nyuuzyou/wikis huggingface dataset:
Complex Function is Entire iff it has Everywhere Convergent Power Series

== Theorem ==
Let $f: \C \to \C$ be a complex function.
Then $f$ is an entire function $f$ can be given by an everywhere convergent power series:
:$\ds \map f z = \sum_{n \mathop = 0}^\infty a_n z^n; \quad \lim_{n \mathop \to \infty} \sqrt [n] {\size {a_n} } = 0$
== Proof ==
=== Sufficient Condition ===
From Holomorphic Function is Analytic, if $f$ is holomorphic on some open ball $D$ in $\C$, then $f$ is complex analytic on $D$. 
However, as, by the definition of an entire function, $f$ is holomorphic everywhere, the radius of $D$ can be made arbitrarily large, meaning that $f$ is complex analytic everywhere.
This implies that an entire function is complex analytic everywhere.
=== Necessary Condition ===
It remains to prove that if $f$ is complex analytic everywhere, $f$ is entire. 
By Power Series is Termwise Differentiable within Radius of Convergence, if $f$ is complex analytic everywhere, $f$ is holomorphic everywhere. 
That is, a function complex analyticeverywhere is entire. 
Note that by the Cauchy-Hadamard theorem: 
:$\ds \lim_{n \mathop \to \infty} \sqrt [n] {\size {a_n} } = 0$
 this series has an infinite radius of convergence. 
That is, the series converges for all $z \in \C$, satisfying our second demand. 
Category:Entire Functions
Complex General Harmonic Numbers extend Integer version

== Theorem ==
=== Valid over Integers ===
=== Complex Extension ===
== Proof ==
For $x \in \C$ the harmonic numbers order $r$ are defined as follows:
:$\ds \harm r x = \sum_{k \mathop = 1}^\infty \paren {\frac 1 {k^r} - \frac 1 {\paren {k + x}^r} }$
Category:General Harmonic Numbers
Complex Integration by Substitution

== Theorem ==
Let $\closedint a b$ be a closed real interval.
Let $\phi: \closedint a b \to \R$ be a real function which has a derivative on $\closedint a b$.
Let $f: A \to \C$ be a continuous complex function, where $A$ is a subset of the image of $\phi$.
If $\map \phi a \le \map \phi b$, then:
:$\ds \int_{\map \phi a}^{\map \phi b} \map f t \rd t = \int_a^b \map f {\map \phi u} \map {\phi'} u \rd u$
If $\map \phi a > \map \phi b$, then:
:$\ds \int_{\map \phi b}^{\map \phi a} \map f t \rd t = -\int_a^b \map f {\map \phi u} \map {\phi'} u \rd u$
== Proof ==
Let $\Re$ and $\Im$ denote real parts and imaginary parts respectively.
Let $\map \phi a \le \map \phi b$.
Then:
Let $\map \phi a > \map \phi b$.
Then:
== Sources ==
* $\S 2.2$
Category:Complex Contour Integrals
Complex Inverse Sine/Examples/2

== Examples of Complex Inverse Sine Function ==
:$\map {\sin^{-1} } 2 = \dfrac {4 k + 1} 2 \pi - i \map \ln {2 \pm \sqrt 3}$
for $k \in \Z$.
== Proof ==
By definition of complex inverse sine:
:$\map {\sin^{-1} } 2 := \set {z \in \C: \sin z = 2}$
Thus:
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $7$
Category:Examples of Complex Inverse Sine
Complex Logarithm Function is Transcendental

== Theorem ==
Let $a \in \R_{>0}$ be a strictly positive real number such that $a \ne 1$.
Let $\log_a: \C \to \C$ denote the complex general logarithm base $a$.
$\log_a$ is a transcendental function.
== Proof ==
Category:Logarithms
Category:Examples of Transcendental Functions
Complex Modulus/Examples

