hcju/proofwikips huggingface dataset:
Fibonacci String/Examples/S4
Fibonacci String/Examples/S4 0 69981 360466 2018-07-14T08:45:06Z Prime.mover 59 Created page with "== Example of Fibonacci Strings ==  The Fibonacci string $S_4$ is $\text{bab}$.   ==..." wikitext text/x-wiki == Example of Fibonacci Strings ==  The Fibonacci string $S_4$ is $\text{bab}$.  == Proof == {{begin-eqn}} {{eqn | l = S_4       | r = S_3 S_2       | c = {{Defof|Fibonacci String}} }} {{eqn | r = \text{ba} \ \text{b}       | c = Definition of Fibonacci String $S_3$ }} {{end-eqn}}  == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Fibonacci String/Examples/S3|next = Fibonacci String/Examples/S5}}: $\S 1.2.8$: Fibonacci Numbers: Exercise $36$
Complex Modulus/Examples/i
== Example of Complex Modulus ==  :$\cmod i = \cmod {-i} = 1$  == Proof == {{begin-eqn}} {{eqn | l = \cmod i       | r = \cmod {0 + 1 i}       | c =  }} {{eqn | r = \sqrt {0^2 + 1^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | r = \sqrt 1       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \cmod {-i}       | r = \cmod {0 + \left({-1}\right) i}       | c =  }} {{eqn | r = \sqrt {0^2 + \left({-1}\right)^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | r = \sqrt 1       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Power of Complex Modulus equals Complex Modulus of Power|next = Complex Modulus/Examples/-5}}: $\S 1.2$. The Algebraic Theory: Examples
Complex Modulus/Examples/1+i
== Example of Complex Modulus ==  :$\cmod {1 + i} = \sqrt 2$  == Proof == {{begin-eqn}} {{eqn | l = \cmod {1 + i}       | r = \cmod {1 + 1 i}       | c =  }} {{eqn | r = \sqrt {1^2 + 1^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | r = \sqrt 2       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Complex Modulus/Examples/-5|next = Power of Complex Modulus equals Complex Modulus of Power/Examples/(1+i)^4}}: $\S 1.2$. The Algebraic Theory: Examples
Polar Form of Complex Number/Examples/i
== Example of Polar Form of Complex Number ==  The imaginary unit $i$ can be expressed in polar form as $\polar {1, \dfrac \pi 2}$.  == Proof == {{begin-eqn}} {{eqn | l = \cmod i       | r = \sqrt {0^2 + 1^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | r = 1       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \map \cos {\map \arg i}       | r = \dfrac 0 1       | c = {{Defof|Argument of Complex Number}} }} {{eqn | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \map \arg i       | r = \pm \dfrac \pi 2       | c = Cosine of Half-Integer Multiple of Pi }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \sin {\map \arg i}       | r = \dfrac 1 1       | c = {{Defof|Argument of Complex Number}} }} {{eqn | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \map \arg i       | r = \dfrac \pi 2       | c = Sine of Half-Integer Multiple of Pi }} {{end-eqn}} Hence: :$\map \arg i = \dfrac \pi 2$ and hence the result.  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Argument of Quotient equals Difference of Arguments|next = Multiplication by Imaginary Unit is Equivalent to Rotation through Right Angle}}: $\S 2$. Geometrical Representations
Euler's Formula/Examples/e^i pi by 2
