nyuuzyou/wikis huggingface dataset:
Complex Numbers form Preordered Vector Space

== Theorem ==
Consider the complex numbers $\C$ as a vector space over itself.
Define the relation $\ge^\C$ by:
:$z \ge^\C w$
:
:$z - w \in \hointr 0 \infty$
for each $z, w \in \C$.
Then $\struct {\C, \ge^\C}$ is a preordered vector space.
== Proof ==
From Characterization of Preordered Vector Spaces, it is enough to show that $\hointr 0 \infty$ is a convex cone.
Let $x \in \hointr 0 \infty$ and $\alpha \in \R_{\ge 0}$. 
Then $\alpha x \ge 0$, so $\alpha x \in \hointr 0 \infty$.
So $\hointr 0 \infty$ is a cone.
Now let $x, y \in \hointr 0 \infty$. 
Then we have $x + y \ge 0$, so $x + y \in \hointr 0 \infty$.
So $\hointr 0 \infty$ is a convex cone.
Category:Preordered Vector Spaces
Complex Numbers form Ring

== Theorem ==
The set of complex numbers $\C$ forms a ring under addition and multiplication: $\struct {\C, +, \times}$.
== Proof ==
From Complex Numbers under Addition form Infinite Abelian Group, $\struct {\C, +}$ is an abelian group.
We also have that:
:Complex Multiplication is Closed:
::$\forall x, y \in \C: x \times y \in \C$
:Complex Multiplication is Associative:
::$\forall x, y, z \in \C: x \times \paren {y \times z} = \paren {x \times y} \times z$
Thus $\struct{\C, +}$ is a semigroup.
Finally we have that Complex Multiplication Distributes over Addition:
Hence the result, by definition of ring.
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $4$
* : $\S 54$. The definition of a ring and its elementary consequences
Category:Examples of Rings
Category:Complex Numbers
Complex Numbers form Subfield of Quaternions

== Theorem ==
The field of complex numbers $\left({\C, +, \times}\right)$ is isomorphic to the subfields of the quaternions $\left({\mathbb H, +, \times}\right)$ whose underlying subsets are:
:$(1): \quad \mathbb H_\mathbf i = \left\{{a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k \in \mathbb H: c = d = 0}\right\}$
:$(2): \quad \mathbb H_\mathbf j = \left\{{a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k \in \mathbb H: b = d = 0}\right\}$
:$(3): \quad \mathbb H_\mathbf k = \left\{{a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k \in \mathbb H: b = c = 0}\right\}$
That is:
:$(1): \quad \mathbb H_\mathbf i = \left\{{a \mathbf 1 + b \mathbf i \in \mathbb H}\right\}$
:$(2): \quad \mathbb H_\mathbf j = \left\{{a \mathbf 1 + c \mathbf j \in \mathbb H}\right\}$
:$(3): \quad \mathbb H_\mathbf k = \left\{{a \mathbf 1 + d \mathbf k \in \mathbb H}\right\}$
== Proof ==
Let $\phi_i: \mathbb H_\mathbf i \to \C$ be defined as:
:$\forall \mathbf x = \mathbf a \mathbf 1 + b \mathbf i \in \mathbb H_\mathbf i: \phi_i \left({\mathbf x}\right) = a + b i$
where in this context $i$ is the imaginary unit.
Similarly we can define $\phi_j$ and $\phi_k$:
:$\forall \mathbf x = \mathbf a \mathbf 1 + c \mathbf j \in \mathbb H_\mathbf j: \phi_j \left({\mathbf x}\right) = a + c i$
:$\forall \mathbf x = \mathbf a \mathbf 1 + d \mathbf k \in \mathbb H_\mathbf k: \phi_k \left({\mathbf x}\right) = a + d i$
=== Proof of Bijectivity ===
First note that each of $\phi_i, \phi_j, \phi_k$ are bijections, as follows:
* Injections:
Let $\mathbf x_1 = \mathbf a_1 \mathbf 1 + b_1 \mathbf i, \mathbf x_2 = \mathbf a_2 \mathbf 1 + b_2 \mathbf i$. Then:
:$\phi_i \left({x_1}\right) = \phi_i \left({x_2}\right) \implies a_1 + b_1 i = a_2 + b_2 i \implies a_1 = a_2, b_1 = b_2 \implies \mathbf x_1 = \mathbf x_2$
and similarly with $\phi_j$ and $\phi_k$.
* Surjections:
Let $x = a + b i \in \C$.
Then:
: $\exists \mathbf x = \mathbf a \mathbf 1 + b \mathbf i \in \mathbb H_\mathbf i: \phi_i \left({\mathbf x}\right) = x$
and similarly with $\phi_j$ and $\phi_k$.
Thus it is established that $\phi_i, \phi_j, \phi_k$ are bijections.
=== Proof of Morphism Property ===
Let $\mathbf x_1 = \mathbf a_1 \mathbf 1 + b_1 \mathbf i, \mathbf x_2 = \mathbf a_2 \mathbf 1 + b_2 \mathbf i$.
First we show that $+$ has the morphism property under $\phi_i$:
and similarly for $\phi_j$ and $\phi_k$.
Next we show that $\times$ has the morphism property under $\phi_i$:
Similarly for $\phi_j$ and $\phi_k$.
So we have shown that $\phi_i, \phi_j, \phi_k$ are ring homomorphisms.
So all of these mappings are bijective homomorphisms, that is, isomorphisms.
== Sources ==
* : $\S 2.2$: Homomorphisms
Category:Complex Numbers
Category:Quaternions
Category:Subfields
Complex Numbers form Unital C*-Algebra

