nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Complex Inverse Cosecant Function

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
The proof strategy is to show that for all $z \in \C$:
:$\set {w \in \C: z = \csc w} = \set {\dfrac 1 i \map \ln {\dfrac {i + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } z} + 2 k \pi: k \in \Z}$
Thus let $z \in \C$.
=== Definition 1 implies Definition 2 ===
It will be demonstrated that:
:$\set {w \in \C: z = \csc w} \subseteq \set {\dfrac 1 i \map \ln {\dfrac {i + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } z} + 2 k \pi: k \in \Z}$
Let $w \in \set {w \in \C: z = \csc w}$.
From Euler's Cosecant Identity:
:$(1): \quad z = \dfrac {2 i} {e^{i w} - e^{-i w}}$
Let $v = e^{i w}$.
Then:
Let $s = z^2 - 1$.
Then:
We have that:
Thus from $(2)$ and $(3)$:
Thus by definition of subset:
:$\set {w \in \C: z = \csc w} \subseteq \set {\dfrac 1 i \map \ln {\dfrac {i + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } z} + 2 k \pi: k \in \Z}$
=== Definition 2 implies Definition 1 ===
It will be demonstrated that:
:$\set {w \in \C: z = \csc w} \supseteq \set {\dfrac 1 i \map \ln {\dfrac {i + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } z} + 2 k \pi: k \in \Z}$
Let $w \in \set {\dfrac 1 i \map \ln {\dfrac {i + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } z} + 2 k \pi: k \in \Z}$.
Then:
Thus by definition of superset:
:$\set {w \in \C: z = \csc w} \supseteq \set {\dfrac 1 i \map \ln {\dfrac {i + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } z} + 2 k \pi: k \in \Z}$
Thus by definition of set equality:
:$\set {w \in \C: z = \csc w} = \set {\dfrac 1 i \map \ln {\dfrac {i + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } z} + 2 k \pi: k \in \Z}$
Category:Inverse Cosecant
Equivalence of Definitions of Complex Inverse Cosine Function

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
The proof strategy is to show that for all $z \in \C$:
:$\set {w \in \C: z = \cos w} = \set {\dfrac 1 i \map \ln {z + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } + 2 k \pi: k \in \Z}$
Thus let $z \in \C$.
=== Definition 1 implies Definition 2 ===
It will be demonstrated that:
:$\set {w \in \C: z = \cos w} \subseteq \set {\dfrac 1 i \map \ln {z + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } + 2 k \pi: k \in \Z}$
Let $w \in \set {w \in \C: z = \cos w}$.
From Euler's Cosine Identity:
:$(1): \quad z = \dfrac {e^{i w} + e^{-i w} } 2$
Let $v = e^{i w}$.
Then:
Let $s = z^2 - 1$.
Then:
We have that:
Thus from $(2)$ and $(3)$:
Thus by definition of subset:
:$\set {w \in \C: z = \cos w} \subseteq \set {\dfrac 1 i \map \ln {z + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } + 2 k \pi: k \in \Z}$
=== Definition 2 implies Definition 1 ===
It will be demonstrated that:
:$\set {w \in \C: z = \cos w} \supseteq \set {\dfrac 1 i \map \ln {z + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } + 2 k \pi: k \in \Z}$
Let $w \in \set {\dfrac 1 i \map \ln {z + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } + 2 k \pi: k \in \Z}$.
Then:
Thus by definition of superset:
:$\set {w \in \C: z = \cos w} \supseteq \set {\dfrac 1 i \map \ln {z + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } + 2 k \pi: k \in \Z}$
Thus by definition of set equality:
:$\set {w \in \C: z = \cos w} = \set {\dfrac 1 i \map \ln {z + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } + 2 k \pi: k \in \Z}$
Category:Inverse Cosine
Equivalence of Definitions of Complex Inverse Cotangent Function