== Example of Use of Euler's Formula ==  :$e^{i \pi / 2} = i$  == Proof == {{begin-eqn}} {{eqn | l = e^{i \pi / 2}       | r = \cos \frac \pi 2 + i \sin \frac \pi 2       | c = Euler's Formula }} {{eqn | r = 0 + i \times 1       | c = Cosine of $\dfrac \pi 2$, Sine of $\dfrac \pi 2$ }} {{eqn | r = i       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Euler's Formula/Examples/e^i pi by 4|next = Euler's Formula/Examples/e^-i pi by 2}}: $\S 2$. Geometrical Representations: $(2.19)$
Euler's Formula/Examples/e^-i pi by 2
== Example of Use of Euler's Formula ==  :$e^{-i \pi / 2} = -i$  == Proof == {{begin-eqn}} {{eqn | l = e^{-i \pi / 2}       | r = \cos \frac {-\pi} 2 + i \sin \frac {-\pi} 2       | c = Euler's Formula }} {{eqn | r = \cos \frac {3 \pi} 2 + i \sin \frac {3 \pi} 2       | c = Cosine of Angle plus Full Angle, Sine of Angle plus Full Angle }} {{eqn | r = 0 + i \times \paren {-1}       | c = Cosine of $\dfrac {3 \pi} 2$, Sine of $\dfrac {3 \pi} 2$ }} {{eqn | r = -i       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Euler's Formula/Examples/e^i pi by 2|next = Euler's Formula/Examples/e^i pi}}: $\S 2$. Geometrical Representations: $(2.19)$
Euler's Formula/Examples/e^2 i pi
== Example of Use of Euler's Formula ==  :$e^{2 i \pi} = 1$  == Proof == {{begin-eqn}} {{eqn | l = e^{2 i \pi}       | r = \cos 2 \pi + i \sin 2 \pi       | c = Euler's Formula }} {{eqn | r = 1 + i \times 0       | c = Cosine of $2 \pi$, Sine of $2 \pi$ }} {{eqn | r = 1       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Euler's Formula/Examples/e^i pi|next = Exponential of Complex Number plus 2 pi i}}: $\S 2$. Geometrical Representations: $(2.19)$ * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Euler's Formula/Proof|next = Period of Complex Exponential Function}}: $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$: $\text{(ii)}$
Euler's Formula/Examples/e^2 k i pi
Euler's Formula/Examples/e^2 k i pi 0 70480 363057 2018-08-20T07:07:01Z Prime.mover 59 Created page with "== Example of Use of Euler's Formula ==  :$\forall k \in \Z: e^{2 k i \pi} = 1$   == Proof ==  {{begin-eqn}} {{eqn | l = e^{2 k i \pi}       | r..." wikitext text/x-wiki == Example of Use of Euler's Formula ==  :$\forall k \in \Z: e^{2 k i \pi} = 1$  == Proof == {{begin-eqn}} {{eqn | l = e^{2 k i \pi}       | r = \cos 2 k \pi + i \sin 2 k \pi       | c = Euler's Formula }} {{eqn | r = 1 + i \times 0       | c = Cosine of Multiple of $\pi$, Sine of Multiple of $\pi$ }} {{eqn | r = 1       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Exponential of Complex Number plus 2 pi i|next = Period of Complex Exponential Function}}: $\S 2$. Geometrical Representations
Euler's Theorem/Corollary 1
== Corollary to Euler's Theorem ==  Let $p^n$ be a prime power for some prime number $p > 1$. Let $a$ be an integer not divisible by $p: p \nmid a$. Then: : $a^{\paren {p - 1} p^{n - 1} } \equiv 1 \pmod {p^n}$  == Proof == We have that Divisor Relation is Transitive. Since $p \divides p^n$, it follows that $p^n \nmid a$. From Euler's Theorem: : $a^{\map \phi {p^n} } \equiv 1 \pmod {p^n}$ From Euler Phi Function of Prime Power: :$\map \phi {p^n} = \paren {p - 1} p^{n - 1}$ Then: : $a^{\paren {p - 1} p^{n - 1} } \equiv 1 \pmod {p^n}$
Argument of Negative Real Number is Pi
