nyuuzyou/wikis huggingface dataset:
Modulo Multiplication is Well-Defined/Examples/2x3 equiv -6x15 mod 4

== Example of Use of Modulo Multiplication is Well-Defined ==
To confirm:
That is:
:$\eqclass 2 4 \eqclass 3 4 = \eqclass 2 4$
== Sources ==
* : Chapter $1$: Integral Domains: $\S 6$. The Residue Classes: Example $8$
Category:Examples of Modulo Arithmetic
Modulo Multiplication is Well-Defined/Proof 1

== Theorem ==
== Proof ==
We need to show that if:
:$\eqclass {x'} m = \eqclass x m$
and:
:$\eqclass {y'} m = \eqclass y m$
then:
:$\eqclass {x' y'} m = \eqclass {x y} m$
We have that:
:$\eqclass {x'} m = \eqclass x m$
and:
:$\eqclass {y'} m = \eqclass y m$
It follows from the definition of residue class modulo $m$ that:
:$x \equiv x' \pmod m$
and:
:$y \equiv y' \pmod m$
By definition, we have:
:$x \equiv x' \pmod m \implies \exists k_1 \in \Z: x = x' + k_1 m$
:$y \equiv y' \pmod m \implies \exists k_2 \in \Z: y = y' + k_2 m$
which gives us:
:$x y = \paren {x' + k_1 m} \paren {y' + k_2 m} = x' y' + \paren {x' k_2 + y' k_1} m + k_1 k_2 m^2$
Thus by definition:
:$x y \equiv \paren {x' y'} \pmod m$
Therefore, by the definition of residue class modulo $m$:
:$\eqclass {x' y'} m = \eqclass {x y} m$
== Examples ==
== Sources ==
* : $\S 1$: Some examples of groups
* : Chapter $1$: Integral Domains: $\S 6$. The Residue Classes: Theorem $4$
* : $\text {4-1}$ Basic Properties of Congruences: Theorem $\text {4-2}$
* : $\S 18.4$: Congruence classes
* : $\S 2.3$: Congruences
Category:Modulo Multiplication is Well-Defined
Modulo Multiplication is Well-Defined/Proof 2

== Theorem ==
== Proof ==
The equivalence class $\eqclass a m$ is defined as:
:$\eqclass a m = \set {x \in \Z: x = a + k m: k \in \Z}$
that is, the set of all integers which differ from $a$ by an integer multiple of $m$.
Thus the notation for multiplication of two residue classes modulo $z$ is not usually $\eqclass a m \times_m \eqclass b m$.
What is more normally seen is:
:$a b \pmod m$
Using this notation:
== Warning ==
This result does not hold when $a, b, x, y, m \notin \Z$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Example $11.2$
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $17$
Category:Modulo Multiplication is Well-Defined
Modulo Multiplication is Well-Defined/Warning

== Theorem ==
Let $z \in \R$ be a real number.
Let:
:$a \equiv b \pmod z$
and:
:$x \equiv y \pmod z$
where $a, b, x, y \in \R$.
Then it does not necessarily hold that:
:$a x \equiv b y \pmod z$
== Proof ==
But it is not necessarily the case that:
:$b k_2 + y k_1 + k_1 k_2 z$
is an integer.
In fact, $b k_2 + y k_1 + k_1 k_2 z$ can only be guaranteed to be an integer if each of $b, y, z \in \Z$.
Hence $a b$ is not necessarily congruent to $x y$.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $24$
Category:Modulo Multiplication
Modulo Multiplication on Reduced Residue System is Cancellable

