nyuuzyou/wikis huggingface dataset:
Pointwise Supremum of Measurable Functions is Measurable

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space, and let $I$ be a countable set.
Let $\family {f_i}_{i \mathop \in I}$, $f_i: X \to \overline \R$ be an $I$-indexed collection of $\Sigma$-measurable functions.
Then the pointwise supremum $\ds \sup_{i \mathop \in I} f_i: X \to \overline \R$ is also $\Sigma$-measurable.
== Proof ==
Let $a \in \R$; for all $i \in I$, we have by Characterization of Measurable Functions that:
:$\set {f_i > a} \in \Sigma$
and as $\Sigma$ is a $\sigma$-algebra and $I$ is countable, also:
:$\ds \bigcup_{i \mathop \in I} \set {f_i > a} \in \Sigma$
We will now show that:
:$\ds \set {\sup_{i \mathop \in I} f_i > a} = \bigcup_{i \mathop \in I} \set {f_i > a}$
First, observe that for all $i \in I$:
:$\map {f_i} x \le \ds \sup_{i \mathop \in I} \map {f_i} x$
and hence:
:$\set {f_i > a} \subseteq \ds \set {\sup_{i \mathop \in I} f_i > a}$
From Union is Smallest Superset: Family of Sets:
:$\ds \bigcup_{i \mathop \in I} \set {f_i > a} \subseteq \set {\sup_{i \mathop \in I} f_i > a}$
Next, suppose that:
:$x \notin \ds \bigcup_{i \mathop \in I} \set {f_i > a}$
Then, by definition of union:
:$\forall i \in I: \map {f_i} x \le a$
which is to say that $a$ is an upper bound for the $\map {f_i} x$.
Hence, by definition of supremum, it follows that:
:$\ds \sup_{i \mathop \in I} \map {f_i} x \le a$
and therefore:
:$x \notin \ds \set {\sup_{i \mathop \in I} f_i > a}$
Thus, we have shown:
:$\ds \set {\sup_{i \mathop \in I} f_i > a} = \bigcup_{i \mathop \in I} \set {f_i > a} \in \Sigma$
and by Characterization of Measurable Functions, it follows that $\ds \sup_{i \mathop \in I} f_i$ is measurable.
== Sources ==
* : $8.9$
Category:Measurable Functions
Pointwise Supremum of Stopping Times is Stopping Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {T_n}_{n \in \N}$ be a sequence of stopping times with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let:
:$\ds T = \sup_{n \in \N} T_n$
be the pointwise supremum of the $\sequence {T_n}_{n \in \N}$.
Then $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
== Proof ==
We have, for $t \in \Z_{\ge 0}$ and $\omega \in \Omega$: 
:$\map T \omega \le t$ $\map {T_n} \omega \le t$ for all $n \in \N$.
That is: 
:$\ds \set {\omega \in \Omega : \map T \omega \le t} = \bigcap_{n \in \N} \set {\omega \in \Omega : \map {T_n} \omega \le t}$
for each $t \in \Z_{\ge 0}$.
Now fix $t \in \Z_{\ge 0}$. 
Since each $T_n$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$, we have: 
:$\set {\omega \in \Omega : \map {T_n} \omega \le t} \in \FF_t$
for each $n \in \N$.
Then, since $\FF_t$ is a $\sigma$-algebra and $\sigma$-algebras are closed under countable intersection, we have: 
:$\ds \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
So $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Category:Stopping Times
Pointwise Upper Limit of Measurable Functions is Measurable

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\sequence {f_n}_{n \mathop \in \N}$, $f_n: X \to \overline \R$ be a sequence of $\Sigma$-measurable functions.
Then the pointwise upper limit $\ds \limsup_{n \mathop \to \infty} f_n: X \to \overline \R$ is also $\Sigma$-measurable.
== Proof ==
By definition of upper limit, we have:
:$\ds \limsup_{n \mathop \to \infty} f_n = \inf_{m \mathop \in \N} \sup_{n \mathop \ge m} f_n$
The result follows from combining:
:Pointwise Supremum of Measurable Functions is Measurable
:Pointwise Infimum of Measurable Functions is Measurable
== Sources ==
* : $8.9$
Category:Measurable Functions
Poisson's Differential Equation/Examples/Electric Field of Volume Distribution of Charges