Argument of Negative Real Number is Pi 0 70516 363220 2018-08-22T21:43:48Z Prime.mover 59 Created page with "== Theorem ==  Let $x \in \R_{>0}$ be a positive real number.  Then: :$\arg \paren {-x} = \pi$  where $\arg$ denotes the Def..." wikitext text/x-wiki == Theorem ==  Let $x \in \R_{>0}$ be a positive real number. Then: :$\arg \paren {-x} = \pi$ where $\arg$ denotes the argument of a complex number.  == Proof == We have that: :$-x = -x + 0 i$ and so: {{begin-eqn}} {{eqn | l = \cmod {-x}       | r = \sqrt {\paren {-x}^2 + 0^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | r = x       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \cos \paren {\arg \paren {-x} }       | r = \dfrac {-x} x       | c = {{Defof|Argument of Complex Number}} }} {{eqn | r = -1       | c =  }} {{eqn | ll= \leadsto       | l = \arg \paren {-x}       | r = \pi       | c = Cosine of Multiple of Pi }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sin \paren {\arg \paren {-x} }       | r = \dfrac 0 x       | c = {{Defof|Argument of Complex Number}} }} {{eqn | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \arg \paren {-x}       | r = 0 \text { or } \pi       | c = Sine of Multiple of Pi }} {{end-eqn}} Hence: : $\arg \paren {-x} = \pi$
Equation of Unit Circle in Complex Plane/Corollary 1
Equation of Unit Circle in Complex Plane/Corollary 1 0 70543 363328 2018-08-24T22:57:32Z Prime.mover 59 Created page with "== Corollary to Equation of Unit Circle in Complex Plane ==  Consider the unit circle $C$ whose center is at $\l..." wikitext text/x-wiki == Corollary to Equation of Unit Circle in Complex Plane == Consider the unit circle $C$ whose center is at $\left({0, 0}\right)$ on the complex plane.  The equation of $C$ can be given by: :$z \overline z = 1$ where $\overline z$ denotes the complex conjugate of $z$.  == Proof == From Equation of Unit Circle in Complex Plane, the equation of $C$ can also be given by: :$\cmod z = 1$ where $\cmod z$ denotes the complex modulus of $z$. Thus: {{begin-eqn}} {{eqn | l = \cmod z       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \cmod z^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = z \overline z       | r = 1       | c = Modulus in Terms of Conjugate }} {{end-eqn}}  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Equation of Unit Circle|next = Equation of Unit Circle in Complex Plane/Corollary 2}}: $\S 3$. Roots of Unity
Complex Roots of Unity/Examples/4th Roots
== Example of Complex Roots of Unity ==  The '''complex $4$th roots of unity''' are the elements of the set: :$U_n = \set {z \in \C: z^4 = 1}$ They are: {{begin-eqn}} {{eqn | l = e^{0 i \pi / 4}       | r = 1 }} {{eqn | l = e^{i \pi / 2}       | r = i }} {{eqn | l = e^{i \pi}       | r = -1 }} {{eqn | l = e^{3 i \pi / 2}       | r = -i }} {{end-eqn}}  == Proof == By definition, the first complex $4$th root of unity $\alpha$ is given by: {{begin-eqn}} {{eqn | l = \alpha       | r = e^{2 i \pi / 4}       | c =  }} {{eqn | r = e^{i \pi / 2}       | c =  }} {{eqn | r = \cos \frac \pi 2 + i \sin \frac \pi 2       | c =  }} {{eqn | r = 0 + i \times 1       | c = Cosine of $\dfrac \pi 2$, Sine of $\dfrac \pi 2$ }} {{eqn | r = i       | c =  }} {{end-eqn}} We have that: :$e^{0 i \pi / 4} = e^0 = 1$ which gives us, as always, the zeroth complex $n$th root of unity for all $n$. The remaining complex $4$th roots of unity can be expressed as $e^{4 i \pi / 4} = e^{i \pi}$ and $e^{6 i \pi / 4} = e^{3 i \pi / 