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

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

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:

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

$\blacksquare$