== Theorem ==
Let $\C$ be the complex numbers realised as a normed algebra (this is possible from Complex Numbers form Algebra).
Let $\overline {\, \cdot \,}$ be complex conjugation.
Then $\tuple {\C, \overline {\, \cdot \,} }$ is a unital $\text C^\ast$-algebra.
== Proof ==
From Complex Plane is Banach Space, $\C$ is a Banach space.
Further, we have:
:$\cmod 1 = 1$
Since $\C$ is also a normed algebra, it therefore a unital Banach algebra.
We prove that complex conjugation satisfies the properties of an involution as well as the $\text C^\ast$ identity.
=== Proof of $(\text C^\ast 1)$ ===
From Complex Conjugation is Involution, we have:
:$\overline {\overline z} = z$ 
for each $z \in \C$.
=== Proof of $(\text C^\ast 2)$ ===
From Sum of Complex Conjugates, we have:
:$\overline {z + w} = \overline z + \overline w$
for each $z, w \in \C$.
=== Proof of $(\text C^\ast 3)$ ===
From Product of Complex Conjugates, we have:
:$\overline {z w} = \overline z \overline w = \overline w \overline z$
for each $z, w \in \C$.
=== Proof of $(\text C^\ast 4)$ ===
From Product of Complex Conjugates again, we have:
:$\overline {\alpha z} = \overline \alpha \overline w$
for each $\alpha, z \in \C$.
=== Proof of $(\text C^\ast 5)$ ===
From Product of Complex Number with Conjugate, we have:
:$z \overline z = \cmod z^2$
Hence, we have:
:$\cmod {z \overline z} = \cmod z^2$
for each $z \in \C$.
Hence $\tuple {\C, \overline {\, \cdot \,} }$ is a unital Banach $\ast$-algebra satisfying the $\text C^\ast$ identity.
Hence $\tuple {\C, \overline {\, \cdot \,} }$ is a unital $\text C^\ast$-algebra.
Category:C*-Algebras
Category:Complex Numbers
Complex Numbers form Vector Space over Reals