== Example of Poisson's Differential Equation ==
Let $R$ be a region of ordinary space.
Let $R$ contain a volume distribution of particles bearing electric charge.
Let $\mathbf V$ be the vector field arising from these electric charge.
Then $\mathbf V$ satisfies Poisson's differential equation.
An example of such a scenario is a thermionic tube.
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $7$. The Classification of Vector Fields: $\text {(ii)}$
Category:Poisson's Differential Equation
Category:Electrostatics
Poisson's Differential Equation/Examples/General Scenario

== Example of Poisson's Differential Equation ==
Let $R$ be a region of ordinary space.
Let $R$ contain a spatial distribution of sources of flux.
Let $\mathbf V$ be the vector field arising from the sources in $R$.
Then $\mathbf V$ satisfies Poisson's differential equation.
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $7$. The Classification of Vector Fields: $\text {(ii)}$
Category:Poisson's Differential Equation
Poisson's Differential Equation/Examples/Gravitational Force inside Mass

== Example of Poisson's Differential Equation ==
Let $B$ be a body situated in ordinary space.
$B$ consists of a volume distribution of particles with mass.
Let $\mathbf V$ be the vector field arising from these electric charge.
Then $\mathbf V$ satisfies Poisson's differential equation.
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $7$. The Classification of Vector Fields: $\text {(ii)}$
Category:Poisson's Differential Equation
Category:Mechanics
Poisson's Differential Equation for Rotational and Solenoidal Field

== Theorem ==
Let $R$ be a region of ordinary space.
Let $\mathbf V$ be a vector field over $R$.
Let $\mathbf V$ be both rotational and solenoidal.
Let $\mathbf A$ be a vector field such that $\mathbf V = \curl \mathbf A$.
Then $\mathbf V$ satisfies this version of Poisson's differential equation:
:$\curl \mathbf V = -\nabla^2 \mathbf A \ne \bszero$
== Proof ==
As $\mathbf V$ is rotational it is not conservative.
Hence from Vector Field is Expressible as Gradient of Scalar Field iff Conservative $\mathbf V$ cannot be the gradient of some scalar field.
However, by definition of rotational vector field:
:$\curl \mathbf V \ne \bszero$
As $\mathbf V$ is solenoidal:
:$\operatorname {div} \mathbf V = 0$
Hence from Divergence of Curl is Zero, for some vector field $\mathbf A$ over $R$:
:$\operatorname {div} \mathbf V = \operatorname {div} \curl \mathbf A = 0$
and so:
:$\mathbf V = \curl \mathbf A$
Then we have:
== Examples ==
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $7$. The Classification of Vector Fields: $\text {(iii)}$
Category:Poisson's Differential Equation
Category:Rotational Vector Fields
Category:Solenoidal Vector Fields
Category:Poisson's Differential Equation for Rotational and Solenoidal Field
Poisson's Differential Equation for Rotational and Solenoidal Field/Examples/Magnetic Field in Conductor carrying Steady Current

== Example of Poisson's Differential Equation for Rotational and Solenoidal Field ==
Consider a conductor of electricity $C$.
Let $C$ be carrying a steady current $I$.
From Curl Operator: Magnetic Field of Conductor, the curl of the magneto-motive force per unit area $\mathbf H$ is given by:
:$\curl \mathbf H = \mathbf J$
Hence this satisfies Poisson's Differential Equation for Rotational and Solenoidal Field:
:$\curl \mathbf J = -\nabla^2 \mathbf J$
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $7$. The Classification of Vector Fields: $\text {(iii)}$
Category:Poisson's Differential Equation for Rotational and Solenoidal Field
Category:Electromagnetism
Poisson's Differential Equation for Rotational and Solenoidal Field/Examples/Rotational Motion of Incompressible Fluid

== Example of Poisson's Differential Equation for Rotational and Solenoidal Field ==
Let $B$ be a body of incompressible fluid.
Let $B$ be undergoing rotational motion.
Let $\mathbf V$ be the vector field which describes the rotational motion of $B$.
Then $\mathbf V$ satisfies Poisson's Differential Equation for Rotational and Solenoidal Field.
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $7$. The Classification of Vector Fields: $\text {(iii)}$
Category:Poisson's Differential Equation for Rotational and Solenoidal Field
Category:Fluid Mechanics
Poisson Brackets of Classical Particle in Radial Potential on Plane