== Theorem ==
Let $S$ be the subset of the complex plane:
:$S = \C \setminus \set {0 + i, 0 - i}$
=== Definition 1 ===
=== Definition 2 ===
== 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 $z = 0 - i$:
Similarly, when $z = 0 + i$:
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:
Thus by definition of subset:
:$\set {w \in \C: \cot w = z} \subseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {z + i} {z - i} } + k \pi: k \in \Z}$
=== 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:
Thus by definition of superset:
:$\set {w \in \C: \cot w = z} \supseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {z + i} {z - i} } + k \pi: k \in \Z}$
Thus by definition of set equality:
:$set {w \in \C: \cot w = z} = \set {\dfrac 1 {2 i} \map \ln {\dfrac {z + i} {z - i} } + k \pi: k \in \Z}$
Category:Inverse Cotangent
Equivalence of Definitions of Complex Inverse Hyperbolic Cosecant

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
The proof strategy is to show that for all $z \in \C$:
:$\set {w \in \C: z = \csch w} = \set {\map \ln {\dfrac {1 + \sqrt {\cmod {z^2 + 1} } e^{\paren {i / 2} \map \arg {z^2 + 1} } } z} + 2 k \pi i: k \in \Z}$
Thus let $z \in \C$.
=== Definition 1 implies Definition 2 ===
It will be demonstrated that:
:$\set {w \in \C: z = \csch w} \subseteq \set {\map \ln {\dfrac {1 + \sqrt {\cmod {z^2 + 1} } e^{\paren {i / 2} \map \arg {z^2 + 1} } } z} + 2 k \pi i: k \in \Z}$
Let $w \in \set {w \in \C: z = \csch w}$.
From the definition of hyperbolic cosecant:
:$(1): \quad z = \dfrac 2 {e^w - e^{- w} }$
Let $v = e^w$.
Then:
Let $s = z^2 + 1$.
Then:
We have that:
Thus from $(2)$ and $(3)$:
Thus by definition of subset:
:$\set {w \in \C: z = \csch w} \subseteq \set {\map \ln {\dfrac {1 + \sqrt {\cmod {z^2 + 1} } e^{\paren {i / 2} \map \arg {z^2 + 1} } } z} + 2 k \pi i: k \in \Z}$
=== Definition 2 implies Definition 1 ===
It will be demonstrated that:
:$\set {w \in \C: z = \csch w} \supseteq \set {\map \ln {\dfrac {1 + \sqrt {\cmod {z^2 + 1} } e^{\paren {i / 2} \map \arg {z^2 + 1} } } z} + 2 k \pi i: k \in \Z}$
Let $w \in \set {\map \ln {\dfrac {1 + \sqrt {\cmod {z^2 + 1} } e^{\paren {i / 2} \map \arg {z^2 + 1} } } z} + 2 k \pi i: k \in \Z}$.
Then:
Thus by definition of superset:
:$\set {w \in \C: z = \csch w} \supseteq \set {\map \ln {\dfrac {1 + \sqrt {\cmod {z^2 + 1} } e^{\paren {i / 2} \map \arg {z^2 + 1} } } z} + 2 k \pi i: k \in \Z}$
Thus by definition of set equality:
:$\set {w \in \C: z = \csch w} = \set {\map \ln {\dfrac {1 + \sqrt {\cmod {z^2 + 1} } e^{\paren {i / 2} \map \arg {z^2 + 1} } } z} + 2 k \pi i: k \in \Z}$
Category:Inverse Hyperbolic Cosecant
Equivalence of Definitions of Complex Inverse Hyperbolic Cosine

