Equivalence of Definitions of Complex Inverse Tangent Function

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Theorem

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

Let $S$ be the subset of the complex plane:

$S = \C \setminus \set {0 + i, 0 - i}$

Definition 1

The inverse tangent is a multifunction defined on $S$ as:

$\forall z \in S: \inv \tan z := \set {w \in \C: \map \tan w = z}$

where $\map \tan w$ is the tangent of $w$.

Definition 2

The inverse tangent is a multifunction defined on $S$ as:

$\forall z \in S: \inv \tan z := \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$

where $\ln$ denotes the complex natural logarithm as a multifunction.


Proof

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

$\set {w \in \C: \tan w = z} = \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$


Note that when $z = 0 + i$:

\(\ds i - z\) \(=\) \(\ds 0 + 0 i\)
\(\ds \leadsto \ \ \) \(\ds \frac {i - z} {i + z}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \map \ln {\dfrac {i - z} {i + z} }\) \(\) \(\ds \text {is undefined}\)


Similarly, when $z = 0 - i$:

\(\ds i + z\) \(=\) \(\ds 0 + 0 i\)
\(\ds \leadsto \ \ \) \(\ds \frac {i - z} {i + z}\) \(\) \(\ds \text {is undefined}\)


Thus let $z \in \C \setminus \set {0 + i, 0 - i}$.


Definition 1 implies Definition 2

It is demonstrated that:

$\set {w \in \C: \tan w = z} \subseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$


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


Then:

\(\ds z\) \(=\) \(\ds i \frac {1 - e^{2 i w} } {1 + e^{2 i w} }\) Euler's Tangent Identity
\(\ds \leadsto \ \ \) \(\ds e^{2 i w}\) \(=\) \(\ds \frac {1 + i z} {1 - i z}\) solving for $e^{2 i w}$
\(\ds \) \(=\) \(\ds \frac {i - z} {i + z}\) multiplying top and bottom by $i$, noting $i^2 = -1$
\(\ds \leadsto \ \ \) \(\ds \map \ln {e^{2 i w} }\) \(=\) \(\ds \ln \frac {i - z} {i + z}\)
\(\ds \leadsto \ \ \) \(\ds 2 i w + 2 k' \pi i: k' \in \Z\) \(=\) \(\ds \ln \frac {i - z} {i + z}\) Definition of Complex Natural Logarithm
\(\ds \leadsto \ \ \) \(\ds w\) \(=\) \(\ds \frac 1 {2 i} \ln \frac {i - z} {i + z} + k \pi: k \in \Z\) putting $k = -k'$


Thus by definition of subset:

$\set {w \in \C: \tan w = z} \subseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$

$\Box$


Definition 2 implies Definition 1

It is demonstrated that:

$\set {w \in \C: \tan w = z} \supseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$

Let $w \in \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$.

Then:

\(\ds \exists k \in \Z: \, \) \(\ds w\) \(=\) \(\ds \dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi\)
\(\ds \leadsto \ \ \) \(\ds \exists k \in \Z: \, \) \(\ds 2 i w + 2 \paren {-k} \pi i\) \(=\) \(\ds \map \ln {\dfrac {i - z} {i + z} }\)
\(\ds \leadsto \ \ \) \(\ds e^{2 i w + 2 \paren {-k} \pi i}\) \(=\) \(\ds \dfrac {i - z} {i + z}\) Definition of Complex Natural Logarithm
\(\ds \leadsto \ \ \) \(\ds e^{2 i w}\) \(=\) \(\ds \dfrac {i - z} {i + z}\) Complex Exponential Function has Imaginary Period
\(\ds \leadsto \ \ \) \(\ds z\) \(=\) \(\ds i \dfrac {1 - e^{2 i w} } {1 + e^{2 i w} }\)
\(\ds \leadsto \ \ \) \(\ds z\) \(=\) \(\ds \tan w\) Euler's Tangent Identity
\(\ds \leadsto \ \ \) \(\ds w\) \(\in\) \(\ds \set {w \in \C: \tan w = z}\)


Thus by definition of superset:

$\set {w \in \C: \tan w = z} \supseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$

$\Box$


Thus by definition of set equality:

$\set {w \in \C: \tan w = z} = \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$

$\blacksquare$

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