== Theorem ==
Let $m \in \Z_{> 0}$ be a (strictly) positive integer.
Let $\Z'_m$ be the reduced residue system modulo $m$:
:$\Z'_m = \set {\eqclass k m \in \Z_m: k \perp m}$
Let $S = \struct {\Z'_m, \times_m}$ be the algebraic structure consisting of $\Z'_m$ under modulo multiplication.
Then $\times_m$ is cancellable, in the sense that:
:$\forall a, b, c \in \Z'_m: a \times_m c = b \times_m c \implies a = b$
and:
:$\forall a, b, c \in \Z'_m: c \times_m a = c \times_m b \implies a = b$
== Proof ==
Let $a, b, c \in \Z'_m$ such that $a \times_m c = b \times_m c$
Let $p, q, r$ be integers such that:
:$p \in a$
:$q \in b$
:$r \in c$
By definition of residue class, this means:
Thus as $p \in a$ and $q \in b$ it follows that $a = b$.
Hence the result.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(iii)}$
Category:Modulo Multiplication
Category:Reduced Residue Systems
Modulo Multiplication on Reduced Residue System is Closed

== Theorem ==
Let $m \in \Z_{> 0}$ be a (strictly) positive integer.
Let $\Z'_m$ be the reduced residue system modulo $m$:
:$\Z'_m = \set {\eqclass k m \in \Z_m: k \perp m}$
Let $S = \struct {\Z'_m, \times_m}$ be the algebraic structure consisting of $\Z'_m$ under the modulo multiplication.
Then $S$ is closed, in the sense that:
:$\forall a, b \in \Z'_m: a \times_m b \in \Z'_m$
== Proof ==
Let $\eqclass r m, \eqclass s m \in \Z'_m$.
Then by definition of reduced residue system:
:$r, s \perp m$
By Bézout's Identity:
:$\exists u_1, v_1 \in \Z: u_1 r + v_1 m = 1$
:$\exists u_2, v_2 \in \Z: u_2 s + v_2 m = 1$
Then:
So, again by Bézout's Identity, $r s$ is coprime to $m$.
So the product of two elements of $\struct {\Z'_m, \times_m}$ is again in $\struct {\Z'_m, \times_m}$.
That is, $\struct {\Z'_m, \times_m}$ is closed.
== Proof ==
* : Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(iii)}$: $(1.31)$
Category:Modulo Multiplication
Category:Reduced Residue Systems
Modulo Operation/Examples

== Examples of Use of Modulo Operation ==
=== Modulo Operation: $5 \bmod 3$ ===
=== Modulo Operation: $18 \bmod 3$ ===
=== Modulo Operation: $-2 \bmod 3$ ===
=== Modulo Operation: $100 \bmod 3$ ===
=== Modulo Operation: $100 \bmod 7$ ===
=== Modulo Operation: $-100 \bmod 7$ ===
=== Modulo Operation: $-100 \bmod 0$ ===
=== Modulo Operation: $5 \bmod -3$ ===
=== Modulo Operation: $18 \bmod -3$ ===
=== Modulo Operation: $-2 \bmod -3$ ===
=== Modulo Operation: $1 \cdotp 1 \bmod 1$ ===
=== Modulo Operation: $0 \cdotp 11 \bmod 0 \cdotp 1$ ===
=== Modulo Operation: $0 \cdotp 11 \bmod -0 \cdotp 1$ ===
=== Modulo Operation: $x \bmod 3 = 2$ and $x \bmod 5 = 3$ ===
Category:Examples of Modulo Operation
Modulo Operation/Examples/-100 mod 0

== Theorem ==
:$-100 \bmod 0 = -100$
where $\bmod$ denotes the modulo operation.
== Proof ==
By definition of modulo $0$:
:$\forall x \in \R: x \bmod 0 = x$
Hence:
:$-100 \bmod 0 = -100$
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $8$
Category:Examples of Modulo Operation
Modulo Operation/Examples/-100 mod 7

== Theorem ==
:$-100 \bmod 7 = 5$
where $\bmod$ denotes the modulo operation.
== Proof ==
By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {-100} 7 = -15 + \dfrac 5 7$
and so:
:$\floor {\dfrac {-100} 7} = -15$
Thus:
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $8$
Category:Examples of Modulo Operation
Modulo Operation/Examples/-2 mod -3