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
The proof strategy is to how that for all $z \in \C$:
:$\set {w \in \C: z = \cosh w} = \set {\map \ln {z + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } + 2 k \pi i: k \in \Z}$
Thus let $z \in \C$.
=== Definition 1 implies Definition 2 ===
It is demonstrated that:
:$\set {w \in \C: z = \cosh w} \subseteq \set {\map \ln {z + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } + 2 k \pi i: k \in \Z}$
Let $w \in \set {w \in \C: z = \cosh w}$.
By definition of hyperbolic cosine:
:$(1): \quad z = \dfrac {e^w + e^{-w} } 2$
Let $v = e^w$.
Then:
Let $s = z^2 - 1$.
Then:
We have that:
Thus from $(2)$ and $(3)$:
Thus by definition of subset:
:$\set {w \in \C: z = \cos w} \subseteq \set {\map \ln {z + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } + 2 k \pi i: k \in \Z}$
=== Definition 2 implies Definition 1 ===
It is demonstrated that:
:$\set {w \in \C: z = \cos w} \supseteq \set {\map \ln {z + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } + 2 k \pi i: k \in \Z}$
Let $w \in \set {\map \ln {z + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } + 2 k \pi i: k \in \Z}$.
Then:
Thus by definition of superset:
:$\set {w \in \C: z = \cosh w} \supseteq \set {\map \ln {z + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } + 2 k \pi i: k \in \Z}$
Thus by definition of set equality:
:$\set {w \in \C: z = \cosh w} = \set {\map \ln {z + \sqrt {\cmod {z^2 - 1} } e^{\paren {i / 2} \map \arg {z^2 - 1} } } + 2 k \pi i: k \in \Z}$
Category:Inverse Hyperbolic Cosine
Equivalence of Definitions of Complex Inverse Hyperbolic Cotangent

== Theorem ==
Let $S$ be the subset of the complex plane:
:$S = \C \setminus \set {-1 + 0 i, 1 + 0 i}$
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
The proof strategy is to how that for all $z \in S$:
:$\set {w \in \C: z = \coth w} = \set {\dfrac 1 2 \map \ln {\dfrac {z + 1} {z - 1} } + k \pi i: k \in \Z}$
Note that when $z = -1 + 0 i$:
Similarly, when $z = 1 + 0 i$:
Thus let $z \in \C \setminus \set {-1 + 0 i, 1 + 0 i}$.
=== Definition 1 implies Definition 2 ===
It is demonstrated that:
:$\set {w \in \C: z = \coth w} \subseteq \set {\dfrac 1 2 \map \ln {\dfrac {z + 1} {z - 1} } + k \pi i: k \in \Z}$
Let $w \in \set {w \in \C: z = \coth w}$.
Then:
Thus by definition of subset:
:$\set {w \in \C: z = \coth w} \subseteq \set {\dfrac 1 2 \map \ln {\dfrac {z + 1} {z - 1} } + k \pi i: k \in \Z}$
=== Definition 2 implies Definition 1 ===
It is demonstrated that:
:$\set {w \in \C: z = \coth w} \supseteq \set {\dfrac 1 2 \map \ln {\dfrac {z + 1} {z - 1} } + k \pi i: k \in \Z}$
Let $w \in \set {\dfrac 1 2 \map \ln {\dfrac {z + 1} {z - 1} } + k \pi i: k \in \Z}$.
Then:
Thus by definition of superset:
:$\set {w \in \C: z = \coth w} \supseteq \set {\dfrac 1 2 \map \ln {\dfrac {z + 1} {z - 1} } + k \pi i: k \in \Z}$
Thus by definition of set equality:
:$\set {w \in \C: z = \coth w} = \set {\dfrac 1 2 \map \ln {\dfrac {z + 1} {z - 1} } + k \pi i: k \in \Z}$
Category:Inverse Hyperbolic Cotangent
Equivalence of Definitions of Complex Inverse Hyperbolic Secant

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
The proof strategy is to show that for all $z \in \C_{\ne 0}$:
:$\set {w \in \C: z = \sech w} = \set {\map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi i: k \in \Z}$
Thus let $z \in \C_{\ne 0}$.
=== Definition 1 implies Definition 2 ===
It will be demonstrated that:
:$\set {w \in \C: z = \sech w} \subseteq \set {\map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi i: k \in \Z}$
Let $w \in \set {w \in \C: z = \sech w}$.
From the definition of hyperbolic secant:
:$(1): \quad z = \dfrac 2 {e^w + e^{- w}}$
Let $v = e^w$.
Then:
Let $s = 1 - z^2$.
Then:
We have that:
Thus from $(2)$ and $(3)$:
Thus by definition of subset:
:$\set {w \in \C: z = \sech w} \subseteq \set {\map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi i: k \in \Z}$
=== Definition 2 implies Definition 1 ===
It will be demonstrated that:
:$\set {w \in \C: z = \sech w} \supseteq \set {\map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi i: k \in \Z}$
Let $w \in \set {\map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi i: k \in \Z}$.
Then:
Thus by definition of superset:
:$\set {w \in \C: z = \sech w} \supseteq \set {\map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi i: k \in \Z}$
Thus by definition of set equality:
:$\set {w \in \C: z = \sech w} = \set {\map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi i: k \in \Z}$
Category:Inverse Hyperbolic Secant
Equivalence of Definitions of Complex Inverse Hyperbolic Sine

