Equivalence of Definitions of Complex Inverse Secant Function

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:

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

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

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

Thus let $z \in \C$.

Definition 1 implies Definition 2

It will be demonstrated that:

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

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

From Secant Exponential Formulation:

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

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

Then:

\(\ds z \paren {v + \frac 1 v}\)

\(=\)

\(\ds 2\)

multiplying $(1)$ by $v + \dfrac 1 v$

\(\ds \leadsto \ \ \)

\(\ds z v^2 - 2 v + z\)

\(=\)

\(\ds 0\)

multiplying by $v$ and rearranging

\(\ds \leadsto \ \ \)

\(\ds v\)

\(=\)

\(\ds \frac {1 + \paren {1 - z^2}^{1/2} } z\)

Quadratic Formula

Let $s = 1 - z^2$.

Then:

\(\ds v\)

\(=\)

\(\ds \frac {1 + s^{1/2} } z\)

\(\ds \)

\(=\)

\(\ds \frac {1 + \sqrt {\cmod s} \paren {\map \cos {\dfrac {\map \arg s} 2} + i \map \sin {\dfrac {\map \arg s} 2} } } z\)

Definition of Complex Square Root

\(\text {(2)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds \ln v\)

\(=\)

\(\ds \map \ln {\frac {1 + \sqrt {\cmod s} \paren {\map \cos {\dfrac {\map \arg s} 2} + i \map \sin {\dfrac {\map \arg s} 2} } } z}\)

where $\ln$ denotes the Complex Natural Logarithm

We have that:

\(\ds v\)

\(=\)

\(\ds e^{i w}\)

\(\ds \leadsto \ \ \)

\(\ds \ln v\)

\(=\)

\(\ds \map \ln {e^{i w} }\)

\(\text {(3)}: \quad\)

\(\ds \)

\(=\)

\(\ds i w + 2 k' \pi i: k' \in \Z\)

Definition of Complex Natural Logarithm

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

\(\ds i w + 2 k' \pi i\)

\(=\)

\(\ds \map \ln {\frac {1 + \sqrt {\cmod s} \paren {\map \cos {\dfrac {\map \arg s} 2} + i \map \sin {\dfrac {\map \arg s} 2} } } z}\)

\(\ds \leadsto \ \ \)

\(\ds w\)

\(=\)

\(\ds \frac 1 i \map \ln {\frac {1 + \sqrt {\cmod s} \paren {\map \cos {\dfrac {\map \arg s} 2} + i \map \sin {\dfrac {\map \arg s} 2} } } z} + 2 k \pi\)

putting $k = -k'$

\(\ds \leadsto \ \ \)

\(\ds w\)

\(=\)

\(\ds \dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi\)

Definition of Exponential Form of Complex Number

Thus by definition of subset:

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

$\Box$

Definition 2 implies Definition 1

It will be demonstrated that:

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

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:

\(\ds \exists k \in \Z: \, \)

\(\ds i w + 2 \paren {-k} \pi i\)

\(=\)

\(\ds \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z}\)

\(\ds \leadsto \ \ \)

\(\ds e^{i w + 2 \paren {-k} \pi i}\)

\(=\)

\(\ds \dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z\)

Definition of Complex Natural Logarithm

\(\ds \leadsto \ \ \)

\(\ds e^{i w}\)

\(=\)

\(\ds \dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z\)

Complex Exponential Function has Imaginary Period

\(\ds \leadsto \ \ \)

\(\ds z e^{i w} - 1\)

\(=\)

\(\ds \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} }\)

\(\ds \leadsto \ \ \)

\(\ds \paren {z e^{i w} - 1}^2\)

\(=\)

\(\ds \cmod {1 - z^2} e^{i \arg \paren {1 - z^2} }\)

Roots of Complex Number

\(\ds \leadsto \ \ \)

\(\ds \paren {z e^{i w} - 1}^2\)

\(=\)

\(\ds 1 - z^2\)

Definition of Exponential Form of Complex Number

\(\ds \leadsto \ \ \)

\(\ds z^2 e^{2 i w} - 2 z e^{i w} + 1\)

\(=\)

\(\ds 1 - z^2\)

Square of Difference

\(\ds \leadsto \ \ \)

\(\ds z^2 e^{2 i w} - 2 z e^{i w}\)

\(=\)

\(\ds - z^2\)

\(\ds \leadsto \ \ \)

\(\ds z e^{2 i w} + z\)

\(=\)

\(\ds 2 e^{i w}\)

\(\ds \leadsto \ \ \)

\(\ds z \paren {e^{i w} + \frac 1 {e^{i w} } }\)

\(=\)

\(\ds 2\)

\(\ds \leadsto \ \ \)

\(\ds z\)

\(=\)

\(\ds \frac 2 {e^{i w} + e^{-i w} }\)

\(\ds \leadsto \ \ \)

\(\ds z\)

\(=\)

\(\ds \sec w\)

Secant Exponential Formulation

\(\ds \leadsto \ \ \)

\(\ds w\)

\(\in\)

\(\ds \set {w \in \C: z = \sec w}\)

Thus by definition of superset:

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

$\Box$

Thus by definition of set equality:

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

$\blacksquare$

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