Equivalence of Definitions of Complex Inverse Cotangent Function
Theorem
The following definitions of the concept of Complex Inverse Cotangent are equivalent:
Let $S$ be the subset of the complex plane:
- $S = \C \setminus \set {0 + i, 0 - i}$
Definition 1
The inverse cotangent is a multifunction defined on $S$ as:
- $\forall z \in S: \cot^{-1} \left({z}\right) := \left\{{w \in \C: \cot \left({w}\right) = z}\right\}$
where $\cot \left({w}\right)$ is the cotangent of $w$.
Definition 2
The inverse cotangent is a multifunction defined on $S$ as:
- $\forall z \in S: \cot^{-1} \left({z}\right) := \left\{{\dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i}}\right) + k \pi: k \in \Z}\right\}$
where $\ln$ denotes the complex natural logarithm as a multifunction.
Proof
The proof strategy is to how that for all $z \in S$:
- $\set {w \in \C: \cot w = z} = \set {\dfrac 1 {2 i} \map \ln {\dfrac {z + i} {z - i} } + k \pi: k \in \Z}$
Note that when $z = 0 - i$:
\(\ds z + i\) | \(=\) | \(\ds 0 + 0 i\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {z + i} {z - i}\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \ln {\dfrac {z + i} {z - i} }\) | \(\) | \(\ds \text {is undefined}\) |
Similarly, when $z = 0 + i$:
\(\ds z - i\) | \(=\) | \(\ds 0 + 0 i\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {z + i} {z - i}\) | \(\) | \(\ds \text {is undefined}\) |
Thus let $z \in \C \setminus \set {0 + i, 0 - i}$.
Definition 1 implies Definition 2
It is demonstrated that:
- $\set {w \in \C: \cot w = z} \subseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {z + i} {z - i} } + k \pi: k \in \Z}$
Let $w \in \set {w \in \C: z = \cot w}$.
Then:
\(\ds z\) | \(=\) | \(\ds i \frac {e^{i w} + e^{-i w} } {e^{i w} - e^{-i w} }\) | Euler's Cotangent Identity | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{2 i w}\) | \(=\) | \(\ds \frac {i z - 1} {i z + 1}\) | solving for $e^{2 i w}$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac {z + i} {z - i}\) | multiplying top and bottom by $-i$, noting $i^2 = -1$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \ln {e^{2 i w} }\) | \(=\) | \(\ds \map \ln {\frac {z + i} {z - i} }\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 i w + 2 k' \pi i: k' \in \Z\) | \(=\) | \(\ds \map \ln {\frac {z + i} {z - i} }\) | Definition of Complex Natural Logarithm | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists k \in \Z: \, \) | \(\ds w\) | \(=\) | \(\ds \frac 1 {2 i} \map \ln {\frac {z + i} {z - i} } + k \pi\) | putting $k = -k'$ |
Thus by definition of subset:
- $\set {w \in \C: \cot w = z} \subseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {z + i} {z - i} } + k \pi: k \in \Z}$
$\Box$
Definition 2 implies Definition 1
It is demonstrated that:
- $\set {w \in \C: \cot w = z} \supseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {z + i} {z - i} } + k \pi: k \in \Z}$
Let $w \in \set {\dfrac 1 {2 i} \map \ln {\dfrac {z + i} {z - i} } + k \pi: k \in \Z}$.
Then:
\(\ds \exists k \in \Z: \, \) | \(\ds w\) | \(=\) | \(\ds \dfrac 1 {2 i} \map \ln {\dfrac {z + i} {z - i} } + k \pi\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists k \in \Z:: \, \) | \(\ds 2 i w + 2 \paren {-k} \pi i\) | \(=\) | \(\ds \map \ln {\frac {z + i} {z - i} }\) | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{2 i w + 2 \paren {-k} \pi i}\) | \(=\) | \(\ds \frac {z + i} {z - i}\) | Definition of Complex Natural Logarithm | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{2 i w}\) | \(=\) | \(\ds \frac {z + i} {z - i}\) | Complex Exponential Function has Imaginary Period | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds i \frac {e^{2 i w} + 1} {e^{2 i w} - 1}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds \cot w\) | Euler's Cotangent Identity | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds w\) | \(\in\) | \(\ds \set {w \in \C: \cot w = z}\) |
Thus by definition of superset:
- $\set {w \in \C: \cot w = z} \supseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {z + i} {z - i} } + k \pi: k \in \Z}$
$\Box$
Thus by definition of set equality:
- $set {w \in \C: \cot w = z} = \set {\dfrac 1 {2 i} \map \ln {\dfrac {z + i} {z - i} } + k \pi: k \in \Z}$
$\blacksquare$