nyuuzyou/wikis huggingface dataset:
Complex Conjugate Coordinates/Examples/2x + y = 5

== Example of Complex Conjugate Coordinates ==
The equation of the straight line in the plane:
:$2 x + y = 5$
can be expressed in complex conjugate coordinates as:
:$\paren {2 i + 1} z + \paren {2 i - 1} \overline z = 10 i$
== Proof ==
We have that:
and so:
Hence:
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Complex Conjugate Coordinates: $43 \ \text {(a)}$
Category:Examples of Complex Conjugate Coordinates
Complex Conjugate Coordinates/Examples/2x - 3y = 5

== Example of Complex Conjugate Coordinates ==
The equation of the straight line in the plane:
:$2 x - 3 y = 5$
can be expressed in complex conjugate coordinates as:
:$\paren {2 i - 3} z + \paren {2 i + 3} \overline z = 10 i$
== Proof ==
We have that:
and so:
Hence:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Conjugate Coordinates: $117 \ \text{(b)}$
Category:Examples of Complex Conjugate Coordinates
Complex Conjugate Coordinates/Examples/4x^2 + 16y^2 = 25

== Example of Complex Conjugate Coordinates ==
The equation of the ellipse:
:$4 x^2 + 16 y^2 = 25$
can be expressed in complex conjugate coordinates as:
:$3 \paren {z^2 + \overline z^2} - 10 z \overline z + 25 = 0$
== Proof ==
This can be written in the form:
:$a \paren {x^2 + y^2} + 2 b \paren {x^2 - y^2} = 25$
where:
This is because:
So:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Conjugate Coordinates: $117 \ \text{(c)}$
Category:Examples of Complex Conjugate Coordinates
Complex Conjugate Coordinates/Examples/x^2 + y^2 = 36

== Example of Complex Conjugate Coordinates ==
The equation of the circle:
:$x^2 + y^2 = 36$
can be expressed in complex conjugate coordinates as:
:$z \overline z = 36$
== Proof 1 ==
== Proof 2 ==
Category:Examples of Complex Conjugate Coordinates
Complex Conjugate Coordinates/Examples/x^2 + y^2 = 36/Proof 2

== Example of Complex Conjugate Coordinates ==
== Proof ==
We have that:
and so:
Hence:
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Complex Conjugate Coordinates: $43 \ \text {(b)}$
Category:Examples of Complex Conjugate Coordinates
Complex Conjugate of Complex-Valued Function Vanishing at Infinity Vanishes at Infinity

== Theorem ==
Let $X$ be a locally compact Hausdorff space. 
Let $f : X \to \C$ be a complex-valued function vanishing at infinity.
Define $\overline f : X \to \C$ by:
:$\map {\overline f} x = \overline {\map f x}$
for each $x \in X$. 
Then $\overline f$ is a complex-valued function vanishing at infinity.
== Proof ==
Let $\epsilon > 0$.
Then since $f$ vanishes at infinity, there exists a compact set $K \subseteq X$ such that:
:$\cmod {\map f x} < \epsilon$ for all $x \in X \setminus K$. 
From Complex Modulus equals Complex Modulus of Conjugate, we then have:
:$\cmod {\overline {\map f x} } < \epsilon$ for all $x \in X \setminus K$. 
Since $\epsilon$ was arbitrary: 
:$\overline f$ is a complex-valued function vanishing at infinity.
Category:Complex-Valued Functions Vanishing at Infinity
Complex Conjugate of Gamma Function

== Theorem ==
Let $\Gamma$ denote the gamma function.
Then:
:$\forall z \in \C \setminus \set {0, -1, -2, \ldots}: \map \Gamma {\overline z} = \overline {\map \Gamma z}$
== Proof ==
This is immediate from, say, the Euler form of $\Gamma$ and the fact that complex conjugation preserves products and sums.
Category:Gamma Function
Category:Complex Conjugates
Complex Conjugation is Automorphism