== Theorem ==
Let $P$ be a classical particle embedded in a 2-dimensional Euclidean manifold.
Let the real-valued functions $\map r t$, $\map \theta t$ denote the position of $P$ in polar coordinates, where $t$ is time.
Suppose, the potential energy of $P$ depends only on $r$.
Then $P$ has the following Poisson brackets:
== Proof ==
The standard Lagrangian of $P$ in polar coordinates is:
:$L = \dfrac 1 2 m \paren { {\dot r}^2 + r^2 {\dot \theta}^2 } - \map U r$
The canonical momenta are:
:$p_r = \dfrac {\partial L} {\partial \dot r} = m \dot r$
:$p_\theta = \dfrac {\partial L} {\partial \dot \theta} = m r^2 \dot \theta$
The Hamiltonian associated to $L$ in canonical coordinates reads:
:$H = \dfrac {p_r^2} {2 m} + \dfrac 1 2 \dfrac {p_\theta^2} {m r^2} + \map U r$
Then:
== Sources ==
* : $\S 4.23$: The Hamilton-Jacobi Equation. Jacobi's Theorem
Category:Lagrangian Mechanics
Category:Applied Mathematics
Category:Mathematical Physics
Poisson Brackets of Harmonic Oscillator

== Theorem ==
Let $P$ be a classical harmonic oscillator.
Let the real-valued function $\map x t$ be the position of $P$, where $t$ is time.
Then $P$ has the following Poisson brackets:
== Proof ==
The standard Lagrangian of $P$ is:
:$L = \dfrac 1 2 \paren {m {\dot x}^2 - k x^2}$
The canonical momentum is:
:$p = \dfrac {\partial L} {\partial \dot x} = m \dot x$
The Hamiltonian associated to $L$ in canonical coordinates reads:
:$H = \dfrac {p^2} {2 m} + \dfrac k 2 x^2$
Then:
== Sources ==
* : $\S 4.23$: The Hamilton-Jacobi Equation. Jacobi's Theorem
Category:Lagrangian Mechanics
Category:Applied Mathematics
Category:Mathematical Physics
Poisson Distribution Approximated by Hat-Check Distribution

== Theorem ==
Let $X$ be a discrete random variable which has the hat-check distribution with parameter $n$.
Then $X$ can be approximated by a Poisson distribution with parameter $\lambda = 1$.
== Proof ==
Let $X$ be as described.
Let $k \ge 0$ be fixed.
Then:
Hence the result.
== Examples ==
Category:Hat-Check Distribution
Category:Poisson Distribution
Poisson Distribution Approximated by Hat-Check Distribution/Examples/N equals 8

== Example of Poisson Distribution Approximated by Hat-Check Distribution ==
Let $X$ be a discrete random variable which has the hat-check distribution with parameter $n = 8$.
Then $X$ can be approximated by a Poisson distribution with parameter $\lambda = 1$.
== Proof ==
From , we have:
:$\begin{array}{r|cccccccccc}
k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
!k \dbinom n k & 1 & 0 & 28 & 112 & 630 & 2464 & 7420 & 14832 & 14833 \\
\hline
\dfrac {!k \dbinom n k} {8!} & \dfrac 1 {40320} & \dfrac 0 {40320} & \dfrac {28} {40320} & \dfrac {112} {40320} & \dfrac {630} {40320} & \dfrac {2464} {40320} & \dfrac {7420} {40320} & \dfrac {14832} {40320} & \dfrac {14833} {40320} \\
\hline
\dfrac {!k \dbinom n k} {8!} & 0.0000 & 0.0000 & 0.0007 & 0.0028 & 0.0156 & 0.0611 & 0.1840 & 0.3679 & 0.3679 \\
\hline
\end{array}$
If we let $k = n - k$ and compare the hat-check probabilities from the table above to the probabilities from a Poisson distribution where $\lambda = 1$, we get:
:$\begin{array}{r|cccccccccc}
k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
\text {Hat-Check} & 0.3679 & 0.3679 & 0.1840 & 0.0611 & 0.0156 & 0.0028 & 0.0007 & 0.0000 & 0.0000 \\
\hline
\text {Poisson} & 0.3679 & 0.3679 & 0.1839 & 0.0613 & 0.0153 & 0.0031 & 0.0005 & 0.0000 & 0.0000 \\
\hline
\end{array}$
Category:Hat-Check Distribution
Category:Poisson Distribution
Poisson Distribution Approximated by Normal Distribution