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
The proof strategy is to how that for all $z \in \C$:
:$\set {w \in \C: z = \map \sinh w} = \set {\map \ln {z + \sqrt {\size {z^2 + 1} } e^{\paren {i / 2} \map \arg {z^2 + 1} } } + 2 k \pi i}$
Thus let $z \in \C$.
=== Definition 1 implies Definition 2 ===
It is demonstrated that:
:$\set {w \in \C: z = \map \sinh w} \subseteq \set {\map \ln {z + \sqrt {\size {z^2 + 1} } e^{\paren {i / 2} \map \arg {z^2 + 1} } } + 2 k \pi i}$
Let $w \in \set {w \in \C: z = \map \sinh w}$.
Then by definition of the hyperbolic sine function:
:$(1): \quad z = \dfrac {e^w - e^{-w} } 2$
Let $v = e^w$.
Then:
Let $s = z^2 + 1$.
Then:
We have that:
Thus from $(2)$ and $(3)$:
Thus by definition of subset:
:$\set {w \in \C: z = \map \sinh w} \subseteq \set {\map \ln {z + \sqrt {\size {z^2 + 1} } e^{\paren {i / 2} \map \arg {z^2 + 1} } } + 2 k \pi i}$
=== Definition 2 implies Definition 1 ===
It is demonstrated that:
:$\set {w \in \C: z = \map \sinh w} \supseteq \set {\map \ln {z + \sqrt {\size {z^2 + 1} } e^{\paren {i / 2} \map \arg {z^2 + 1} } } + 2 k \pi i}$
Let $w \in \set {\map \ln {z + \sqrt {\size {z^2 + 1} } e^{\paren {i / 2} \map \arg {z^2 + 1} } } + 2 k \pi i}$.
Then:
Thus by definition of superset:
:$\set {w \in \C: z = \map \sinh w} \supseteq \set {\map \ln {z + \sqrt {\size {z^2 + 1} } e^{\paren {i / 2} \map \arg {z^2 + 1} } } + 2 k \pi i}$
Thus by definition of set equality:
:$\set {w \in \C: z = \map \sinh w} = \set {\map \ln {z + \sqrt {\size {z^2 + 1} } e^{\paren {i / 2} \map \arg {z^2 + 1} } } + 2 k \pi i}$
Category:Inverse Hyperbolic Sine
Equivalence of Definitions of Complex Inverse Hyperbolic Tangent

== Theorem ==
Let $S$ be the subset of the complex plane:
:$S = \C \setminus \set {-1 + 0 i, 1 + 0 i}$
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
The proof strategy is to how that for all $z \in S$:
:$\set {w \in \C: z = \tanh w} = \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$
Note that when $z = -1 + 0 i$:
Similarly, when $z = 1 + 0 i$:
Thus let $z \in \C \setminus \set {-1 + 0 i, 1 + 0 i}$.
=== Definition 1 implies Definition 2 ===
It is demonstrated that:
:$\set {w \in \C: z = \tanh w} \subseteq \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$
Let $w \in \set {w \in \C: z = \tanh w}$.
Then:
Thus by definition of subset:
:$\set {w \in \C: z = \tanh w} \subseteq \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$
=== Definition 2 implies Definition 1 ===
It is demonstrated that:
:$\set {w \in \C: z = \tanh w} \supseteq \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$
Let $w \in \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$.
Then:
Thus by definition of superset:
:$\set {w \in \C: z = \tanh w} \supseteq \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$
Thus by definition of set equality:
:$\set {w \in \C: z = \tanh w} = \set {\dfrac 1 2 \map \ln {\dfrac {1 + z} {1 - z} } + k \pi i: k \in \Z}$
Category:Inverse Hyperbolic Tangent
Equivalence of Definitions of Complex Inverse Secant Function