== Theorem ==
Consider the field of complex numbers $\C$.
The operation of complex conjugation:
:$\forall z \in \C: z \mapsto \overline z$
is a field automorphism.
== Proof ==
Let:
:$z_1 = x_1 + i y_1$
and:
:$z_2 = x_2 + i y_2$.
Let us define the mapping $\phi: \C \to \C$ defined as:
:$\forall z \in \C: \map \phi z = \overline z$
We check that $\phi$ has the morphism property:
By Sum of Complex Conjugates:
:$\map \phi {z_1 + z_2} = \map \phi {z_1} + \map \phi {z_2}$
By Product of Complex Conjugates:
:$\map \phi {z_1 z_2} = \map \phi {z_1} \map \phi {z_2}$
So the morphism property holds for both complex addition and complex multiplication.
Hence we can say that complex conjugation is a field homomorphism.
We note that:
:$\overline z_1 = \overline z_2 \implies z_1 = z_2$
and so complex conjugation is injective.
Also, complex conjugation is trivially surjective, and hence bijective.
The result then follows by definition of field automorphism.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms: Example $5$
* 
* 
Category:Complex Conjugates
Category:Field Automorphisms
Complex Conjugation is Involution

== Theorem ==
Let $z = x + i y$ be a complex number.
Let $\overline z$ denote the complex conjugate of $z$.
Then the operation of complex conjugation is an involution:
:$\overline {\paren {\overline z} } = z$
== Proof ==
== Sources ==
* : $1$ The complex plane: Complex numbers $\S 1.3$ Complex conjugation: $(1)$
* 
Category:Complex Conjugates
Category:Involutions
Complex Conjugation is not Linear Mapping

== Theorem ==
Let $\overline \cdot: \C \to \C: z \mapsto \overline z$ be the complex conjugation over the field of complex numbers.
Then complex conjugation is not a linear mapping.
== Proof ==
By definition, it is not a linear mapping.
== Sources ==
* : Chapter $\S 2.1$: Continuous and linear maps. Linear transformations
Category:Complex Conjugates
Category:Linear Transformations
Complex Contour Integral as Contour Integrals

== Theorem ==
Let $f: D \to \C$ be a complex-differentiable function, where $D \subseteq \C$ is a connected domain.
Let $u, v: \R^2 \to \R$ be defined by:
:$\map f {x + i y} = \map u {x, y} + i \map v {x, y}$
Let $C$ be a contour in $D$.
Let $\phi: \R^2 \to \C$ be defined by:
:$\map \phi {x, y} = x + i y$
Then there exists a piecewise continuously differentiable function $\gamma: \closedint a b \to \R^2$ such that:
:$\ds \int_C \map f z \rd z = \ds \int_\gamma \tuple {u, -v} \cdot \rd \mathbf l + i \int_\gamma \tuple {v, u} \cdot \rd \mathbf l$
where the integral on the is a complex contour integral, and the two integrals on the are general contour integrals.
The function $\phi \circ \gamma : \closedint a b \to \C$ is a parameterization of $C$.
== Proof ==
First, suppose that $C$ consists of one directed smooth curve $C_1$.
Let $\gamma_1 : \closedint a b \to D$ be a smooth path that is a parameterization of $C_1$.
Define $x, y: \closedint a b \to \R$ by:
:$\map {\gamma_1} t = \map x t + i \map y t$
Then:
where $\gamma: \closedint a b \to \R^2$ is defined by:
:$\map \gamma t = \tuple {\map x t, \map y t}$
By definition of smooth path, $\gamma$ is continuously differentiable.
In the general case, $C$ is a concatenation of $n$ directed smooth curves $C_1, \ldots, C_n$.
Reparameterization of Directed Smooth Curve with Given Domain shows that we can find a parameterization $\gamma_k : \closedint {\dfrac {k - 1} n} {\dfrac k n} \to D$ of $C_k$ for all $k \in \set {1, \ldots, n}$.
Define $x_k, y_k: \closedint {\dfrac {k - 1} n} {\dfrac k n} \to \R$ by:
:$\map {\gamma_k} t = \map {x_k} t + i \map {y_k} t$
Then:
where $\gamma : \closedint 0 1 \to \R^2$ is defined by:
:$\map \gamma t = \tuple {\map {x_k} t, \map {y_k} t}$ for all $ t \in \closedint {\dfrac {k - 1} n} {\dfrac k n}$
Pasting Lemma for Finite Union of Closed Sets shows that $\gamma$ is continuous.
By definition of smooth path, $\gamma$ is continuously differentiable in the intervals $\openint {\dfrac {k - 1} n} {\dfrac k n}$ for all $k \in \set {1, \ldots, n}$.
It follows that $\gamma$ is piecewiese continuously differentiable.
As:
:$\map \phi {\map {x_k} t, \map {y_k} t } = \map {x_k} t + i \map {y_k} t = \map {\gamma_k} t$
it follows that $\phi \circ \gamma {\restriction_{ \closedint {\paren {k - 1} / n} {k / n} } }$ is a parameterization of $C_k$.
By definition of parameterization of contour, it follows that $\phi \circ \gamma$ is a parameterization of $C$.
== Sources ==
* : $\S 2.2$
Category:Complex Contour Integrals
Category:Contour Integrals
Complex Cosine Function/Examples/4 cos z = 3+i