== Examples of Complex Modulus ==
=== Complex Modulus of $i$ ===
=== Complex Modulus of $-5$ ===
=== Complex Modulus of $1 + i$ ===
=== Complex Modulus of $4 + 3 i$ ===
=== Complex Modulus of $-4 + 2 i$ ===
=== Complex Modulus of $7 + 24 i$ ===
=== Complex Modulus of $3iz - z^2$ ===
=== Complex Modulus of $\tan \theta + i$ ===
=== Complex Modulus of $\dfrac {1 + 2 i t - t^2} {1 + t^2}$ ===
Let $z_1 = 4 - 3 i$ and $z_2 = -1 + 2 i$.
=== Complex Modulus of $z_1 + z_2$ ===
=== Complex Modulus of $z_1 - z_2$ ===
=== Complex Modulus of $2 \overline z_1 - 3 \overline z_2 - 2$ ===
Category:Examples of Complex Modulus
Complex Modulus/Examples/1+2it-t^2 over 1+t^2

== Example of Complex Modulus ==
:$\cmod {\dfrac {1 + 2 i t - t^2} {1 + t^2} } = 1$
where:
:$t \in \R$ is a real number.
== Proof ==
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $4 \ \text{(iv)}$
Category:Examples of Complex Modulus
Complex Modulus/Examples/2 conj z1 - 3 conj z2 - 2

== Example of Complex Modulus ==
Let $z_1 = 4 - 3 i$ and $z_2 = -1 + 2 i$.
Then:
:$\cmod {2 \overline {z_1} - 3 \overline {z_2} - 2} = 15$
== Proof 1 ==
== Proof 2 ==
Category:Examples of Complex Modulus
Complex Modulus/Examples/2 conj z1 - 3 conj z2 - 2/Proof 1

== Example of Complex Modulus ==
== Proof ==
An illustration of the evaluation of $\cmod {2 \overline {z_1} - 3 \overline {z_2} - 2}$, where:
:$z_1 = 4 - 3 i$
:$z_2 = -1 + 2 i$
is given below:
:
The modulus is seen to be the radius of the circle.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $63 \ \text {(e)}$
Category:Examples of Complex Modulus
Complex Modulus/Examples/3iz - z^2

== Example of Complex Modulus ==
Let:
:$w = 3 i z - z^2$
where $z = x + i y$.
Then:
:$\cmod w^2 = x^4 + y^4 + 2 x^2 y^2 - 6 x^2 y - 6 y^3 + 9 x^2 + 9 y^2$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Fundamental Operations with Complex Numbers: $60$
Category:Examples of Complex Modulus
Complex Modulus/Examples/Tangent of Angle + i

== Example of Complex Modulus ==
:$\left\vert{\tan \theta + i}\right\vert = \left\vert{\sec \theta}\right\vert$
where:
:$\theta \in \R$ is a real number
:$\tan \theta$ denotes the tangent function
:$\sec \theta$ denotes the secant function.
== Proof ==
== Sources ==
* : $\S 1.2$. The Algebraic Theory: Examples
Category:Examples of Complex Modulus
Complex Modulus/Examples/z1 + z2

== Example of Complex Modulus ==
Let $z_1 = 4 - 3 i$ and $z_2 = -1 + 2 i$.
Then:
:$\cmod {z_1 + z_2} = \sqrt {10}$
== Proof 1 ==
== Proof 2 ==
Category:Examples of Complex Modulus
Complex Modulus/Examples/z1 + z2/Proof 1

== Example of Complex Modulus ==
== Proof ==
An illustration of the modulus of the sum of the complex numbers:
:$z_1 = 4 - 3 i$
:$z_2 = -1 + 2 i$
is given below:
:
The modulus is seen to be the radius of the circle.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $63 \ \text {(a)}$
Category:Examples of Complex Modulus
Complex Modulus/Examples/z1 - z2

== Example of Complex Modulus ==
Let $z_1 = 4 - 3 i$ and $z_2 = -1 + 2 i$.
Then:
:$\cmod {z_1 - z_2} = 5 \sqrt 2$
== Proof 1 ==
== Proof 2 ==
Category:Examples of Complex Modulus
Complex Modulus/Examples/z1 - z2/Proof 1