== Theorem ==
Let $X$ be a discrete random variable which has the Poisson distribution $\Poisson \lambda$.
Then for large $\lambda$:
:$\Poisson \lambda \approx \Gaussian \lambda \lambda$
where $\Gaussian \lambda \lambda$ denotes the normal distribution.
== Proof ==
== Sources ==
* 
Category:Poisson Distribution
Category:Normal Distribution
Poisson Distribution Gives Rise to Probability Mass Function

== Theorem ==
Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have the poisson distribution with parameter $\lambda$ (where $\lambda > 0$).
Then $X$ gives rise to a probability mass function.
== Proof ==
By definition:
:$\Img X = \N$
:$\map \Pr {X = k} = \dfrac 1 {k!} \lambda^k e^{-\lambda}$
Then:
So $X$ satisfies $\map \Pr \Omega = 1$, and hence the result.
== Sources ==
* : $\S 2.2$: Examples
Category:Poisson Distribution
Category:Probability Mass Functions
Poisson Summation Formula

== Theorem ==
Let $f: \R \to \C$ be a Schwarz function.
Let $\hat f$ be its Fourier transform.
Then:
:$\ds \sum_{n \mathop \in \Z} \map f n = \sum_{m \mathop \in \Z} \map {\hat f} m$
== Proof ==
Let:
:$\ds \map F x = \sum_{n \mathop \in \Z} \map f {x + n}$
Then $\map F x$ is $1$-periodic (because of absolute convergence), and has Fourier coefficients:
Therefore by the definition of the Fourier series of $f$:
:$\ds \map F x = \sum_{k \mathop \in \Z} \map {\hat f} k e^{i k x}$
Choosing $x = 0$ in this formula:
:$\ds \sum_{n \mathop \in \Z} \map f n = \sum_{k \mathop \in \Z} \map {\hat f} k$
as required.
== Sources ==
* : $\S 19$: The Poisson Summation Formula: $19.46$
* : $\S 21$: Series of Constants: The Poisson Summation Formula: $21.46.$
Category:Harmonic Analysis
Category:Fourier Analysis
Polar Equation of Conchoid of Nicomedes

== Theorem ==
Let $a \in \R$, $b \in \R_{>0}$ be real constants.
Let the focus point of a conchoid of Nicomedes $\KK$ be located at the origin of a polar plane.
Let the directrix of $\KK$ be the straight line through $\polar {a, 0}$ perpendicular to the polar axis.
Then $\KK$ can be expressed in polar coordinates as:
:$r = b + a \sec \theta$
== Proof ==
== Also see ==
* Cartesian Equation of Conchoid of Nicomedes
== Sources ==
* 
* 
* 
* 
Category:Conchoid of Nicomedes
Polar Form of Complex Conjugate

== Theorem ==
Let $z := r \paren {\cos \theta + i \sin \theta} \in \C$ be a complex number expressed in polar form.
Then:
:$\overline z = r \paren {\cos \theta - i \sin \theta}$
where $\overline z$ denotes the complex conjugate of $z$.
== Proof ==
== Sources ==
* : $\S 2$. Geometrical Representations
Category:Complex Conjugates
Category:Polar Form of Complex Number
Polar Form of Complex Number/Examples