== Theorem ==
Let $\R$ be the set of real numbers.
Let $\C$ be the set of complex numbers.
Then the $\R$-module $\C$ is a vector space.
== Proof ==
Recall that Real Numbers form Field.
Thus by definition, $\R$ is also a division ring.
Thus we only need to show that $\R$-module $\C$ is a unitary module, by demonstrating the module properties:
$\forall x, y, \in \C, \forall \lambda, \mu \in \R$:
:$(1): \quad \lambda \paren {x + y} = \paren {\lambda x} + \paren {\lambda y}$
:$(2): \quad \paren {\lambda + \mu} x = \paren {\lambda x} + \paren {\mu x}$
:$(3): \quad \paren {\lambda \mu} x = \lambda \paren {\mu x}$
:$(4): \quad 1 x = x$
As $\lambda, \mu \in \R$ it follows that $\lambda, \mu \in \C$.
Thus from Complex Multiplication Distributes over Addition, $(1)$ and $(2)$ immediately follow.
$(3)$ follows from Complex Multiplication is Associative.
$(4)$ follows from Complex Multiplication Identity is One, as $1 + 0 i$ is the unity of $\C$.
== Also see ==
* Properties of Complex Numbers
* Complex Numbers form Vector Space over Themselves
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.2$
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $4$. LINEAR VECTOR SPACE: Example $2$
Category:Linear Algebra
Category:Complex Analysis
Category:Examples of Vector Spaces
Complex Numbers form Vector Space over Themselves

== Theorem ==
The set of complex numbers $\C$, with the operations of addition and multiplication, forms a vector space.
== Proof ==
Let the field of complex numbers be denoted $\struct {\C, +, \times}$.
By Complex Numbers under Addition form Infinite Abelian Group, $\struct {\C, +}$ is an abelian group.
From Complex Multiplication Distributes over Addition:
From Complex Multiplication is Associative:
:$\forall x, y, z \in \C: x \times \paren {y \times z} = \paren {x \times y} \times z$
From Complex Multiplication Identity is One:
:$\forall x \in \C: 1 \times x = x$
Therefore $\struct {\C, +, \times}$ forms a vector space.
== Also see ==
* Properties of Complex Numbers
* Complex Numbers form Vector Space over Reals
== Sources ==
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $4$. LINEAR VECTOR SPACE: Example $2$
Category:Linear Algebra
Category:Complex Numbers
Category:Examples of Vector Spaces
Complex Numbers of Type Rational a plus b root 2 form Field

== Theorem ==
Let $\Q \sqbrk {\sqrt 2}$ denote the set:
:$\Q \sqbrk {\sqrt 2} := \set {x \in \C: x = a + b \sqrt 2: a, b \in \Q}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are rational numbers.
Then the algebraic structure:
:$\struct {\Q \sqbrk {\sqrt 2}, +, \times}$
where $+$ and $\times$ are conventional addition and multiplication on real numbers, is a number field.
== Proof ==
From Real Numbers of Type Rational a plus b root 2 form Field, the set:
:$\Q \sqbrk {\sqrt 2} := \set {x \in \R: x = a + b \sqrt 2: a, b \in \Q}$
forms a field.
As $\Q \sqbrk {\sqrt 2} \subseteq \R$ and $\R \subseteq \C$ it follows that $\struct {\Q \sqbrk {\sqrt 2}, +, \times}$ is a subfield of $\C$.
Hence the result by definition of number field.
== Sources ==
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 88 \beta$
Category:Number Fields
Complex Numbers under Addition form Group

== Theorem ==
Let $\C$ be the set of complex numbers.
The structure $\struct {\C, +}$ is a group.
== Proof ==
Taking the group axioms in turn:
=== ===
Complex Addition is Closed.
=== ===
Complex Addition is Associative.
=== ===
From Complex Addition Identity is Zero, we have that the identity element of $\struct {\C, +}$ is the complex number $0 + 0 i$:
:$\paren {x + i y} + \paren {0 + 0 i} = \paren {x + 0} + i \paren {y + 0} = x + i y$
and similarly for $\paren {0 + 0 i} + \paren {x + i y}$.
=== ===
From Inverse for Complex Addition, the inverse of $x + i y \in \struct {\C, +}$ is $-x - i y$.
== Sources ==
* : $\S 1$: Some examples of groups: Example $1.04$
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $2$. GROUP: Example $5$
Category:Additive Group of Complex Numbers
Category:Complex Addition
Category:Examples of Groups
Complex Numbers under Addition form Infinite Abelian Group