== Example of Complex Cosine Function ==
Let:
: $4 \cos z = 3 + i$
Then:
:$z = \dfrac {\paren {8 n + 1} \pi} 4 - \dfrac {i \ln 2} 2$ for $n \in \Z$
or:
:$z = \dfrac {\paren {8 m - 1} i \pi} 4 + \dfrac {i \ln 2} 2$ for $m \in \Z$
== Proof ==
Thus we have:
and:
== Sources ==
* : $\S 4.6$. The Logarithm: Examples: $\text {(iii)}$
Category:Examples of Cosine Function
Complex Cosine Function is Entire/Proof 1

== Theorem ==
== Proof ==
By the definition of the complex cosine function, $\cos$ admits a power series expansion about $0$: 
:$\ds \cos z = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!}$
By Complex Function is Entire iff it has Everywhere Convergent Power Series, to show that $\cos$ is entire it suffices to show that this series is everywhere convergent.
From Radius of Convergence from Limit of Sequence: Complex Case, it is sufficient to show that: 
:$\ds \lim_{n \mathop \to \infty} \size {\frac {\paren {-1}^{n + 1} } {\paren {2 n + 2}!} \times \frac {\paren {2 n}!} {\paren {-1}^n} } = 0$
We have: 
Hence the result.
Category:Complex Cosine Function is Entire
Complex Cosine Function is Entire/Proof 2

== Theorem ==
== Proof ==
Let: 
for all $z \in \C$. 
By Complex Exponential Function is Entire, we have that $f$ is entire. 
By Polynomial is Entire, we have that $g$ and $h$ are entire. 
Therefore, by Composition of Entire Functions is Entire, we have that both $f \circ g$ and $f \circ h$ are entire.
By Linear Combination of Entire Functions is Entire, we then have that: 
:$\dfrac 1 2 \paren {f \circ g + f \circ h}$
is entire.
Then: 
Therefore $\cos$ is an entire function.
Category:Complex Cosine Function is Entire
Complex Cosine Function is Unbounded/Proof 1

== Theorem ==
== Proof ==
Let $K \in \R_{>0}$ be an arbitrary real number.
Let $p = \ln {2 K}$.
Let $z = i p$, where $i$ denotes the imaginary unit.
Then:
Thus for any $K$ we can find $z \in \C$ such that $\cmod {\cos z} > K$.
Hence the result by definition of unbounded.
== Sources ==
* : $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$
Category:Complex Cosine Function is Unbounded
Complex Cosine Function is Unbounded/Proof 2

== Theorem ==
== Proof ==
By Complex Cosine Function is Entire, we have that $\cos$ is an entire function. 
 that $\cos$ is a bounded function.
By Liouville's Theorem, we have that $\cos$ is a constant function.
However, by Cosine of Zero is One: 
:$\cos 0 = 1$
and by Cosine of Right Angle: 
:$\cos \dfrac \pi 2 = 0$
Therefore, $\cos$ is not a constant function.
This contradicts the statement that $\cos$ is a constant function.
We hence conclude, by Proof by Contradiction, that our assumption that $\cos$ is a bounded function is false.
Hence $\cos$ is unbounded.
Category:Complex Cosine Function is Unbounded
Complex Cross Product/Examples/2+5i cross 3-i

== Example of Complex Cross Product ==
Let:
:$z_1 = 2 + 5 i$
:$z_2 = 3 - i$
Then:
:$z_1 \times z_2 = -17$
where $\times$ denotes (complex) cross product.
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $110 \ \text {(b)}$
Category:Examples of Complex Cross Product
Complex Cross Product/Examples/3-4i cross -4+3i

== Example of Complex Cross Product ==
Let:
:$z_1 = 3 - 4 i$
:$z_2 = -4 + 3 i$
Then:
:$z_1 \times z_2 = -7$
where $\times$ denotes (complex) cross product.
== Proof 1 ==
== Proof 2 ==
Category:Examples of Complex Cross Product
Complex Cross Product/Examples/3-i cross 2+5i