== Examples of Polar Form of Complex Number ==
=== Example: $i$ ===
=== Example: $-i$ ===
=== Example: $-1$ ===
=== Example: $-4$ ===
=== Example: $\sqrt 2 i$ ===
=== Example: $2 - 2 i$ ===
=== Example: $-1 + \sqrt 3 i$ ===
=== Example: $2 + 2 \sqrt 3 i$ ===
=== Example: $2 \sqrt 2 + 2 \sqrt 2 i$ ===
=== Example: $\dfrac {\sqrt 3} 2 - \dfrac {3 i} 2$ ===
=== Example: $-5 + 5 i$ ===
=== Example: $-\sqrt 6 - \sqrt 2 i$ ===
=== Example: $-3 i$ ===
=== Example: $-2 \sqrt 3 - 2 i$ ===
=== Example: $2 + i$ ===
=== Example: $-3 - 4 i$ ===
=== Example: $1 - 2 i$ ===
=== Example: $6 \paren {\cos 240 \degrees + i \sin 240 \degrees}$ ===
=== Example: $4 \cis \dfrac {3 \pi} 5$ ===
=== Example: $2 \cis \dfrac {-\pi} 4$ ===
=== Example: $6 \cis 135 \degrees$ ===
=== Example: $12 \cis 90 \degrees$ ===
=== Example: $4 \cis 315 \degrees$ ===
=== Example: $2 \cis \dfrac {5 \pi} 4$ ===
=== Example: $5 \cis \dfrac {7 \pi} 6$ ===
=== Example: $3 \cis \dfrac {-2 \pi} 3$ ===
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/-1

== Example of Polar Form of Complex Number ==
The real number $-1$ can be expressed as a complex number in polar form as $\polar {1, \pi}$.
== Proof ==
Then:
Hence:
:$\map \arg {-1} = \pi$
and hence the result.
== Sources ==
* : $\S 2$. Geometrical Representations
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/-1 + root 3 i

== Example of Polar Form of Complex Number ==
The complex number $-1 + \sqrt 3 i$ can be expressed as a complex number in polar form as $\polar {2, \dfrac {2 \pi} 3}$.
== Proof ==
Then:
Hence:
:$\map \arg {-1 + \sqrt 3 i} = \dfrac {2 \pi} 3$
and hence the result.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $81 \ \text {(b)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/-2 root 3 - 2 i

== Example of Polar Form of Complex Number ==
The complex number $-2 \sqrt 3 - 2 i$ can be expressed as a complex number in polar form as $\polar {4, \dfrac {7 \pi} 6}$.
== Proof ==
Then:
Hence:
:$\map \arg {-2 \sqrt 3 - 2 i} = \dfrac {7 \pi} 6$
and hence the result.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $81 \ \text {(f)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/-3 - 4i

== Example of Polar Form of Complex Number ==
The complex number $-3 - 4 i$ can be expressed as a complex number in polar form as $\polar {5, \pi + \arctan {\dfrac 4 3} }$.
== Proof ==
Then:
Hence:
:$\map \arg {-3 - 4 i} = \pi + \arctan {\dfrac 4 3}$
as $-3 - 4 i$ is in Quadrant III.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $83 \ \text {(a)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/-3i

== Example of Polar Form of Complex Number ==
The complex number $-3 i$ can be expressed as a complex number in polar form as $\polar {3, \dfrac {3 \pi} 2}$.
== Proof ==
:
Then:
Hence:
:$\map \arg {-3 i} = \dfrac {3 \pi} 2$
and hence the result.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Polar Form of Complex Numbers: $16 \ \text {(d)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/-4

== Example of Polar Form of Complex Number ==
The real number $-4$ can be expressed as a complex number in polar form as $\polar {4, \pi}$.
== Proof ==
Then:
Hence:
:$\map \arg {-4} = \pi$
and hence the result.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $81 \ \text {(e)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/-5 + 5i

== Example of Polar Form of Complex Number ==
The complex number $-5 + 5 i$ can be expressed as a complex number in polar form as $\polar {5 \sqrt 2, \dfrac {3 \pi} 4}$.
== Proof ==
:
Then:
Hence:
:$\map \arg {-5 + 5 i} = \dfrac {3 \pi} 4$
and hence the result.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Polar Form of Complex Numbers: $16 \ \text {(b)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/-i

== Example of Polar Form of Complex Number ==
The imaginary number $-i$ can be expressed in polar form as $\polar {1, \dfrac {3 \pi} 2}$.
== Proof ==
Then:
Hence:
:$\map \arg {-i} = \dfrac {3 \pi} 2$
and hence the result.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $81 \ \text {(d)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/-root 6 - root 2 i

