# Equivalence of Definitions of Complex Inverse Tangent Function

## 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:

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

where $\tan \left({w}\right)$ is the tangent of $w$.

### Definition 2

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

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

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

## Proof

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

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

Note that when $z = 0 + i$:

 $\displaystyle i - z$ $=$ $\displaystyle 0 + 0 i$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {i - z} {i + z}$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle \map \ln {\dfrac {i - z} {i + z} }$  $\displaystyle \text {is undefined}$

Similarly, when $z = 0 - i$:

 $\displaystyle i + z$ $=$ $\displaystyle 0 + 0 i$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {i - z} {i + z}$  $\displaystyle \text {is undefined}$

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

### Definition 1 implies Definition 2

It is demonstrated that:

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

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

Then:

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

Thus by definition of subset:

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

$\Box$

### Definition 2 implies Definition 1

It is demonstrated that:

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

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

Then:

 $\displaystyle \exists k \in \Z: \ \$ $\displaystyle w$ $=$ $\displaystyle \dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi$ $\displaystyle \leadsto \ \$ $\displaystyle \exists k \in \Z: \ \$ $\displaystyle 2 i w + 2 \paren {-k} \pi i$ $=$ $\displaystyle \map \ln {\dfrac {i - z} {i + z} }$ $\displaystyle \leadsto \ \$ $\displaystyle e^{2 i w + 2 \paren {-k} \pi i}$ $=$ $\displaystyle \dfrac {i - z} {i + z}$ Definition of Complex Natural Logarithm $\displaystyle \leadsto \ \$ $\displaystyle e^{2 i w}$ $=$ $\displaystyle \dfrac {i - z} {i + z}$ Complex Exponential Function has Imaginary Period $\displaystyle \leadsto \ \$ $\displaystyle z$ $=$ $\displaystyle i \dfrac {1 - e^{2 i w} } {1 + e^{2 i w} }$ $\displaystyle \leadsto \ \$ $\displaystyle z$ $=$ $\displaystyle \tan w$ Tangent Exponential Formulation $\displaystyle \leadsto \ \$ $\displaystyle w$ $\in$ $\displaystyle \set {w \in \C: \map \tan w = z}$

Thus by definition of superset:

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

$\Box$

Thus by definition of set equality:

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

$\blacksquare$