# Equivalence of Definitions of Complex Inverse Cosine Function

## Theorem

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

### Definition 1

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

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

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

where $\cos \left({w}\right)$ is the cosine of $w$.

### Definition 2

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

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

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

where:

$\sqrt{\left|{z^2 - 1}\right|}$ denotes the positive square root of the complex modulus of $z^2 - 1$
$\arg \left({z^2 - 1}\right)$ denotes the argument of $z^2 - 1$
$\ln$ denotes the complex natural logarithm considered as a multifunction.

## Proof

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

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

Thus let $z \in \C$.

### Definition 1 implies Definition 2

It will be demonstrated that:

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

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

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

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

Then:

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

Let $s = z^2 - 1$.

Then:

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

We have that:

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

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

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

Thus by definition of subset:

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

$\Box$

### Definition 2 implies Definition 1

It will be demonstrated that:

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

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

Then:

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

Thus by definition of superset:

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

$\Box$

Thus by definition of set equality:

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

$\blacksquare$