== Example of Complex Modulus ==
== Proof ==
An illustration of the modulus of the sum of the complex numbers:
:$z_1 = 4 - 3 i$
:$z_2 = -1 + 2 i$
is given below:
:
The modulus is seen to be the radius of the circle.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $63 \ \text {(b)}$
Category:Examples of Complex Modulus
Complex Modulus Function is Continuous

== Theorem ==
Let $z_0 \in \C$ be a complex number.
Then the complex modulus function is continuous at $z_0$.
== Proof ==
Let $\epsilon>0$.
Let $z \in \C$ be a complex number satisfying $\left\vert{z - z_0}\right\vert < \epsilon$.
By the Reverse Triangle Inequality:
: $\left\vert{ \left\vert{z}\right\vert - \left\vert{z_0}\right\vert }\right\vert \le \left\vert{z - z_0}\right\vert < \epsilon$
Hence the result, by the $\epsilon$-$\delta$ definition of continuity (taking $\delta = \epsilon$).
Category:Complex Modulus
Category:Continuous Functions
Complex Modulus equals Complex Modulus of Conjugate

== Theorem ==
Let $z \in \C$ be a complex number.
Let $\overline z$ denote the complex conjugate of $z$.
Let $\cmod z$ denote the modulus of $z$.
Then:
:$\cmod z = \cmod {\overline z}$
== Proof ==
Let $z = a + b i$.
Then:
and:
== Sources ==
* : $\S 1.2$. The Algebraic Theory
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Complex Conjugate of $z$: $3.7.8$
* : $1$ The complex plane: Complex numbers $\S 1.3$ Complex conjugation: $(3)$
Category:Complex Modulus
Category:Complex Conjugates
Complex Modulus equals Zero iff Zero

== Theorem ==
Let $z = a + i b$ be a complex number.
Let $\cmod z$ be the modulus of $z$.
Then:
:$\cmod z = 0 \iff z = 0$
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 1$. Complex Numbers: $\text {(ii)}$
* : $\S 1.2$. The Algebraic Theory
* : $1$ The complex plane: Complex numbers $\S 1.1$ Complex numbers and their representation
Category:Complex Modulus
Complex Modulus is Norm

== Theorem ==
The complex modulus is a norm on the set of complex numbers $\C$.
== Proof ==
We prove the norm axioms.
=== Positive Definiteness ===
This is demonstrated in:
: Complex Modulus equals Zero iff Zero
: Complex Modulus is Non-Negative
=== Multiplicativity ===
Follows from Modulus of Product.
=== Triangle Inequality ===
Follows from Triangle Inequality for Complex Numbers.
Category:Examples of Norms
Category:Complex Modulus
Complex Modulus of Additive Inverse

== Theorem ==
Let $z \in \C$ be a complex number.
Let $-z$ be the negative of $z$:
:$z + \paren {-z} = 0$
Then:
:$\cmod z = \cmod {\paren {-z} }$
where $\cmod z$ denotes the modulus of $z$.
== Proof ==
Let $z = a + i b$.
Category:Complex Modulus
Complex Modulus of Difference of Complex Numbers

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $\theta_1$ and $\theta_2$ be arguments of $z_1$ and $z_2$, respectively.
Then:
:$\cmod {z_1 - z_2}^2 = \cmod {z_1}^2 + \cmod {z_2}^2 - 2 \cmod {z_1} \cmod {z_2} \map \cos {\theta_1 - \theta_2}$
== Proof ==
By Complex Argument of Additive Inverse, $\theta_2 + \pi$ is an argument of $-z_2$.
We have:
== Also see ==
* Complex Modulus of Sum of Complex Numbers
Category:Complex Modulus
Complex Modulus of Measurable Function is Measurable