== Theorem ==
:$-2 \bmod -3 = -2$
where $\bmod$ denotes the modulo operation.
== Proof ==
By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {-2} {-3} = 0 + \dfrac 2 3$
and so:
:$\floor {\dfrac {-2} {-3} } = 0$
Thus:
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $9$
Category:Examples of Modulo Operation
Modulo Operation/Examples/-2 mod 3

== Theorem ==
:$-2 \bmod 3 = 1$
where $\bmod$ denotes the modulo operation.
== Proof ==
By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {-2} 3 = -1 + \dfrac 1 3$
and so:
:$\floor {\dfrac {-2} 3} = -1$
Thus:
:$-2 \bmod 3 = -2 - 3 \times \floor {\dfrac {-2} 3} = -2 - 3 \times \paren {-1} = 1$
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: $(3)$
Category:Examples of Modulo Operation
Modulo Operation/Examples/0.11 mod -0.1

== Theorem ==
:$0 \cdotp 11 \bmod -0 \cdotp 1 = -0 \cdotp 09$
where $\bmod$ denotes the modulo operation.
== Proof ==
By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
and so:
:$\floor {\dfrac {0 \cdotp 11} {-0 \cdotp 1} } = -2$
Thus:
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $10$
Category:Examples of Modulo Operation
Modulo Operation/Examples/0.11 mod 0.1

== Theorem ==
:$0 \cdotp 11 \bmod 0 \cdotp 1 = 0 \cdotp 01$
where $\bmod$ denotes the modulo operation.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $10$
Category:Examples of Modulo Operation
Modulo Operation/Examples/0.11 mod 0.1/Proof 1

== Theorem ==
== Proof ==
By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
and so:
:$\floor {\dfrac {0 \cdotp 11} {0 \cdotp 1} } = 1$
Thus:
Category:Examples of Modulo Operation
Modulo Operation/Examples/0.11 mod 0.1/Proof 2

== Theorem ==
== Proof ==
From Modulo Operation: $1 \cdotp 1 \bmod 1$:
:$1 \cdotp 1 \bmod 1 = 0 \cdotp 1$
From Product Distributes over Modulo Operation:
:$z \left({x \bmod y}\right) = \left({z x}\right) \bmod \left({z y}\right)$
and so:
Category:Examples of Modulo Operation
Modulo Operation/Examples/1.1 mod 1

== Theorem ==
:$1 \cdotp 1 \bmod 1 = 0 \cdotp 1$
where $\bmod$ denotes the modulo operation.
== Proof ==
By definition of modulo $1$:
:$x \bmod 1 = x - \floor x$
Thus:
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $10$
Category:Examples of Modulo Operation
Modulo Operation/Examples/100 mod 3

== Theorem ==
:$100 \bmod 3 = 1$
where $\bmod$ denotes the modulo operation.
== Proof ==
By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {100} 3 = 33 + \dfrac 1 3$
and so:
:$\floor {\dfrac {100} 3} = 33$
Thus:
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $8$
Category:Examples of Modulo Operation
Modulo Operation/Examples/100 mod 7

== Theorem ==
:$100 \bmod 7 = 2$
where $\bmod$ denotes the modulo operation.
== Proof ==
By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {100} 7 = 14 + \dfrac 2 7$
and so:
:$\floor {\dfrac {100} 7} = 14$
Thus:
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $8$
Category:Examples of Modulo Operation
Modulo Operation/Examples/18 mod -3

== Theorem ==
:$18 \bmod -3 = 0$
where $\bmod$ denotes the modulo operation.
== Proof ==
By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {18} {-3} = -6 + \dfrac 0 3$
and so:
:$\floor {\dfrac {18} {-3} } = -6$
Thus:
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $9$
Category:Examples of Modulo Operation
Modulo Operation/Examples/18 mod 3

== Theorem ==
:$18 \bmod 3 = 0$
where $\bmod$ denotes the modulo operation.
== Proof ==
By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {18} 3 = 6 + \dfrac 0 3$
and so:
:$\floor {\dfrac {18} 3} = 6$
Thus:
:$18 \bmod 3 = 18 - 3 \times \floor {\dfrac {18} 3} = 18 - 3 \times 6 = 0$
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: $(3)$
Category:Examples of Modulo Operation
Modulo Operation/Examples/5 mod -3