2}$, but it is simpler to calculate them as follows: {{begin-eqn}} {{eqn | l = \alpha^2       | r = i^2       | c =  }} {{eqn | r = -1       | c = {{Defof|Imaginary Unit}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \alpha^3       | r = \alpha^2 \times \alpha       | c =  }} {{eqn | r = \paren {-1} \times i       | c =  }} {{eqn | r = -i       | c =  }} {{end-eqn}} :500px  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Complex Roots of Unity are Vertices of Regular Polygon Inscribed in Circle|next = Complex 5th Roots of Unity}}: $\S 3$. Roots of Unity: Example $1$. * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Viète's Formulas/Examples/Sum 4, Product 8|next = Complex Roots of Unity/Examples/7th Roots}}: $1$: Complex Numbers: Supplementary Problems: The $n$th Roots of Unity: $105 \ \text {(a)}$ * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Four Squares Theorem|next = Definition:Fractal|entry = fourth root of unity}}
Complex Roots of Unity/Examples/5th Roots
== Example of Complex Roots of Unity ==  The '''complex $5$th roots of unity''' are the elements of the set: :$U_n = \set {z \in \C: z^5 = 1}$ They are: {{begin-eqn}} {{eqn | l = e^{0 \pi / 5}       | r = 1 }} {{eqn | l = e^{2 \pi / 5}       | r = \dfrac {\sqrt 5 - 1} 4 + i \dfrac {\sqrt {10 + 2 \sqrt 5} } 4 }} {{eqn | l = e^{4 \pi / 5}       | r = -\dfrac {1 + \sqrt 5} 4 + i \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8} }} {{eqn | l = e^{6 \pi / 5}       | r = -\dfrac {1 + \sqrt 5} 4 - i \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8} }} {{eqn | l = e^{8 \pi / 5}       | r = \dfrac {\sqrt 5 - 1} 4 - i \dfrac {\sqrt {10 + 2 \sqrt 5} } 4 }} {{end-eqn}}  == Proof == By definition, the first complex $5$th root of unity $\alpha$ is given by: {{begin-eqn}} {{eqn | l = \alpha       | r = e^{2 i \pi / 5}       | c =  }} {{eqn | r = \cos \frac {2 \pi} 5 + i \sin \frac {2 \pi} 5       | c =  }} {{eqn | r = \dfrac {\sqrt 5 - 1} 4 + i \dfrac {\sqrt {10 + 2 \sqrt 5} } 4       | c = Cosine of $\dfrac {2 \pi} 5$, Sine of $\dfrac {2 \pi} 5$ }} {{end-eqn}} We have that: :$e^{0 i \pi / 5} = e^0 = 1$ which gives us, as always, the zeroth complex $n$th root of unity for all $n$. The remaining complex $5$th roots of unity follow: {{begin-eqn}} {{eqn | l = \alpha^2       | r = e^{4 i \pi / 5}       | c =  }} {{eqn | r = -\dfrac {1 + \sqrt 5} 4 + i \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8}       | c =  }} {{eqn | r = \dfrac {\sqrt 5 - 1} 4 + i \dfrac {\sqrt {10 + 2 \sqrt 5} } 4       | c = Cosine of $\dfrac {4 \pi} 5$, Sine of $\dfrac {4 \pi} 5$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \alpha^3       | r = \overline {\alpha^{5 - 3} }       | c = Complex Roots of Unity occur in Conjugate Pairs }} {{eqn | r = \overline {\alpha^2}       | c =  }} {{eqn | r = \overline {-\dfrac {1 + \sqrt 5} 4 + i \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8} }       | c =  }} {{eqn | r = -\dfrac {1 + \sqrt 5} 4 - i \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8}       | c = {{Defof|Complex Conjugate}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \alpha^4       | r = \overline {\alpha^{4 - 1} }       | c = Complex Roots of Unity occur in Conjugate Pairs }} {{eqn | r = \overline \alpha       | c =  }} {{eqn | r = \overline {\dfrac {\sqrt 5 - 