Equivalence of Definitions of Complex Inverse Cosine Function

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

From Cosine Exponential Formulation:

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


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:

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

$\blacksquare$