== Theorem ==
:$5 \bmod -3 = -1$
where $\bmod$ denotes the modulo operation.
== Proof ==
By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac 5 {-3} = -2 + \dfrac 1 3$
and so:
:$\floor {\dfrac 5 {-3} } = -2$
Thus:
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $9$
Category:Examples of Modulo Operation
Modulo Operation/Examples/5 mod 3

== Theorem ==
:$5 \bmod 3 = 2$
where $\bmod$ denotes the modulo operation.
== Proof ==
By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac 5 3 = 1 + \dfrac 2 3$
and so:
:$\floor {\dfrac 5 3} = 1$
Thus:
:$5 \bmod 3 = 5 - 3 \times \floor {\dfrac 5 3} = 5 - 3 \times 1 = 2$
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: $(3)$
Category:Examples of Modulo Operation
Modulo Operation/Examples/x mod 3 = 2 and x mod 5 = 3

== Example of use of Modulo Operation ==
Let $x \bmod 3 = 2$ and $x \bmod 5 = 3$.
Then:
:$x \bmod 15 = 8$
== Proof ==
Hence by definition of congruence by the modulo operation:
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $14$
Category:Examples of Modulo Operation
Modulo Operation as Integer Difference by Quotient

== Theorem ==
Let $x, y, z \in \R$ be real numbers.
Let $y > 0$.
Let $0 \le z < y$.
Let:
:$\dfrac {x - z} y = k$
for some integer $k$.
Then:
:$z = x \bmod y$
where $\bmod$ denotes the modulo operation.
== Proof ==
We have:
We also have:
:$0 \le z < y$
Hence:
:$0 \le \dfrac z y < 1$
and so by definition of floor function:
:$(2): \quad \floor {\dfrac z y} = 0$
Thus:
The result follows.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $16$
Category:Modulo Arithmetic
Modulo Subtraction is Well-Defined

== Corollary to Modulo Addition is Well-Defined ==
Let $m \in \Z$ be an integer.
Let $\Z_m$ be the set of integers modulo $m$.
The modulo subtraction operation on $\Z_m$, defined by the rule:
:$\eqclass a m -_m \eqclass b m = \eqclass {a - b} m$
is a well-defined operation.
That is:
:If $a \equiv b \pmod m$ and $x \equiv y \pmod m$, then $a - x \equiv b - y \pmod m$.
== Proof ==
We have:
The result follows from the fact that Modulo Addition is Well-Defined for all integers.
== Examples ==
== Sources ==
* : $\S 2.6$. Algebra of congruences: $\text{(ii)}$
* : $\text {4-1}$ Basic Properties of Congruences: Theorem $\text {4-2}$
* : $\S 14.3 \ \text {(ii)}$: Congruence modulo $m$ ($m \in \N$)
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Law $\text{A}$
* 
Category:Modulo Subtraction
Category:Modulo Addition
Category:Modulo Addition is Well-Defined
Category:Examples of Well-Defined Mappings‎
Modulus 1 Rational Argument Complex Numbers under Multiplication form Infinite Abelian Group

== Theorem ==
Let $S$ be the set defined as:
:$S = \set {\cos \theta + i \sin \theta: \theta \in \Q}$
Then the algebraic structure $\struct {S, \times}$ is an infinite abelian group.
== Proof ==
By definition of polar form of complex numbers, the elements of $S$ are also elements of the circle group $\struct {K, \times}$:
:$K = \set {z \in \C: \cmod z = 1}$
$S$ is infinite by construction.
Thus $S \subseteq C$ and trivially $S \ne \O$.
Let $a, b \in S$.
Then:
:$a = \cos \theta_1 + i \sin \theta_1$
and:
:$b = \cos \theta_2 + i \sin \theta_2$
for some $\theta_1, \theta_2 \in \Q$.
We have that:
and:
Hence by the Two-Step Subgroup Test, $\struct {S, \times}$ is a subgroup of $\struct {K, \times}$.
It has been established that $S$ is an infinite set.
Hence by definition $\struct {S, \times}$ is an infinite group.
Finally, from Subgroup of Abelian Group is Abelian, $\struct {S, \times}$ is an abelian group.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: Examples: $(1) \ \text {(c)}$
Category:Examples of Abelian Groups
Category:Examples of Infinite Groups
Modulus Larger than Imaginary Part