== Theorem ==
Let $\C$ be the set of complex numbers.
The structure $\struct {\C, +}$ is an infinite abelian group.
== Proof ==
From Complex Numbers under Addition form Group, $\struct {\C, +}$ is a group.
Then:
:Complex Addition is Commutative
and:
:Complex Numbers are Uncountable.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.1$
* : $\S 34$. Examples of groups: $(1)$
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(i)}$
* : Chapter $1$: Definitions and Examples: Example $1.4$
Category:Additive Group of Complex Numbers
Category:Complex Addition
Category:Examples of Abelian Groups
Category:Examples of Infinite Groups
Complex Numbers under Addition form Monoid

== Theorem ==
The set of complex numbers under addition $\left({\C, +}\right)$ forms a monoid.
== Proof ==
Taking the monoid axioms in turn:
=== ===
Complex Addition is Closed.
=== ===
Complex Addition is Associative.
=== ===
From Complex Addition Identity is Zero, we have that the identity element of $\struct {\C, +}$ is the complex number $0 + 0 i$:
:$\paren {x + i y} + \paren {0 + 0 i} = \paren {x + 0} + i \paren {y + 0} = x + i y$
and similarly for $\paren {0 + 0 i} + \paren {x + i y}$.
Hence the result.
== Sources ==
* : $\S 3.1$: Monoids
Category:Complex Addition
Category:Examples of Monoids
Complex Numbers under Multiplication do not form Group

== Theorem ==
The algebraic structure $\struct {\C, \times}$ consisting of the set of complex numbers $\C$ under multiplication $\times$ is not a group.
== Proof ==
 that $\struct {\C, \times}$ is a group.
By the definition of the number $0 \in \C$:
:$\forall x \in \C: x \times 0 = 0 = 0 \times x$
Thus $0$ is a zero in the abstract algebraic sense.
From Group with Zero Element is Trivial, $\struct {\C, \times}$ is the trivial group.
But $\C$ contains other elements besides $0$.
From this contradiction it follows that $\struct {\C, \times}$ is not a group.
== Also see ==
* Non-Zero Complex Numbers under Multiplication form Infinite Abelian Group
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.2$
Category:Complex Multiplication
Category:Examples of Groups
Complex Numbers under Multiplication form Monoid

== Theorem ==
The set of complex numbers under multiplication $\struct {\C, \times}$ forms a monoid.
== Proof ==
Taking the monoid axioms in turn:
=== ===
Complex Multiplication is Closed.
=== ===
Complex Multiplication is Associative.
=== ===
Complex Multiplication Identity is $1$.
Hence the result.
== Sources ==
* : $\S 3.1$: Monoids
Category:Complex Multiplication
Category:Examples of Monoids
Complex Numbers with Complex Modulus form Normed Vector Space

== Theorem ==
Let $\C$ be the set of complex numbers.
Let $\size {\, \cdot \,}$ be the complex modulus.
Then $\struct {\C, \size {\, \cdot \,}}$ is a normed vector space.
== Proof ==
We have that:
:Complex Numbers form Vector Space over Themselves
:Complex Modulus is Norm
By definition, $\struct {\C, \size {\, \cdot \,}}$ is a normed vector space.
== Sources ==
Category:Examples of Normed Vector Spaces
Category:Complex Numbers
Complex Plane is Banach Space