== Example of Complex Cross Product ==
Let:
:$z_1 = 3 - i$
:$z_2 = 2 + 5 i$
Then:
:$z_1 \times z_2 = 17$
where $\times$ denotes (complex) cross product.
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $110 \ \text {(d)}$
Category:Examples of Complex Cross Product
Complex Cross Product/Examples/Size of 2+5i cross 3-i

== Example of Complex Cross Product ==
Let:
:$z_1 = 2 + 5 i$
:$z_2 = 3 - i$
Then:
:$\size {z_1 \times z_2} = 17$
where:
:$\circ$ denotes (complex) cross product
:$\size {\, \cdot \,}$ denotes the absolute value function.
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $110 \ \text {(f)}$
Category:Examples of Complex Cross Product
Complex Cross Product/Examples/Size of 3-i cross 2+5i

== Example of Complex Cross Product ==
Let:
:$z_1 = 3 - i$
:$z_2 = 2 + 5 i$
Then:
:$\size {z_1 \times z_2} = 17$
where:
:$\circ$ denotes (complex) cross product
:$\size {\, \cdot \,}$ denotes the absolute value function.
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $110 \ \text {(h)}$
Category:Examples of Complex Cross Product
Complex Cross Product Distributes over Addition

== Theorem ==
Let $z_1, z_2, z_3 \in \C$ be complex numbers.
Then:
:$z_1 \times \paren {z_2 + z_3} = z_1 \times z_2 + z_1 \times z_3$
where $\times$ denotes cross product.
== Proof ==
Let:
:$z_1 = x_1 + i y_1$
:$z_2 = x_2 + i y_2$
:$z_3 = x_3 + i y_3$
Then:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $113 \ \text{(b)}$
Category:Complex Cross Product
Category:Complex Addition
Complex Cross Product in Exponential Form

== Theorem ==
Let $z_1 := r_1 e^{i \theta_1}, z_2 := r_2 e^{i \theta_2} \in \C$ be complex numbers expressed in exponential form.
Then:
:$z_1 \times z_2 = r_1 r_2 \map \sin {\theta_2 - \theta_1}$
where $z_1 \times z_2$ denotes the dot product of $z_1$ and $z_2$.
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $112 \ \text{(b)}$
Category:Complex Cross Product
Complex Cross Product is Anticommutative/Examples

== Examples of Use of Complex Cross Product is Anticommutative ==
=== Example: $\paren {2 + 5 i} \times \paren {3 - i} = -\paren {\paren {3 - i} \times \paren {2 + 5 i} }$ ===
Category:Examples of Use of Complex Cross Product is Anticommutative
Complex Cross Product is Anticommutative/Examples/2+5i cross 3-i

== Example of Use of Complex Cross Product is Anticommutative ==
==== Example: $\paren {2 + 5 i} \times \paren {3 - i}$ ====
==== Example: $\paren {3 - i} \times \paren {2 + 5 i}$ ====
As can be seen:
:$\paren {2 + 5 i} \times \paren {3 - i} = -\paren {\paren {3 - i} \times \paren {2 + 5 i} }$
Category:Examples of Use of Complex Cross Product is Anticommutative
Complex Division/Examples

== Examples of Complex Division ==
=== Example: $\dfrac {3 + 4 i} {1 + 2 i}$ ===
=== Example: $\dfrac {1 - i} {1 + i}$ ===
=== Example: $\dfrac {3 - 2 i} {-1 + i}$ ===
=== Example: $\dfrac {1 + \sin \theta + i \cos \theta} {1 + \sin \theta - i \cos \theta}$ ===
=== Example: $\dfrac {2 - 3 i} {4 - i}$ ===
=== Example: $\dfrac {5 - 2 i} {2 + 3 i}$ ===
Category:Examples of Complex Division
Complex Division/Examples/(1 + sin theta + i cos theta) (1 + sin theta - i cos theta)^-1

== Theorem ==
:$\dfrac {1 + \sin \theta + i \cos \theta} {1 + \sin \theta - i \cos \theta} = \sin \theta + i \cos \theta$
== Proof ==
== Sources ==
* : $\S 2$. Geometrical Representations: Exercise $11$
Category:Examples of Complex Division
Complex Division/Examples/(2 - 3i) (4 - i)^-1

== Example of Complex Division ==
:$\dfrac {2 - 3 i} {4 - i} = \dfrac {11} {17} - \dfrac {10} {17} i$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Fundamental Operations with Complex Numbers: $53 \ \text {(e)}$
Category:Examples of Complex Division
Complex Division/Examples/(3+4i) (1+2i)^-1