== Theorem ==
Let $z \in \C$ be a complex number.
Then the modulus of $z$ is larger than the imaginary part $\map \Im z$ of $z$: 
:$\cmod z \ge \size {\map \Im z}$
== Proof ==
By the definition of a complex number, we have:
:$z = \map \Re z + i \map \Im z$
== Also see ==
* Modulus Larger than Real Part
== Sources ==
* : $1$ The complex plane: Complex numbers $\S 1.4$ Inequalities: $(1)$
Category:Complex Modulus
Modulus Larger than Real Part

== Theorem ==
Let $z \in \C$ be a complex number.
Then the modulus of $z$ is larger than the real part $\map \Re z$ of $z$: 
:$\cmod z \ge \size {\map \Re z}$
== Proof ==
By the definition of a complex number, we have:
:$z = \map \Re z + i \map \Im z$
Then:
== Also see ==
* Modulus Larger than Imaginary Part
== Sources ==
* : $1$ The complex plane: Complex numbers $\S 1.4$ Inequalities: $(1)$
Category:Complex Modulus
Modulus Larger than Real Part and Imaginary Part

== Theorem ==
Let $z \in \C$ be a complex number.
Let $\operatorname{Re} \left({z}\right)$ denote the real part of $z$, and $\operatorname{Im} \left({z}\right) $ the imaginary part of $z$.
Then:
=== Modulus Larger than Real Part ===
=== Modulus Larger than Imaginary Part ===
Category:Complex Modulus
Modulus and Argument of Complex Exponential

== Theorem ==
Let $z \in \C$ be a complex number.
Let $\hointr a {a + 2 \pi}$ be a half open interval of length $2 \pi$.
Let $r \in \hointr 0 {+\infty}$ and $\theta \in \hointr a {a + 2 \pi}$.
Then:
:$r = \cmod z$ and $\theta = \map \arg z$
:
:$z = r e^{i \theta}$
where:
:$\cmod z$ denotes the modulus of $z$
:$\map \arg z$ denotes the argument of $z$
:$x \mapsto e^x$ is the complex exponential function.
If $z = 0$ or $r = 0$, then $\theta$ may be any number in $\hointr a {a + 2 \pi}$.
== Proof ==
=== Necessary condition ===
Let $r = \cmod z$.
If $z = 0$, we have:
:$z = 0e^{i \theta} = re^{i \theta}$
Suppose $z \ne 0$ and $\theta = \map \arg z$.
By definition of argument, the following two equations hold:
:$(1): \quad \dfrac {\map \Re z} r = \cos \theta$
:$(2): \quad \dfrac {\map \Im z} r = \sin \theta$
where:
:$\map \Re z$ denotes the real part of $z$
:$\map \Im z$ denotes the imaginary part of $z$.
Then:
=== Sufficient condition ===
Let $z = re^{i \theta}$.
From the equations above, we find:
:$\map \Re {re^{i \theta} } = r \cos \theta$
:$\map \Im {re^{i \theta} } = r \sin \theta$
Then:
If $r \ne 0$, we find $\map \arg {r e^{i \theta} }$ by solving the two equations by definition of argument:
:$(1): \quad \dfrac {r \cos \theta} r = \map \cos {\map \arg {r e^{i \theta} } }$
:$(2): \quad \dfrac {r \sin \theta} r = \map \sin {\map \arg {r e^{i \theta} } }$
We find:
:$\map \arg {r e^{i \theta} } = \theta$
== Sources ==
* $\S 1.5$
Category:Complex Analysis
Category:Complex Modulus
Modulus in Terms of Conjugate