== Theorem ==
The complex plane, along with the complex modulus, forms a Banach space over $\C$.
== Proof ==
The complex plane $\C$ is a vector space over $\C$.
That the norm axioms are satisfied is proven in Complex Modulus is Norm.
Then we have Complex Plane is Complete Metric Space.
Hence the result.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Banach Spaces
Complex Plane is Complete Metric Space

== Theorem ==
The complex plane, along with the metric induced by the norm given by the complex modulus, forms a complete metric space.
== Proof ==
Let $z = x + iy$ be a complex number, where $x, y \in \R$.
Now, we can identify the complex number $z$ with the ordered pair $\left( x, y \right) \in \R^2$.
The norm on $\C$ given by the complex modulus is then identical to the Euclidean norm on $\R^2$.
Therefore, metric on $\C$ induced by the norm given by the complex modulus is identical to the Euclidean metric on $\R^2$ induced by the Euclidean norm.
By Euclidean Space is Complete Metric Space, $\C$ is a complete metric space.
== Sources ==
* : $3.11c$
Category:Complete Metric Spaces
Complex Plane is Homeomorphic to Real Plane

== Theorem ==
Let $\C$ denote the complex plane.
Let $\R^2$ denote the Euclidean plane.
Then the function $\phi : \R^2 \to \C$ defined by:
:$\map \phi {x, y} = x + i y$
is a homeomorphism between $\R^2$ and $\C$.
== Proof ==
Define $\phi^{-1} : \C \to \R^2$ by:
:$\map {\phi^{-1} } z = \tuple {\map \Re z, \map \Im z}$
Then $\phi^{-1}$ is the inverse of $\phi$, as:
:$\map \phi {\map { \phi^{-1} } z} = \map \Re z + i \map \Im z = z$
:$\map {\phi^{-1} } {\map \phi {x, y} } = \tuple {\map \Re {x + i y}, \map \Im {x + i y} } = \tuple {x, y}$
By definition of bijection, it follows that $\phi$ is bijective.
We show continuity of $\phi$.
Complex Plane is Metric Space shows that for $\map \phi {x_1, y_1} = x_1 + i y_1, \map \phi {x_2, y_2} = x_2 + i y_2$, we have:
:$\cmod {\map \phi {x_2, y_2} - \map \phi {x_1, y_1} } = \sqrt {\paren {x_2 - x_1}^2 + \paren {y_2 - y_1}^2}$
which is equal to the distance between $\tuple {x_1, x_2}$ and $\tuple {y_1, y_2}$ in the Euclidean metric on the Euclidean plane.
Let $\epsilon \in \R_{>0}$.
With $\delta = \epsilon$, we have:
:$\sqrt {\paren {x_2 - x_1}^2 + \paren {y_2 - y_1}^2} < \delta \implies \cmod {\map \phi {x_2, y_2} - \map \phi {x_1 , y_1} } < \epsilon$
which shows the continuity of $\phi$.
The continuity of $\phi^{-1}$ is proven the same way.
Category:Complex Analysis
Category:Examples of Homeomorphisms
Complex Plane is Metric Space

== Theorem ==
Let $\C$ be the set of all complex numbers.
Let $d: \C \times \C \to \R$ be the function defined as:
:$\map d {z_1, z_2} = \size {z_1 - z_2}$
where $\size z$ is the modulus of $z$.
Then $d$ is a metric on $\C$ and so $\struct {\C, d}$ is a metric space.
== Proof ==
Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$.
From the definition of modulus:
:$\size {z_1 - z_2} = \sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$
This is the euclidean metric on the real number plane.
This is shown in Euclidean Metric on Real Vector Space is Metric to be a metric.
Thus the complex plane is a 2-dimensional Euclidean space.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.4$
Category:Complex Plane
Category:Examples of Metric Spaces
Complex Power/Examples

