e^{i w} - 1}^2\) \(=\) \(\displaystyle \cmod {1 - z^2} e^{i \arg \paren {1 - z^2} }\) Roots of Complex Number
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {z e^{i w} - 1}^2\) \(=\) \(\displaystyle 1 - z^2\) Definition of Exponential Form of Complex Number
\(\displaystyle \leadsto \ \ \) \(\displaystyle z^2 e^{2 i w} - 2 z e^{i w} + 1\) \(=\) \(\displaystyle 1 - z^2\) Square of Difference
\(\displaystyle \leadsto \ \ \) \(\displaystyle z^2 e^{2 i w} - 2 z e^{i w}\) \(=\) \(\displaystyle - z^2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle z e^{2 i w} + z\) \(=\) \(\displaystyle 2 e^{i w}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle z \paren {e^{i w} + \frac 1 {e^{i w} } }\) \(=\) \(\displaystyle 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle z\) \(=\) \(\displaystyle \frac 2 {e^{i w} + e^{-i w} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle z\) \(=\) \(\displaystyle \sec w\) Secant Exponential Formulation
\(\displaystyle \leadsto \ \ \) \(\displaystyle w\) \(\in\) \(\displaystyle \set {w \in \C: z = \map \sec w}\)


Thus by definition of superset:

$\set {w \in \C: \map \sec w = z} \supseteq \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$

$\Box$


Thus by definition of set equality:

$\set {w \in \C: \map \sec w = z} = \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$

$\blacksquare$