== Theorem ==
Let $z = a + i b$ be a complex number.
Let $\cmod z$ be the modulus of $z$.
Let $\overline z$ be the conjugate of $z$.
Then:
:$\cmod z = \sqrt {z \overline z}$
== Proof ==
Let $z = a + i b$.
Then:
== Sources ==
* : $\S 1.2$. The Algebraic Theory: $(1.9)$
* : Chapter $\text{II}$: Groups: Subgroups
* : $\S 1.2.2$: Numbers, Powers, and Logarithms
* : Chapter $2$: Mathematical Background: $2.1$ The Complex Field $C$
* : $\S 4$: Complex Numbers: Complex Plane: $4.6.$
* 
Category:Complex Conjugates
Category:Complex Modulus
Modulus of Change of Base of Logarithm/Examples/Base 10 to Base e

== Example of Modulus of Change of Base of Logarithm ==
Let $\log_{10} x$ be the common (base $10$) logarithm of $x$.
Let $\log_e x$ be the natural (base $e$) logarithm of $x$.
Consider the change from base $10$ to base $e$:
Then $\log_e 10 = 2 \cdotp 30258 \, 5 \ldots$ is the modulus.
== Sources ==
* 
* 
Category:Examples of Change of Base of Logarithm
Modulus of Change of Base of Logarithm/Examples/Base e to Base 10

== Example of Modulus of Change of Base of Logarithm ==
Let $\log_{10} x$ be the common (base $10$) logarithm of $x$.
Let $\log_e x$ be the natural (base $e$) logarithm of $x$.
Consider the change from base $e$ to base $10$:
Then $\log_{10} e = 0 \cdotp 43429 \ldots$ is the modulus.
== Sources ==
* 
* 
Category:Examples of Change of Base of Logarithm
Modulus of Complex Number equals its Distance from Origin

== Theorem ==
The modulus of a complex number equals its distance from the origin on the complex plane.
== Proof ==
Let $z = x + y i$ be a complex number and $O = 0 + 0 i$ be the origin on the complex plane.
We have its modulus:
and its distance from the origin on the complex plane:
The two are seen to be equal.
Category:Complex Analysis
Modulus of Complex Root of Unity equals 1

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer such that $n$ is even.
Let $U_n = \set {z \in \C: z^n = 1}$ be the set of complex $n$th roots of unity.
Let $z \in U_n$.
Then:
:$\cmod z = 1$
where $\cmod z$ denotes the modulus of $z$.
== Proof ==
== Sources ==
* : $\S 3$. Roots of Unity
* : Chapter $2$: Subgroups and Cosets: $\S 44$
Category:Complex Roots of Unity
Modulus of Eigenvalue of Bounded Linear Operator is Bounded Above by Operator Norm

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $T : X \to X$ be a bounded linear operator.
Let $\lambda$ be a eigenvalue of $T$. 
Then: 
:$\cmod \lambda \le \norm T$
where $\norm T$ denotes the norm of $T$.
== Proof ==
Since $\lambda$ is an eigenvalue of $T$, there exists $x \ne 0$ such that: 
:$T x = \lambda x$ 
Then we have: 
while: 
:$\ds \norm {\frac x {\norm x} } = 1$
So we have: 
:$\cmod \lambda \in \set {\norm {T x} : \norm x = 1}$
So, from the definition of supremum and of the norm of a bounded linear transformation, we have:
:$\cmod \lambda \le \norm T$
== Sources ==
* $14.1$: The Resolvent and Spectrum
Category:Eigenvalues of Linear Operators
Category:Bounded Linear Operators
Modulus of Exponential is Exponential of Real Part

== Theorem ==
Let $z \in \C$ be a complex number.
Let $\exp z$ denote the complex exponential function.
Let $\cmod {\, \cdot \,}$ denote the complex modulus
Then:
:$\cmod {\exp z} = \map \exp {\map \Re z}$
where $\map \Re z$ denotes the real part of $z$.
== Proof ==
Let $z = x + iy$.
== Sources ==
* : $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$
Category:Complex Modulus
Category:Exponential Function
Modulus of Exponential of Imaginary Number is One