== Examples of Complex Powers ==
=== Example: $2^i$ ===
=== Example: $\paren {2 + i}^4$ ===
=== Example: $\paren {1 + i \tan \paren {\dfrac {4 m + 1} {4 n} \pi} }^n$ ===
=== Example: $\paren {1 + \sin \dfrac \pi 5 + i \cos \dfrac \pi 5}^5 + i \paren {1 + \sin \dfrac \pi 5 - i \cos \dfrac \pi 5}^5$ ===
=== Example: $\paren {2 \cis 50 \degrees}^6$ ===
Category:Examples of Complex Powers
Complex Power/Examples/(1 + i \tan (4m+1 over 4n) pi)^n

== Example of Complex Power ==
For $m, n \in \Z$ such that $n \ne 0$:
:$\paren {1 + i \map \tan {\dfrac {4 m + 1} {4 n} \pi} }^n = \paren {-1}^m \paren {\sec \dfrac {4 m + 1} {4 n} \pi}^n \paren {\dfrac {1 + i} {\sqrt 2} }$
== Proof ==
First setting $x = \dfrac {4 m + 1} {4 n} \pi$, we have:
Now we have:
The result follows.
== Sources ==
* : $\S 2$. Geometrical Representations: Exercise $10$
Category:Examples of Complex Powers
Complex Power/Examples/(2+i)^4

== Example of Complex Power ==
:$\paren {2 + i}^4 = -7 + 24 i$
== Proof ==
We have that:
:$\paren {2 + i}^4 = \paren {\paren {2 + i}^2}^2$
Hence:
Then:
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $1 \ \text{(iii)}$
Category:Examples of Complex Powers
Complex Power/Examples/(2 cis 50)^6

== Example of Complex Power ==
:$\paren {2 \cis 50 \degrees}^6 = 32 - 32 \sqrt 3 i$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: De Moivre's Theorem: $89 \ \text{(b)}$
Category:Examples of Complex Powers
Complex Power Series/Examples

== Examples of Complex Power Series ==
=== Example: $\ds \sum_{n \mathop \ge 0} n z^n$ ===
=== Example: $\ds \sum_{n \mathop \ge 0} \dfrac {3^n - 1} {2^n + 1} z^n$ ===
=== Example: $\ds \sum_{n \mathop \ge 0} \dfrac {\paren {2 n}!} {\paren {n!}^2} z^n$ ===
=== Example: $\ds \sum_{n \mathop \ge 0} \dfrac {\cos i n} {n^2} z^n$ ===
Category:Examples of Complex Power Series
Complex Power Series/Examples/2n Factorial over n Factorial Squared

== Example of Complex Power Series ==
Let $\sequence {a_n}$ be the sequence defined as:
:$a_n = \dfrac {\paren {2 n}!} {\paren {n!}^2} z^n$
The complex power series:
:$S = \ds \sum_{n \mathop \ge 0} a_n z^n$
has a radius of convergence of $\dfrac 1 4$.
== Proof ==
Let $R$ denote the radius of convergence of $S$.
Thus:
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $3 \ \text {(iii)}$
Category:Examples of Complex Power Series
Complex Power Series/Examples/3^n-1 over 2^n+1

== Example of Complex Power Series ==
Let $\sequence {a_n}$ be the sequence defined as:
:$a_n = \dfrac {3^n - 1} {2^n + 1}$
The complex power series:
:$S = \ds \sum_{n \mathop \ge 0} a_n z^n$
has a radius of convergence of $\dfrac 2 3$.
== Proof ==
Let $R$ denote the radius of convergence of $S$.
By Radius of Convergence from Limit of Sequence:
:$R = \ds \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n - 1} } {a_n} }$
Thus:
Thus:
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $3 \ \text {(ii)}$
Category:Examples of Complex Power Series
Complex Power Series/Examples/cos i n over n^2