== Example of Polar Form of Complex Number ==
The complex number $-\sqrt 6 - \sqrt 2 i$ can be expressed as a complex number in polar form as $\polar {2 \sqrt 2, \dfrac {7 \pi} 6}$.
== Proof ==
:
Then:
Hence:
:$\map \arg {-\sqrt 6 - \sqrt 2 i} = \dfrac {7 \pi} 6$
and hence the result.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Polar Form of Complex Numbers: $16 \ \text {(c)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/12 cis 90

== Example of Polar Form of Complex Number ==
The complex number $\polar {12, 90 \degrees}$ can be expressed in Cartesian form as:
:$12 \cis 90 \degrees = 12 i$
and depicted in the complex plane as:
:
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $84 \ \text {(b)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/1 - 2i

== Example of Polar Form of Complex Number ==
The complex number $1 - 2 i$ can be expressed as a complex number in polar form as $\polar {\sqrt 5, -\arctan 2}$.
== Proof ==
Then:
Hence:
:$\map \arg {1 - 2 i} = \map \arctan {-2} = -\arctan 2$
as $1 - 2 i$ is in Quadrant II.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $83 \ \text {(b)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/2 + 2 root 3 i

== Example of Polar Form of Complex Number ==
The complex number $2 + 2 \sqrt 3 i$ can be expressed as a complex number in polar form as $\polar {4, \dfrac \pi 3}$.
== Proof ==
:
Then:
Hence:
:$\map \arg {2 + 2 \sqrt 3 i} = \dfrac \pi 3$
and hence the result.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Polar Form of Complex Numbers: $16 \ \text {(a)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/2 + i

== Example of Polar Form of Complex Number ==
The complex number $2 + i$ can be expressed as a complex number in polar form as $\polar {\sqrt 5, \arctan {\dfrac 1 2} }$.
== Proof ==
Then:
Hence:
:$\map \arg {2 + i} = \arctan {\dfrac 1 2}$
and hence the result.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $82$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/2 - 2i

== Example of Polar Form of Complex Number ==
The complex number $2 - 2 i$ can be expressed as a complex number in polar form as $\polar {2 \sqrt 2, \dfrac {7 \pi} 4}$.
== Proof ==
Then:
Hence:
:$\map \arg {2 - 2 i} = \dfrac {7 \pi} 4$
and hence the result.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $81 \ \text {(a)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/2 cis -pi 4^-1

== Example of Polar Form of Complex Number ==
The complex number $2 \paren {\cos \dfrac {-\pi} 4 + i \sin \dfrac {-\pi} 4}$ can be expressed as:
:$4 \paren {\cos \dfrac {-\pi} 4 + i \sin \dfrac {-\pi} 4} = 2 \paren {\cos 315 \degrees + i \sin 315 \degrees} = 2 \cis 315 \degrees = 2 \cis \dfrac {-\pi} 4 = 2 e^{-\pi / 4}$
and depicted in the complex plane as:
:
== Sources ==
* : $1$: Solved Problems: Polar Form of Complex Numbers: $17 \ \text {(c)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/2 cis 5 pi 4^-1

== Example of Polar Form of Complex Number ==
The complex number $\polar {2, \dfrac {5 \pi} 4}$ can be expressed in Cartesian form as:
:$2 \cis \dfrac {5 \pi} 4 = -\sqrt 2 - \sqrt 2 i$
and depicted in the complex plane as:
:
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $84 \ \text {(d)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/2 root 2 + 2 root 2 i

== Example of Polar Form of Complex Number ==
The complex number $2 \sqrt 2 + 2 \sqrt 2 i$ can be expressed as a complex number in polar form as $\polar {4, \dfrac \pi 4}$.
== Proof ==
Then:
Hence:
:$\map \arg {2 \sqrt 2 + 2 \sqrt 2 i} = \dfrac \pi 4$
and hence the result.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $81 \ \text {(c)}$
Category:Examples of Polar Form of Complex Number
Polar Form of Complex Number/Examples/3 cis -2 pi 3^-1

== Example of Polar Form of Complex Number ==
The complex number $\polar {3, -\dfrac {2 \pi} 3}$ can be expressed in Cartesian form as:
:$3 \, \map \cis {-\dfrac {2 \pi} 3} = -\dfrac 3 2 - \dfrac {3 \sqrt 3} 2 i$
and depicted in the complex plane as:
:
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $84 \ \text {(f)}$
:::but beware of the mistake in the solution.
Category:Examples of Polar Form of Complex Number