1} 4 + i \dfrac {\sqrt {10 + 2 \sqrt 5} } 4}       | c =  }} {{eqn | r = \dfrac {\sqrt 5 - 1} 4 - i \dfrac {\sqrt {10 + 2 \sqrt 5} } 4       | c = {{Defof|Complex Conjugate}} }} {{end-eqn}}  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Complex 4th Roots of Unity|next = Complex Roots of Unity/Examples/6th Roots/Illustration}}: $\S 3$. Roots of Unity: Example $2$. * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs/Proof 2|next = Sum of Cosines of Fractions of Pi}}: $1$: Complex Numbers: Solved Problems: The $n$th Roots of Unity: $37$
Polynomial Factor Theorem/Corollary/Complex Numbers
== Corollary to Polynomial Factor Theorem ==  Let $\map P z$ be a polynomial in $z$ over the complex numbers $\C$ of degree $n$. Suppose there exists $\zeta \in \C: \map P \xi = 0$. Then: :$\map P z = \paren {x - \zeta} \map Q z$ where $\map Q z$ is a polynomial of degree $n - 1$. Hence, if $\zeta_1, \zeta_2, \ldots, \zeta_n \in \C$ such that all are different, and $\map P {\zeta_1} = \map P {\zeta_2} = \dotsb = \map P {\zeta_n} = 0$, then: :$\displaystyle \map P z = k \prod_{j \mathop = 1}^n \paren {z - \zeta_j}$ where $k \in \C$.  == Proof == Recall that Complex Numbers form Field. The result then follows from the Polynomial Factor Theorem.  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Roots of Complex Number/Corollary/Examples/Cube Roots of 8i|next = Power of Complex Number minus 1}}: $\S 3$. Roots of Unity
Complex Algebra/Examples/z^6 + z^3 + 1
== Example of Complex Algebra ==  :$z^6 + z^3 + 1 = \paren {z^2 - 2 z \cos \dfrac {2 \pi} 9 + 1} \paren {z^2 - 2 z \cos \dfrac {4 \pi} 9 + 1} \paren {z^2 - 2 z \cos \dfrac {8 \pi} 9 + 1}$  == Proof == From Sum of Geometric Sequence or Difference of Two Cubes: :$z^6 + z^3 + 1 = \dfrac {z^9 - 1} {z^3 - 1}$ Then from Factorisation of $x^{2 n + 1} - 1$ in Real Domain: :$z^9 - 1 = \paren {z - 1} \displaystyle \prod_{k \mathop = 1}^4 \paren {z^2 - 2 \cos \dfrac {2 \pi k} 9 + 1}$ and: {{begin-eqn}} {{eqn | l = z^3 - 1       | r = \paren {z - 1} \prod_{k \mathop = 1}^1 \paren {z^2 - 2 \cos \dfrac {2 \pi k} 3 + 1}       | c =  }} {{eqn | r = \paren {z - 1} \paren {z^2 - 2 \cos \dfrac {2 \pi} 3 + 1}       | c =  }} {{eqn | r = \paren {z - 1} \paren {z^2 - 2 \cos \dfrac {6 \pi} 9 + 1}       | c =  }} {{end-eqn}} Thus: :$\paren {z^3 - 1} \paren {z^6 + z^3 + 1} = \paren {\paren {z - 1} \paren {z^2 - 2 \cos \dfrac {6 \pi} 9 + 1} } \paren {\paren {z^2 - 2 \cos \dfrac {2 \pi} 9 + 1} \paren {z^2 - 2 \cos \dfrac {4 \pi} 9 + 1} \paren {z^2 - 2 \cos \dfrac {8 \pi} 9 + 1} }$ from which: :$\paren {z^6 + z^3 + 1} = \paren {z^2 - 2 \cos \dfrac {2 \pi} 9 + 1} \paren {z^2 - 2 \cos \dfrac {4 \pi} 9 + 1} \paren {z^2 - 2 \cos \dfrac {8 \pi} 9 + 1}$  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Factorisation of z^n-a|next = Triple Angle Formulas/Cosine/2 cos 3 theta + 1}}: $\S 3$. Roots of Unity: Example $6$: $(3.14)$
Roots of Complex Number/Examples/z^8 + 1 = 0/Illustration
Roots of Complex Number/Examples/z^8 + 1 = 0/Illustration 0 70649 364035 2018-09-02T06:13:38Z Prime.mover 59 Created page with "== Illustration of Roots of $z^8 + 1 = 0$ ==  The roots of the Definition:Pol..." wikitext text/x-wiki == Illustration of Roots of $z^8 + 1 = 0$ ==  The roots of the polynomial: :$z^8 + 1 = 0$ are illustrated below: :620px  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Roots of Complex Number/Examples/z^8 + 1 = 0|next = Complex Algebra/Examples/z^8 + 1}}: $\S 3$. Roots of Unity: Exercise $8$