== Example of Complex Power Series ==
Let $\sequence {a_n}$ be the sequence defined as:
:$a_n = \dfrac {\cos i n} {n^2} z^n$
The complex power series:
:$S = \ds \sum_{n \mathop \ge 0} a_n z^n$
has a radius of convergence of $\dfrac 1 e$.
== Proof ==
Let $R$ denote the radius of convergence of $S$.
Thus:
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $3 \ \text {(iv)}$
Category:Examples of Complex Power Series
Complex Power Series/Examples/n

== Example of Complex Power Series ==
The complex power series:
:$S = \ds \sum_{n \mathop \ge 0} n z^n$
has a radius of convergence of $1$.
== Proof ==
Let $R$ denote the radius of convergence of $S$.
Thus:
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $3 \ \text {(i)}$
Category:Examples of Complex Power Series
Complex Power by Complex Exponential is Analytic

== Lemma ==
Let $\psi, \eta \in \C$ be constant.
Let $\map f z = z^\psi \, \map \exp {-\eta z}$, where:
:$z^\psi$ denotes $z$ to the power of $\psi$, defined on its principal branch
:$\map \exp {-\eta z}$ denotes the complex exponential function.
Then $f$ is analytic on any simply connected domain that does not contain the origin nor any points on the negative real axis.
== Proof ==
Let $z$ be written in exponential form:
:$z = r \map \exp {i \theta}$ 
where:
:$r > 0$
:$\theta \in \hointl {-\pi} \pi$
Let $\psi = a + i b, \eta = c + i d$.
By the definition of $f$:
Define:
Then:
Define:
Then:
We check the Polar Form of Cauchy-Riemann Equations.
As a preliminary: 
Then:
Category:Complex Analysis
Complex Power is of Exponential Order Epsilon

== Theorem == 
Let: 
:$f: \hointr 0 \to \to \C: t \mapsto t^\phi$
be $t$ to the power of $\phi$, for $\phi \in \C$, defined on its principal branch.
Let $\map \Re \phi > -1$.
Then $f$ is of exponential order $\epsilon$ for any $\epsilon > 0$ arbitrarily small in magnitude. 
== Proof ==
The result follows from Real Power is of Exponential Order Epsilon.
Category:Exponential Order
Complex Riemann Integral is Contour Integral

== Theorem ==
Let $f: \R \to \C$ be a complex Riemann integrable function over some closed real interval $\closedint a b$.
Then:
:$\ds \int_a^b \map f t \rd t = \int_\CC \map f t \rd t$
where:
:the integral on the is a complex Riemann integral
:the integral on the is a contour integral
:$\CC$ is a straight line segment along the real axis, connecting $a$ to $b$.
== Proof ==
Category:Integral Calculus
Category:Complex Contour Integrals
Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs

== Theorem ==
Let $\map f z = a_n z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z + a_0$ be a polynomial over complex numbers where $a_0, \ldots, a_n$ are real numbers.
Let $\alpha \in \C$ be a root of $f$.
Then $\overline \alpha$ is also a root of $f$, where $\overline \alpha$ denotes the complex conjugate of $\alpha$.
That is, all complex roots of $f$ appear as conjugate pairs.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2$. The Algebraic Theory
* 
Category:Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs
Category:Polynomial Theory
Category:Complex Conjugates
Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs/Proof 1

== Theorem ==
== Proof ==
Let $\alpha \in \C$ be a root of $f$.
Then $\map f \alpha = 0$ by definition.
Suppose $\alpha$ is wholly real.
Then by Complex Number equals Conjugate iff Wholly Real:
:$\alpha = \overline \alpha$
and so $\overline \alpha$ is a root of $f$.
Now let $\alpha \in \C$ not be wholly real.
By definition of complex conjugate, we have that:
:$\overline 0 = 0$
and so:
:$\map f \alpha = \overline {\map f \alpha}$
From Conjugate of Polynomial is Polynomial of Conjugate:
:$\overline {\map f \alpha} = \map f {\overline \alpha}$
from which it follows that:
:$\map f {\overline \alpha} = 0$
That is, $\overline \alpha$ is also a root of $f$.
Category:Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs