# Equivalence of Definitions of Complex Inverse Hyperbolic Cosine

## Theorem

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

### Definition 1

The inverse hyperbolic cosine is a multifunction defined as:

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

where $\cosh \left({w}\right)$ is the hyperbolic cosine function.

### Definition 2

The inverse hyperbolic cosine is a multifunction defined as:

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

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 how that for all $z \in \C$:

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

Thus let $z \in \C$.

### Definition 1 implies Definition 2

It is demonstrated that:

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

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

By definition of hyperbolic cosine:

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

Let $v = e^w$.

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:

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

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

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

Thus by definition of subset:

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

$\Box$

### Definition 2 implies Definition 1

It is demonstrated that:

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

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

Then:

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

Thus by definition of superset:

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

$\Box$

Thus by definition of set equality:

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

$\blacksquare$