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
The proof strategy is to show that for all $z \in \C_{\ne 0}$:
:$\set {w \in \C: \sec w = z} = \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$
Thus let $z \in \C$.
=== Definition 1 implies Definition 2 ===
It will be demonstrated that:
:$\set {w \in \C: \sec w = z} \subseteq \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$
Let $w \in \set {w \in \C: \sec w = z}$.
From Euler's Secant Identity:
:$(1): \quad z = \dfrac 2 {e^{i w} + e^{-i w} }$
Let $v = e^{i w}$.
Then:
Let $s = 1 - z^2$.
Then:
We have that:
Thus from $(2)$ and $(3)$:
Thus by definition of subset:
:$\set {w \in \C: \sec w = z} \subseteq \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$
=== Definition 2 implies Definition 1 ===
It will be demonstrated that:
:$\set {w \in \C: \sec w = z} \supseteq \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$
Let $w \in \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$.
Then:
Thus by definition of superset:
:$\set {w \in \C: \sec w = z} \supseteq \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$
Thus by definition of set equality:
:$\set {w \in \C: \sec w = z} = \set {\dfrac 1 i \map \ln {\dfrac {1 + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } z} + 2 k \pi: k \in \Z}$
Category:Inverse Secant
Equivalence of Definitions of Complex Inverse Sine Function

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
The proof strategy is to show that for all $z \in \C$:
:$\set {w \in \C: z = \sin w} = \set {\dfrac 1 i \map \ln {i z + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } + 2 k \pi: k \in \Z}$
Thus let $z \in \C$.
=== Definition 1 implies Definition 2 ===
It will be demonstrated that:
:$\set {w \in \C: z = \sin w} \subseteq \set {\dfrac 1 i \map \ln {i z + \sqrt{\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } + 2 k \pi: k \in \Z}$
Let $w \in \set {w \in \C: z = \sin w}$.
From Euler's Sine Identity:
:$(1): \quad z = \dfrac {e^{i w} - e^{-i w} } {2 i}$
Let $v = e^{i w}$.
Then:
Let $s = 1 - z^2$.
Then:
We have that:
Thus from $(2)$ and $(3)$:
Thus by definition of subset:
:$\set {w \in \C: z = \sin w} \subseteq \set {\frac 1 i \map \ln {i z + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } + 2 k \pi: k \in \Z}$
=== Definition 2 implies Definition 1 ===
It will be demonstrated that:
:$\set {w \in \C: z = \sin w} \supseteq \set {\dfrac 1 i \map \ln {i z + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } + 2 k \pi: k \in \Z}$
Let $w \in \set {\dfrac 1 i \map \ln {i z + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } + 2 k \pi: k \in \Z}$.
Then:
Thus by definition of superset:
:$\set {w \in \C: z = \sin w} \supseteq \set {\dfrac 1 i \map \ln {i z + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } + 2 k \pi: k \in \Z}$
Thus by definition of set equality:
:$\set {w \in \C: z = \sin w} = \set {\dfrac 1 i \map \ln {i z + \sqrt {\cmod {1 - z^2} } e^{\paren {i / 2} \map \arg {1 - z^2} } } + 2 k \pi: k \in \Z}$
Category:Inverse Sine
Equivalence of Definitions of Complex Inverse Tangent Function

== Theorem ==
Let $S$ be the subset of the complex plane:
:$S = \C \setminus \set {0 + i, 0 - i}$
=== Definition 1 ===
=== Definition 2 ===
== 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 $z = 0 + i$:
Similarly, when $z = 0 - i$:
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:
Thus by definition of subset:
:$\set {w \in \C: \tan w = z} \subseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$
=== 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:
Thus by definition of superset:
:$\set {w \in \C: \tan w = z} \supseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$
Thus by definition of set equality:
:$\set {w \in \C: \tan w = z} = \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$
Category:Inverse Tangent
Equivalence of Definitions of Complex Natural Logarithm