== Example of Complex Division ==
:$\dfrac {3 + 4 i} {1 + 2 i} = \dfrac {11} 5 - \dfrac 2 5 i$
== Proof ==
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $1 \ \text{(v)}$
Category:Examples of Complex Division
Complex Division/Examples/(5 - 2i) (2 + 3i)^-1

== Example of Complex Division ==
:$\dfrac {5 - 2 i} {2 + 3 i} = \dfrac 4 {13} - \dfrac {19} {13} i$
== Proof ==
== Sources ==
* : $\S 4$: Complex Numbers: Arithmetic of Complex Numbers: Example
Category:Examples of Complex Division
Complex Division as Product with Conjugate over Square of Modulus

== Theorem ==
Let $z_1$ and $z_2$ be complex numbers.
Then the operation of division can be expressed as:
:$\dfrac {z_1} {z_2} = \dfrac {z_1 \overline {z_2} } {\cmod {z_2}^2}$
where:
:$\overline {z_2}$ denotes the complex conjugate of $z_2$
:$\cmod {z_2}$ denotes the complex modulus of $z_2$.
== Proof ==
== Sources ==
* : $\S 1.2$. The Algebraic Theory: $(1.13)$
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Multiplication and Division: $3.7.13$
Category:Complex Division
Category:Complex Conjugates
Category:Complex Modulus
Complex Dot Product/Examples

== Examples of Complex Dot Product ==
=== Example: $\paren {3 - 4 i} \circ \paren {-4 + 3 i}$ ===
=== Acute Angle Between $\paren {3 - 4 i}$ and $\paren {-4 + 3 i}$ ===
Category:Examples of Complex Dot Product
Complex Dot Product/Examples/2+5i dot 3-i

== Example of Complex Dot Product ==
Let:
:$z_1 = 2 + 5 i$
:$z_2 = 3 - i$
Then:
:$z_1 \circ z_2 = 1$
where $\circ$ denotes (complex) dot product.
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $110 \ \text {(a)}$
Category:Examples of Complex Dot Product
Complex Dot Product/Examples/3-4i dot -4+3i

== Examples of Complex Dot Product ==
Let:
:$z_1 = 3 - 4 i$
:$z_2 = -4 + 3 i$
Then:
:$z_1 \circ z_2 = -24$
where $\circ$ denotes (complex) dot product.
== Proof 1 ==
== Proof 2 ==
Category:Examples of Complex Dot Product
Complex Dot Product/Examples/3-4i dot -4+3i/Acute Angle Between

== Examples of Complex Dot Product ==
Consider:
:$z_1 = 3 - 4 i$
:$z_2 = -4 + 3 i$
expressed as vectors.
Then the acute angle between $z_1$ and $z_2$ is approximately $16 \degrees 16'$
== Proof ==
Let $\theta$ denote the acute angle between $z_1$ and $z_2$.
Then:
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Dot and Cross Product: $40$
Category:Examples of Complex Dot Product
Complex Dot Product/Examples/3-i dot 2+5i

== Examples of Complex Dot Product ==
Let:
:$z_1 = 3 - i$
:$z_1 = 2 + 5 i$
Then:
:$z_1 \circ z_2 = 1$
where $\circ$ denotes (complex) dot product.
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $110 \ \text {(c)}$
Category:Examples of Complex Dot Product
Complex Dot Product/Examples/Size of 2+5i dot 3-i

== Example of Complex Dot Product ==
Let:
:$z_1 = 2 + 5 i$
:$z_2 = 3 - i$
Then:
:$\size {z_1 \circ z_2} = 1$
where:
:$\circ$ denotes (complex) dot product
:$\size {\, \cdot \,}$ denotes the absolute value function.
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $110 \ \text {(e)}$
Category:Examples of Complex Dot Product
Complex Dot Product/Examples/Size of 3-i dot 2+5i

== Example of Complex Dot Product ==
Let:
:$z_1 = 3 - i$
:$z_2 = 2 + 5 i$
Then:
:$\size {z_1 \circ z_2} = 1$
where:
:$\circ$ denotes (complex) dot product
:$\size {\, \cdot \,}$ denotes the absolute value function.
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $110 \ \text {(g)}$
Category:Examples of Complex Dot Product
Complex Dot Product in Exponential Form

== Theorem ==
Let $z_1 := r_1 e^{i \theta_1}, z_2 := r_2 e^{i \theta_2} \in \C$ be complex numbers expressed in exponential form.
Then:
:$z_1 \circ z_2 = r_1 r_2 \, \map \cos {\theta_2 - \theta_1}$
where $z_1 \circ z_2$ denotes the dot product of $z_1$ and $z_2$.
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $112 \ \text{(a)}$
Category:Complex Dot Product
Complex Exponential Function is Entire