== Theorem ==
Let $\cmod z$ denote the modulus of a complex number $z$.
Let $e^z$ be the complex exponential of $z$.
Let $x$ be wholly real.
Then:
:$\cmod {e^{i x} } = 1$
=== Corollary ===
== Proof ==
Category:Complex Modulus
Category:Modulus of Exponential of Imaginary Number is One
Modulus of Exponential of Imaginary Number is One/Corollary

== Corollary to Modulus of Exponential of Imaginary Number is One ==
Let $t > 0$ be wholly real.
Let $t^{i x}$ be $t$ to the power of $i x$ defined on its principal branch.
Then:
:$\cmod {t^{ix} } = 1$
== Proof ==
Category:Modulus of Exponential of Imaginary Number is One
Modulus of Exponential of i z where z is on Circle

== Theorem ==
Let $C$ be the circle embedded in the complex plane given by the equation:
:$z = R e^{i \theta}$
Then:
:$\cmod {e^{i z} } = e^{-R \sin \theta}$
== Proof ==
== Examples ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $87$
Category:Complex Modulus
Category:Exponential Function
Category:Circles
Category:Modulus of Exponential of i z where z is on Circle
Modulus of Exponential of i z where z is on Circle/Examples/6 e^pi i over 3

== Examples of Use of Modulus of Exponential of i z where z is on Circle ==
Let $z = 6 e^{\pi i / 3}$
Then:
:$\cmod {e^{i z} } = e^{-3 \sqrt 3}$
== Proof ==
From Modulus of Exponential of i z where z is on Circle:
:$(1): \quad \cmod {e^{i z} } = e^{-R \sin \theta}$
where:
:$z = R e^{i \theta}$
Here we have:
:$R = 6$
:$\theta = \dfrac \pi 3$
Hence:
:$\sin \theta = \dfrac {\sqrt 3} 2$
and the result follows by substituting for $R$ and $\sin \theta$ in $(1)$.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $131$
Category:Modulus of Exponential of i z where z is on Circle
Modulus of Gamma Function of Imaginary Number

== Theorem == 
Let $t \in \R$ be a real number.
Then: 
:$\cmod {\map \Gamma {i t} } = \sqrt {\dfrac {\pi \csch \pi t} t}$
where:
:$\Gamma$ is the Gamma function
:$\csch$ is the hyperbolic cosecant function.
== Proof == 
By Euler's Reflection Formula:
:$\map \Gamma {i t} \, \map \Gamma {1 - i t} = \pi \, \map \csc {\pi i t}$
From Gamma Difference Equation: 
:$-i t \, \map \Gamma {i t} \, \map \Gamma {-i t} = \pi \, \map \csc {\pi i t}$
Then:
and:
So:
:$\cmod {\map \Gamma {i t} }^2 = \dfrac {\pi \, \map \csch {\pi \cmod t} } {\cmod t}$
As $\cmod z \ge 0$ for all complex numbers $z$, we can take the non-negative square root and write:
:$\cmod {\map \Gamma {i t} } = \sqrt {\dfrac {\pi \, \map \csch {\pi \cmod t} } {\cmod t} }$
However, by Hyperbolic Sine Function is Odd: 
:$\dfrac {\pi \, \map \csch {-\pi t} } {-t} = \dfrac {-\pi \, \map \csch {\pi t} } {-t} = \dfrac {\pi \, \map \csch {\pi t} } t$
Hence we can remove the modulus and simply write: 
:$\cmod {\map \Gamma {i t} } = \sqrt {\dfrac {\pi \csch \pi t} t}$
== Also presented as ==
== Sources ==
* : $15$: Miscellaneous Examples
* : $\S 16$: Miscellaneous Results: $16.17$
* : $\S 25$: The Gamma Function: Miscellaneous Results: $25.17.$
Category:Modulus of Gamma Function of Imaginary Number
Category:Gamma Function
Category:Hyperbolic Cosecant Function