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $z = r e^{i \theta}$ such that $z \ne 0$.
Let $F = \set {\ln r + i \theta + 2 k \pi i: k \in \Z}$.
Let $G = \set {w \in \C: e^w = z}$.
We will demonstrate that $F = G$.
=== Definition 1 implies Definition 2 ===
Let $w = x + i y$ such that $w \in F$.
Then:
:$x + i y = \ln r + i \theta + 2 k \pi i$
for some $k \in \Z$.
Equating real and imaginary parts:
:$x = \ln r$
:$y = \theta + 2 k \pi$
Then:
Thus:
:$w \in G$
and so:
:$f \subseteq G$
=== Definition 2 implies Definition 1 ===
Let $w \in G$.
By definition:
:$\exists z \in \C_{\ne 0}: z = e^w = r e^{i \theta}$
Thus:
Thus $w$ is of the form:
:$\ln r + i \theta + k \pi i$
where $k = 0$.
Therefore:
:$w \in F$
and so:
:$G \subseteq F$
So by definition of set equality:
:$F = G$
== Sources ==
* : $\S 4.6$. The Logarithm: $(4.24)$
* : $0$ The origins of complex analysis, and a modern viewpoint: $1$. The origins of complex numbers
Category:Logarithms
Equivalence of Definitions of Complex Number

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Since:
:$\tuple {x_1, 0} + \tuple {x_2, 0} = \tuple {x_1 + x_2, 0}$
:$\tuple {x_1, 0} \tuple {x_2, 0} = \tuple {x_1 x_2, 0}$
we can identify a complex number (definition 2) $\tuple {x_1, 0}$ with the real number $x_1$.
Specifically, we can define an isomorphism between the set of complex numbers (definition 2) of the form $\tuple {x, 0}$ and the field of real numbers.
Now, we define $i = \tuple {0, 1}$.
Then:
Finally, we see that:
Thus we can say that $i = \sqrt {-1}$.
== Sources ==
* : $1$: Complex Numbers: Axiomatic Foundations of the Complex Number System
* : $1$: Complex Numbers: Solved Problems: Axiomatic Foundations of Complex Numbers: $14$
Category:Complex Numbers
Equivalence of Definitions of Component

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
=== Definition 1: Equivalence Class ===
=== Definition 2: Union of Connected Sets ===
=== Definition 3: Maximal Connected Set ===
== Proof ==
Let $\CC_x = \set {A \subseteq S : x \in A \land A \text{ is connected in } T}$
Let $C = \bigcup \CC_x$
=== Lemma ===
Let $C'$ be the equivalence class containing $x$ of the equivalence relation $\sim$ defined by:
:$y \sim z$ $y$ and $z$ are connected in $T$.
=== Equivalence Class equals Union of Connected Sets ===
It needs to be shown that $C = C'$.
=== Union of Connected Sets is Maximal Connected Set ===
=== Maximal Connected Set is Union of Connected Sets ===
== Also see ==
* Connectedness of Points is Equivalence Relation
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Connected Sets
Equivalence of Definitions of Component/Equivalence Class equals Union of Connected Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
Let $\CC_x = \set {A \subseteq S: x \in A \land A \text{ is connected in } T}$
Let $C = \bigcup \CC_x$
Let $\sim$ be the equivalence relation defined by:
:$y \sim z$ $y$ and $z$ are connected in $T$.
Let $C’$ be the equivalence class of $\sim$ containing $x$.
Then $C = C'$.
== Proof ==
The result follows.
== Also see ==
* Connectedness of Points is Equivalence Relation
Category:Connected Sets
Equivalence of Definitions of Component/Lemma 1

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
Let $\CC_x = \set {A \subseteq S : x \in A \land A \text{ is connected in } T}$
Let $\ds C = \bigcup \CC_x$
Then:
:$C$ is connected in $T$ and $C \in \CC_x$.
== Proof ==
From Singleton is Connected in Topological Space, $\set{x}$ is a connected set of $T$ containing $x$.
It follows that $x \in C$.
From Union of Connected Sets with Common Point is Connected, $C$ is a connected set of $T$.
Hence $C \in \CC_x$.