# Equivalence of Definitions of Complex Inverse Sine Function

## Theorem

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

### Definition 1

Let $z \in \C$ be a complex number.

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

$\sin^{-1} \paren z := \set {w \in \C: \sin \paren w = z}$

where $\sin \paren w$ is the sine of $w$.

### Definition 2

Let $z \in \C$ be a complex number.

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

$\sin^{-1} \paren z := \set {\dfrac 1 i \ln \paren {i z + \sqrt {\cmod {1 - z^2} } \exp \paren {\dfrac i 2 \arg \paren {1 - z^2} } } + 2 k \pi: k \in \Z}$

where:

$\sqrt {\cmod {1 - z^2} }$ denotes the positive square root of the complex modulus of $1 - z^2$
$\arg \paren {1 - z^2}$ denotes the argument of $1 - z^2$
$\ln$ is the complex natural logarithm considered as a multifunction.

## Proof

The proof strategy is to show that for all $z \in \C$:

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

Thus let $z \in \C$.

### Definition 1 implies Definition 2

It will be demonstrated that:

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

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

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

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

Then:

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

Let $s = 1 - z^2$.

Then:

 $\ds v$ $=$ $\ds i z + s^{1/2}$ $\ds$ $=$ $\ds i z + \sqrt {\cmod s} \paren {\map \cos {\frac {\map \arg s} 2} + i \map \sin {\frac {\map \arg s} 2} }$ Definition of Complex Square Root $\text {(2)}: \quad$ $\ds \leadsto \ \$ $\ds \ln v$ $=$ $\ds \map \ln {i z + \sqrt {\cmod s} \paren {\map \cos {\frac {\map \arg s} 2} + i \map \sin {\frac {\map \arg s} 2} } }$ 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 $(3)$:

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

Thus by definition of subset:

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

$\Box$

### Definition 2 implies Definition 1

It will be demonstrated that:

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

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

Then:

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

Thus by definition of superset:

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

$\Box$

Thus by definition of set equality:

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

$\blacksquare$