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\struct {\C, \map \BB \C}$ be the complex numbers made into a measurable space with its Borel $\sigma$-algebra.
Let $f : X \to \C$ be a $\Sigma/\map \BB \C$-measurable mapping.
Let $\struct {\R, \map \BB \R}$ be the real numbers made into a measurable space with its Borel $\sigma$-algebra.
Then $\cmod f : X \to \R$ is $\Sigma/\map \BB \R$-measurable.
== Proof ==
From Complex Modulus Function is Continuous, the mapping $\cmod {\, \cdot \,} : \C \to \R$ is continuous.
From Continuous Mapping is Measurable, $\cmod {\, \cdot \,} : \C \to \R$ is $\map \BB C/\map \BB \R$-measurable.
Hence from Composition of Measurable Mappings is Measurable, $\cmod f : X \to \R$ is $\Sigma/\map \BB \R$-measurable.
Category:Measurable Functions
Complex Modulus of Product of Complex Numbers

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $\cmod z$ be the modulus of $z$.
Then:
:$\cmod {z_1 z_2} = \cmod {z_1} \cdot \cmod {z_2}$
=== General Result ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Absolute Value Function is Completely Multiplicative
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 2$. Conjugate Complex Numbers
* : $\S 1.2$. The Algebraic Theory: $(1.11)$
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Multiplication and Division: $3.7.11$
* : Chapter $\text{II}$: Groups: Morphisms
* : $1$: Complex Numbers: Absolute Value: $1$
* : $1$ The complex plane: Complex numbers $\S 1.3$ Complex conjugation
Category:Complex Modulus
Category:Complex Multiplication
Category:Complex Modulus of Product of Complex Numbers
Complex Modulus of Product of Complex Numbers/General Result

== Theorem ==
Let $z_1, z_2, \ldots, z_n \in \C$ be complex numbers.
Let $\cmod z$ be the modulus of $z$.
Then:
: $\cmod {z_1 z_2 \cdots z_n} = \cmod {z_1} \cdot \cmod {z_2} \cdots \cmod {z_n}$
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
: $\cmod {z_1 z_2 \cdots z_n} = \cmod {z_1} \cdot \cmod {z_2} \cdots \cmod {z_n}$
$P \left({1}\right)$ is trivially true:
:$\cmod {z_1} = \cmod {z_1}$
=== Basis for the Induction ===
$P \left({2}\right)$ is the case:
: $\cmod {z_1 z_2} = \cmod {z_1} \cdot \cmod {z_2}$
which has been proved in Complex Modulus of Product of Complex Numbers.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.
So this is our induction hypothesis:
: $\cmod {z_1 z_2 \cdots z_k} = \cmod {z_1} \cdot \cmod {z_2} \cdots \cmod {z_k}$
Then we need to show:
: $\cmod {z_1 z_2 \cdots z_{k + 1} } = \cmod {z_1} \cdot \cmod {z_2} \cdots \cmod {z_{k + 1} }$
=== Induction Step ===
This is our induction step:
So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
: $\forall n \in \N_{>0}: \cmod {z_1 z_2 \cdots z_n} = \cmod {z_1} \cdot \cmod {z_2} \cdots \cmod {z_n}$
== Sources ==
* : $1$: Complex Numbers: Absolute Value: $1$
Category:Complex Modulus of Product of Complex Numbers
Complex Modulus of Product of Complex Numbers/Proof 1

== Theorem ==
== Proof ==
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$, where $x_1, y_1, x_2, y_2 \in \R$.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Fundamental Operations with Complex Numbers: $4 \ \text{(b)}$
Category:Complex Modulus of Product of Complex Numbers
Complex Modulus of Product of Complex Numbers/Proof 2

== Theorem ==
== Proof ==
Let $\overline z$ denote the complex conjugate of $z$. 
Then:
== Sources ==
* : $\S 1.2$. The Algebraic Theory: $(1.10)$
* : Chapter $\text{II}$: Groups: Subgroups
* : $1$: Complex Numbers: Solved Problems: Fundamental Operations with Complex Numbers: $4 \ \text{(b)}$
Category:Complex Modulus of Product of Complex Numbers
Complex Modulus of Product of Complex Numbers/Proof 3