== Theorem ==
Let $\exp: \C \to \C$ be the complex exponential function. 
Then $\exp$ is entire. 
== Proof ==
By the definition of the complex exponential function, $\exp$ admits a power series expansion about $0$: 
:$\ds \exp z = \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$
By Complex Function is Entire iff it has Everywhere Convergent Power Series, to show that $\exp$ is entire it suffices to show that this series is everywhere convergent.
Note that this power series is of the form: 
:$\ds \sum_{n \mathop = 0}^\infty \frac {\paren {z - \xi}^n} {n!}$
with $\xi = 0$. 
Therefore, by Radius of Convergence of Power Series over Factorial: Complex Case, we have that the former power series is everywhere convergent. 
Hence the result.
== Sources ==
* 
* 
Category:Exponential Function
Category:Entire Functions
Complex Exponential Tends to Zero

== Theorem ==
Let $\exp z$ be the complex exponential function.
Then:
:$\ds \lim_{\map \Re z \mathop \to +\infty} e^{-z} = 0$
where $\map \Re z$ denotes the real part of $z$.
== Proof ==
Let $z = x + iy$.
Let $\epsilon > 0$.
By the definition of limits at infinity, we need to show that there is some $M > 0$ such that:
:$x > M \implies \size {e^{-z} - 0} < \epsilon$
But:
so we need an $M$ such that:
:$x > M \implies \size {e^{-x} - 0} < \epsilon$
This is the definition of the limits at infinity of the real exponential.
The result follows from Exponential Tends to Zero and Infinity.
Category:Exponential Function
Complex Exponential is Uniformly Continuous on Half-Planes

== Theorem ==
Let $a \in \R$.
Then $\exp$ is uniformly continuous on the half-plane $\set {z \in \C : \map \Re z \le a}$.
=== Corollary ===
== Proof ==
Let $\epsilon > 0$.
For $x, y \in \C$ with $\map \Re x, \map \Re y \le a$:
Because Exponential Function is Continuous, there exists $\delta > 0$ such that $\cmod {e^z - 1} < \epsilon$ for $\cmod z < \delta$.
Thus if $\cmod {x - y} < \delta$, $\cmod {e^x - e^y} < e^a \epsilon$.
Thus $\exp$ is uniformly continuous.
Category:Exponential Function
Category:Uniform Continuity
Complex Exponential is Uniformly Continuous on Half-Planes/Corollary

== Corollary to Complex Exponential is Uniformly Continuous on Half-Planes ==
Let $X$ be a set.
Let $\family {g_n}$ be a family of mappings $g_n : X \to \C$.
Let $g_n$ converge uniformly to $g: X \to \C$.
Let there be a constant $a \in \R$ such that $\map \Re {\map g x} \le a$ for all $x \in X$.
Then $\exp g_n$ converges uniformly to $\exp g$.
== Proof ==
By uniform convergence, there exists $N > 0$ such that $\cmod {\map {g_n} x - \map g x} \le 1$ for all $n > N$.
Then $\map \Re {\map {g_n} x} \le a + 1$.
The result now follows from:
:Complex Exponential is Uniformly Continuous on Half-Planes, applied to the half-plane $\set {z \in \C : \map \Re z \le a + 1}$
:Uniformly Continuous Function Preserves Uniform Convergence
== Sources ==
* $\text {VII}$: Compact and Convergence in the Space of Analytic Functions: $\S 5$: Weierstrass Factorization Theorem: Lemma $5.7$
Category:Exponential Function
Category:Uniform Continuity
Complex Function/Examples/Imaginary Part

== Example of Complex Function ==
Let $f: \C \to \C$ be the function defined as:
:$\forall z \in \C: \map f z = \map \Im z$
where $\map \Im z$ denotes the imaginary part of $z$.
$f$ is a complex function whose image is the set of real numbers $\R$.
== Sources ==
* 
* 
Category:Examples of Complex Functions
Complex Function/Examples/Principal Argument

== Example of Complex Function ==
Let $f: \C \to \C$ be the function defined as:
:$\forall z \in \C: \map f z = \Arg z$
where $\Arg z$ denotes the principal argument of $z$.
$f$ is a complex function whose image is the set of real numbers $\R$.
== Sources ==
* 
* 
Category:Examples of Complex Functions