Complex Algebra/Examples/z^8 + 1
== Example of Complex Algebra ==  :$z^8 + 1 = \paren {z^2 - 2 z \cos \dfrac \pi 8 + 1} \paren {z^2 - 2 z \cos \dfrac {3 \pi} 8 + 1} \paren {z^2 - 2 z \cos \dfrac {5 \pi} 8 + 1} \paren {z^2 - 2 z \cos \dfrac {7 \pi} 8 + 1}$  == Proof == From Roots of $z^8 + 1 = 0$ and the corollary to the Polynomial Factor Theorem: :$z^8 + 1 = \displaystyle \prod_{k \mathop = 0}^7 \paren {z - \paren {\cos \dfrac {\paren {2 k + 1} \pi} 8 + i \sin \dfrac {\paren {2 k + 1} \pi} 8} }$ Hence: {{begin-eqn}} {{eqn | l = z^8 + 1       | r = \prod_{k \mathop = 0}^7 \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} 8}       | c = {{Defof|Exponential Form of Complex Number}} }} {{eqn | r = \prod_{k \mathop = 0}^3 \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} 8} \paren {z - \exp \dfrac {-\paren {2 k + 1} i \pi} 8}       | c = Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs }} {{eqn | r = \prod_{k \mathop = 0}^3 \paren {z^2 - z \paren {\exp \dfrac {\paren {2 k + 1} i \pi} 8 + \exp \dfrac {-\paren {2 k + 1} i \pi} 8} + \exp \dfrac {\paren {2 k + 1} i \pi} 8 \exp \dfrac {-\paren {2 k + 1} i \pi} 8}       | c = multiplying out }} {{eqn | r = \prod_{k \mathop = 0}^3 \paren {z^2 - z \paren {\exp \dfrac {\paren {2 k + 1} i \pi} 8 + \exp \dfrac {-\paren {2 k + 1} i \pi} 8} + 1}       | c = simplifying }} {{eqn | r = \prod_{k \mathop = 0}^3 \paren {z^2 - z \paren {\cos \dfrac {\paren {2 k + 1} \pi} 8 + i \sin \dfrac {\paren {2 k + 1} \pi} 8 + \cos \dfrac {\paren {2 k + 1} \pi} 8 - i \sin \dfrac {\paren {2 k + 1} \pi} 8} + 1}       | c = {{Defof|Exponential Form of Complex Number}} }} {{eqn | r = \prod_{k \mathop = 0}^3 \paren {z^2 - 2 z \cos \dfrac {\paren {2 k + 1} \pi} 8 + 1}       | c = simplifying }} {{end-eqn}} Hence the result.  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Roots of Complex Number/Examples/z^8 + 1 = 0/Illustration|next = Quadruple Angle Formulas/Cosine/Factor Form}}: $\S 3$. Roots of Unity: Exercise $8$
Euler's Formula/Corollary
== Corollary to Euler's Formula == Let $z \in \C$ be a complex number. Then:  :$e^{-i z} = \cos z - i \sin z$  where: : $e^{-i z}$ denotes the complex exponential function : $\cos z$ denotes the complex cosine function : $\sin z$ denotes complex sine function : $i$ denotes the imaginary unit. === Corollary === This result is often presented and proved separately for arguments in the real domain: {{:Euler's Formula/Real Domain/Corollary}} == Proof == {{begin-eqn}} {{eqn | l = e^{-i z}       | r = \cos \paren {-z} + i \sin \paren {-z}       | c = Euler's Formula }} {{eqn | r = \cos z + i \sin \paren {-z}       | c = Cosine Function is Even }} {{eqn | r = \cos z - i \sin z       | c = Sine Function is Odd }} {{end-eqn}}  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Sine Function is Odd|next = Cosine Exponential Formulation/Proof 3}}: $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$: $\text{(ii)}$: $(4.16)$
Complex Natural Logarithm/Examples/i