== Theorem ==
== Proof ==
Let:
:$z_1 = r_1 \paren {\cos \theta_1 + i \sin \theta_1}$
:$z_2 = r_2 \paren {\cos \theta_2 + i \sin \theta_2}$
Then:
== Sources ==
* : $\S 2$. Geometrical Representations
Category:Complex Modulus of Product of Complex Numbers
Complex Modulus of Quotient of Complex Numbers

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers such that $z_2 \ne 0$.
Let $\cmod z$ denote the modulus of $z$.
Then:
:$\cmod {\dfrac {z_1} {z_2} } = \dfrac {\cmod {z_1} } {\cmod {z_2} }$
== Proof ==
== Examples ==
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $5$
* : $\S 2$. Geometrical Representations
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Multiplication and Division: $3.7.14$
* : $1$: Complex Numbers: Absolute Value: $2$
* : $1$: Complex Numbers: Supplementary Problems: Fundamental Operations with Complex Numbers: $56 \ \text{(b)}$
Category:Complex Modulus
Category:Complex Division
Complex Modulus of Quotient of Complex Numbers/Examples/(1+2i)^12 (1-2i)^-9

== Example of Complex Modulus of Quotient of Complex Numbers ==
:$\cmod {\dfrac {\paren {1 + 2 i}^{12} } {\paren {1 - 2 i}^9} } = 5 \sqrt 5$
== Proof ==
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $5$
Category:Examples of Complex Modulus
Complex Modulus of Real Number equals Absolute Value

== Theorem ==
Let $x \in \R$ be a real number.
Then the complex modulus of $x$ equals the absolute value of $x$.
== Proof ==
Let $x = x + 0 i \in \R$.
Then:
== Sources ==
* : $\S 1.2$. The Algebraic Theory
Category:Complex Modulus
Category:Absolute Value Function
Complex Modulus of Reciprocal of Complex Number

== Theorem ==
Let $z \in \C$ be a complex number such that $z \ne 0$.
Let $\cmod z$ denote the complex modulus of $z$.
Then:
:$\cmod {\dfrac 1 z} = \dfrac 1 {\cmod z}$
== Proof ==
Let $z = a + i b$.
== Examples ==
Category:Complex Modulus
Category:Reciprocals
Complex Modulus of Reciprocal of Complex Number/Examples/5+12i

== Example of Complex Modulus of Reciprocal of Complex Number ==
:$\left\vert{\dfrac 1 {5 + 12 i} }\right\rvert = \dfrac 1 {13}$
== Proof ==
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $4 \ \text{(iii)}$
Category:Examples of Complex Modulus
Complex Modulus of Sum of Complex Numbers

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $\theta_1$ and $\theta_2$ be arguments of $z_1$ and $z_2$, respectively.
Then:
:$\cmod {z_1 + z_2}^2 = \cmod {z_1}^2 + \cmod {z_2}^2 + 2 \cmod {z_1} \cmod {z_2} \, \map \cos {\theta_1 - \theta_2}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Complex Modulus of Difference of Complex Numbers
* Sum of Complex Numbers in Exponential Form
Category:Complex Modulus
Category:Complex Modulus of Sum of Complex Numbers
Complex Multiplication/Examples

== Examples of Complex Multiplication ==
=== Example: $\paren {1 + 2 i} \paren {3 + 4 i}$ ===
=== Example: $\paren {2 + 3 i} \paren {5 - 2 i}$ ===
=== Example: $\paren {3 + 2 i} \paren {2 - i}$ ===
=== Example: $\paren {4 + i} \paren {3 + 2 i} \paren {1 - i}$ ===
Category:Examples of Complex Multiplication
Complex Multiplication/Examples/((2 - i) (-3 + 2i)) (5 - 4i)