== Example of Complex Natural Logarithm ==  :$\ln \paren i = \paren {4 k + 1} \dfrac {\pi i} 2$ for all $k \in \Z$.  == Proof == {{begin-eqn}} {{eqn | l = i       | r = \exp \paren {\dfrac {i \pi} 2}       | c = Polar Form of  $i$ }} {{eqn | ll= \leadsto       | l = \ln \paren i       | r = \ln \paren {\exp \paren {\dfrac {i \pi} 2 + 2 k \pi i} }       | c =  }} {{eqn | r = \dfrac {i \pi + 4 k \pi i} 2       | c = {{Defof|Complex Natural Logarithm}} }} {{eqn | r = \paren {4 k + 1} \dfrac {\pi i} 2       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Complex Natural Logarithm/Examples/-2|next = Complex Cosine Function/Examples/4 cos z = 3+i}}: $\S 4.6$. The Logarithm: Examples: $\text {(ii)}$
Square Root of Complex Number in Cartesian Form/Examples/-8+6i
== Example of Square Root of Complex Number in Cartesian Form ==  :$\sqrt {-8 + 6 i} = \pm \paren {1 + 3 i}$  == Proof == {{begin-eqn}} {{eqn | l = \paren {x + i y}^2       | r = -8 + 6 i       | c =  }} {{eqn | ll= \leadsto       | l = x^2       | r = \dfrac {-8 + \sqrt {\paren {-8}^2 + 6^2} } 2       | c = Square Root of Complex Number in Cartesian Form }} {{eqn | r = \dfrac {-8 + \sqrt {100} } 2       | c =  }} {{eqn | r = \dfrac {-8 + 10} 2       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \pm 1       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \pm \dfrac 6 {2 \times 1}       | c =  }} {{eqn | r = \pm 3       | c =  }} {{end-eqn}} As $2 x y = 6$ it follows that the two solutions are: :$1 + 3 i$ :$-1 - 3 i$  == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Complex Cosine Function/Examples/4 cos z = 3+i|next = Inverse Tangent of i}}: $\S 4.6$. The Logarithm: Examples: $\text {(iii)}$
Proof by Contraposition
Proof by Contraposition 0 70942 370643 370486 2018-10-12T07:57:47Z KarlFrei 3474 fixed this as well wikitext text/x-wiki == Proof Technique == '''Proof by contraposition''' is a rule of inference used in proofs.   This rule infers a conditional statement from its contrapositive. It is based on the Rule of Transposition, which says that a conditional statement and its contrapositive have the same truth value: :$p \implies q \dashv \vdash \neg q \implies \neg p$ In other words, the conclusion "if A, then B" is drawn from the single premise "if not B, then not A."  == Explanation == '''Proof by Contraposition''' can be expressed in natural language as follows: If we know that by making an assumption  :$\neg q$  we can deduce  :$\neg p$ then it must be the case that  :$p \implies q$. Thus it provides a means of proving a logical implication. This proof is often confused with Reductio ad Absurdum, which also starts with an assumption $\neg q$. Reductio ad Absurdum itself is often confused with Proof by Contradiction. Unlike Reductio ad Absurdum however, Proof by Contraposition ''can'' be a valid proof in intuitionistic logic, just like Proof by Contradiction. Specifically, suppose :$p \implies q$ is true.  Suppose furthermore that we have a proof of :$\neg q$. Then if we had a proof of $p$, it could be turned into a proof of $q$. This would imply :$q\land \neg q$ which is impossible. Therefore :$\neg p$. However, now suppose :$\neg q \implies \neg p$ is true. Suppose furthermore that we have a proof of :$p$. Then if we had a proof of $\neg q$, it could be turned into a proof of $\neg p$. This would imply :$p\land\neg p$ which is impossible. Thus it is not possible to prove $\neg q$. That means in this case we only know :$\neg \neg q$.