== Example of Complex Multiplication ==
:$\paren {\paren {2 - i} \paren {-3 + 2 i} } \paren {5 - 4 i} = 8 + 51 i$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Fundamental Operations with Complex Numbers: $1 \ \text{(i)}$
Category:Examples of Complex Multiplication
Complex Multiplication/Examples/(1+2i)(3+4i)

== Example of Complex Multiplication ==
:$\paren {1 + 2 i} \paren {3 + 4 i} = -5 + 10 i$
== Proof ==
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $1 \ \text{(i)}$
Category:Examples of Complex Multiplication
Complex Multiplication/Examples/(2 + 3i) (5 - 2i)

== Example of Complex Multiplication ==
:$\paren {2 + 3 i} \paren {5 - 2 i} = 16 + 11 i$
== Proof ==
== Sources ==
* : $\S 4$: Complex Numbers: Arithmetic of Complex Numbers: Example
Category:Examples of Complex Multiplication
Complex Multiplication/Examples/(2 - 3i) (4 + 2i)

== Example of Complex Multiplication ==
:$\paren {2 - 3 i} \paren {4 + 2 i} = 14 - 8 i$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Fundamental Operations with Complex Numbers: $1 \ \text {(f)}$
Category:Examples of Complex Multiplication
Complex Multiplication/Examples/(2 - i) ((-3 + 2i) (5 - 4i))

== Example of Complex Multiplication ==
:$\paren {2 - i} \paren {\paren {-3 + 2 i} \paren {5 - 4 i} } = 8 + 51 i$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Fundamental Operations with Complex Numbers: $1 \ \text{(h)}$
Category:Examples of Complex Multiplication
Complex Multiplication/Examples/(3+2i)(2-i)

== Example of Complex Multiplication ==
:$\paren {3 + 2 i} \paren {2 - i} = 8 + i$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Fundamental Operations with Complex Numbers: $53 \ \text {(c)}$
Category:Examples of Complex Multiplication
Complex Multiplication/Examples/(4+i)(3+2i)(1-i)

== Example of Complex Multiplication ==
:$\paren {4 + i} \paren {3 + 2 i} \paren {1 - i} = 21 + i$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Fundamental Operations with Complex Numbers: $53 \ \text {(f)}$
Category:Examples of Complex Multiplication
Complex Multiplication/Examples/(4 + 2i) (2 - 3i)

== Example of Complex Multiplication ==
:$\paren {4 + 2 i} \paren {2 - 3 i} = 14 - 8 i$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Fundamental Operations with Complex Numbers: $1 \ \text {(g)}$
Category:Examples of Complex Multiplication
Complex Multiplication Distributes over Addition

== Theorem ==
The operation of multiplication on the set of complex numbers $\C$ is distributive over the operation of addition.
:$\forall z_1, z_2, z_3 \in \C:$
::$z_1 \paren {z_2 + z_3} = z_1 z_2 + z_1 z_3$
::$\paren {z_2 + z_3} z_1 = z_2 z_1 + z_3 z_1$
== Proof ==
From the definition of complex numbers, we define the following:
where $x_1, x_2, x_3, y_1, y_2, y_3 \in \R$.
Thus:
The result $\paren {z_2 + z_3} z_1 = z_2 z_1 + z_3 z_1$ follows directly from the above, and the fact that Complex Multiplication is Commutative.
== Examples ==
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 1$. Complex Numbers: $\text {(iii)}$ The fundamental operations $\text {(c)}$
* : $1$: Complex Numbers: Axiomatic Foundations of the Complex Number System: $6$
* : $1$: Complex Numbers: Solved Problems: Axiomatic Foundations of Complex Numbers: $15$
* : $1$ The complex plane: Complex numbers $\S 1.2$ The algebraic structure of the complex numbers
Category:Complex Multiplication
Category:Complex Addition
Category:Examples of Distributive Operations
Complex Multiplication Distributes over Addition/Examples

== Examples of Use of Complex Multiplication Distributes over Addition ==
=== Example: $\paren {-1 + 2 i} \paren {\paren {7 - 5 i} + \paren {-3 + 4 i} } = \paren {-1 + 2 i} \paren {7 - 5 i} + \paren {-1 + 2 i} \paren {-3 + 4 i}$ ===
Category:Examples of Use of Complex Multiplication Distributes over Addition
Complex Multiplication Distributes over Addition/Examples/(2 - i) ((-3 + 2i) (5 - 4i))

== Examples of Use of Complex Multiplication Distributes over Addition ==
==== Example: $\paren {-1 + 2 i} \paren {\paren {7 - 5 i} + \paren {-3 + 4 i} }$ ====
==== Example: $\paren {-1 + 2 i} \paren {7 - 5 i} + \paren {-1 + 2 i} \paren {-3 + 4 i}$ ====
As can be seen:
:$\paren {-1 + 2 i} \paren {\paren {7 - 5 i} + \paren {-3 + 4 i} } = \paren {-1 + 2 i} \paren {7 - 5 i} + \paren {-1 + 2 i} \paren {-3 + 4 i}$
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Fundamental Operations with Complex Numbers: $1 \ \text{(j)}$
Category:Examples of Use of Complex Multiplication Distributes over Addition
Complex Multiplication Identity is One

== Theorem ==
Let $\C_{\ne 0}$ be the set of complex numbers without zero.
The identity element of $\struct {\C_{\ne 0}, \times}$ is the complex number $1 + 0 i$.
== Proof ==
and similarly:
== Sources ==
* : $\S 1.2$. The Algebraic Theory
* : Chapter $\text{II}$: Groups: Subgroups
* : $1$: Complex Numbers: Axiomatic Foundations of the Complex Number System: $7$
* : $1$ The complex plane: Complex numbers $\S 1.2$ The algebraic structure of the complex numbers
Category:Complex Multiplication
Category:Examples of Identity Elements
Complex Multiplication as Geometrical Transformation

== Theorem ==
Let $z_1 = \left\langle{r_1, \theta_1}\right\rangle$ and $z_2 = \left\langle{r_2, \theta_2}\right\rangle$ be complex numbers expressed in polar form.
Let $z_1$ and $z_2$ be represented on the complex plane $\C$ in vector form.
Let $z = z_1 z_2$ be the product of $z_1$ and $z_2$.
Then $z$ can be interpreted as the result of:
: rotating $z_1$ about the origin of $\C$ by $\theta_2$ in the positive direction
: multiplying the modulus of $z_1$ by $r_2$.
=== Corollary ===
== Proof ==
:
Let $z = r e^{i \alpha}$.
By Product of Complex Numbers in Exponential Form:
:$z = r_1 r_2 e^{i \left({\theta_1 + \theta_2}\right)}$
Adding $\theta_2$ to $\theta_1$ is equivalent to rotation about the origin of $\C$ by $\theta_2$ in the positive direction.
Similarly, the modulus of $z$ is obtained by multiplying the modulus of $z_1$ by $r_2$.
== Sources ==
* : $\S 2$. Geometrical Representations
Category:Complex Multiplication
Category:Polar Form of Complex Number
Category:Geometry of Complex Plane
Category:Complex Multiplication as Geometrical Transformation
Complex Multiplication as Geometrical Transformation/Corollary

== Corollary to Complex Multiplication as Geometrical Transformation ==
Let $z = \polar {r, \theta}$ be a complex number expressed in polar form.
Let $z$ be represented on the complex plane $\C$ in vector form.
The effect of multiplying $z$ by $e^{i \alpha}$ is to rotate it about the origin of $\C$ by $\alpha$ in the positive direction
== Proof ==
From Complex Multiplication as Geometrical Transformation, the effect of multiplying a complex number by $r e^{i \alpha}$ is:
:to rotate it about the origin of $\C$ by $\alpha$ in the positive direction
:to multiply its modulus by $r$.
In this instance $r = 1$.
Hence the result.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $24$
Category:Complex Multiplication as Geometrical Transformation