hcju/proofwikiqa huggingface dataset:
From Equation of Circle in Cartesian Plane: Corollary 1, the equation for a circle is: :$A \paren {x^2 + y^2} + B x + C y + D = 0$ provided that: : $B^2 + C^2 \ge 4 A D$ : $A > 0$. Thus: {{begin-eqn}} {{eqn | l = A \paren {x^2 + y^2} + B x + C y + D       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = A z \overline z + B x + C y + D       | r = 0       | c = Product of Complex Number with Conjugate }} {{eqn | ll= \leadsto       | l = A z \overline z + \frac B 2 \paren {z + \overline z} + C y + D       | r = 0       | c = Sum of Complex Number with Conjugate }} {{eqn | ll= \leadsto       | l = A z \overline z + \frac B 2 \paren {z + \overline z} + \frac C {2 i} \paren {z - \overline z} + D       | r = 0       | c = Difference of Complex Number with Conjugate }} {{eqn | ll= \leadsto       | l = A z \overline z + \paren {\frac B 2 + \frac C {2 i} } z + \paren {\frac B 2 - \frac C {2 i} } \overline z + D       | r = 0       | c = gathering terms }} {{end-eqn}} By setting: : $\alpha := A$, $\beta := \dfrac B 2 + \dfrac C {2 i}$ and $\gamma := D$ we have: :$\alpha z \overline z + \beta z + \overline \beta \overline z + \gamma = 0$ As $A, B, C, D \in \R$ it follows that both $\alpha$ and $\gamma$ are real. Then we have that $A > 0$ and so $\alpha > 0$. Then note that: {{begin-eqn}} {{eqn | l = \paren {\frac B 2 + \frac C {2 i} } \paren {\frac B 2 - \frac C {2 i} }       | r = \paren {\frac B 2}^2 + \paren {\frac C 2}^2       | c =  }} {{eqn | r = \frac {B^2 + C^2} 4       | c =  }} {{end-eqn}} Given that: :$B^2 + C^2 \ge 4 A D$ it follows that: :$\cmod \beta^2 > \alpha \gamma$ {{qed|lemma}} If $\alpha = 0$ and $\beta \ne 0$ the equation devolves to: :$\beta z + \overline \beta \overline z + \gamma = 0$ which from Equation of Line in Complex Plane: Formulation 1 is the equation of a straight line. The result follows. {{qed}}
{{begin-eqn}} {{eqn | l = \cmod {z - a}       | r = r       | c = Equation of Circle in Complex Plane: Formulation 1 }} {{eqn | ll= \leadsto       | l = \cmod {z - a}^2       | r = r^2       | c =  }} {{eqn | ll= \leadsto       | l = \paren {z - a} \overline {\paren {z - a} }       | r = r^2       | c = Modulus in Terms of Conjugate }} {{eqn | ll= \leadsto       | l = \paren {z - a} {\paren {\overline z - \overline a} }       | r = r^2       | c = Sum of Complex Conjugates }} {{eqn | ll= \leadsto       | l = z \overline z - a z - \overline a \overline z + a \overline a       | r = r^2       | c =  }} {{eqn | ll= \leadsto       | l = \alpha z \overline z - \alpha a z - \alpha \overline a \overline z + \alpha a \overline a - r^2       | r = 0       | c = where $\alpha \in \R_{>0}$ is arbitrary }} {{end-eqn}} By setting: : $\beta := -\alpha a$ and $\gamma = \alpha a \overline a - r^2$ we have: :$\alpha z \overline z + \beta z + \overline \beta \overline z + \gamma = 0$ We have that: {{begin-eqn}} {{eqn | l = r^2       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \alpha a \overline a - \gamma       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\alpha a} \paren {\alpha \overline a}       | o = >       | r = \alpha \gamma       | c =  }} {{eqn | ll= \leadsto       | l = \beta \overline \beta        | o = >       | r = \alpha \gamma       | c =  }} {{eqn | ll= \leadsto       | l = \cmod {\beta}^2       | o = >       | r = \alpha \gamma       | c = Modulus in Terms of Conjugate }} {{end-eqn}} {{qed|lemma}} If $\alpha = 0$ and $\beta \ne 0$ the equation devolves to: :$\beta z + \overline \beta \overline z + \gamma = 0$ which from Equation of Line in Complex Plane: Formulation 1 is the equation of a straight line. The result follows. {{qed}}
Let $z_1 z_2 z_3 z_4$ be a cyclic quadrilateral. By Geometrical Interpretation of Complex Subtraction, the four sides of $z_1 z_2 z_3 z_4$ can be defined as $z_1 - z_2$, $z_3 - z_2$, $z_3 - z_4$ and $z_1 - z_4$. :400px From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, the opposite angles of $z_1 z_2 z_3 z_4$ sum to $\pi$ radians. By Complex Multiplication as Geometrical Transformation, it follows that: {{begin-eqn}} {{eqn | l = \angle \, z_1       | r = \arg \paren {\dfrac {z_1 - z_2} {z_1 - z_4} }       | c =  }} {{eqn | l = \angle \, z_2       | r = \arg \paren {\dfrac {z_3 - z_2} {z_1 - z_2} }       | c =  }} {{eqn | l = \angle \, z_3       | r = \arg \paren {\dfrac {z_3 - z_4} {z_3 - z_2} }       | c =  }} {{eqn | l = \angle \, z_4       | r = \arg \paren {\dfrac {z_1 - z_4} {z_3 - z_4} }       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \angle \, z_1 + \angle \, z_3       | r = \pi       | c =  }} {{eqn | ll= \leadsto       | l = \arg \paren {\dfrac {z_1 - z_2} {z_1 - z_4} } + \arg \paren {\dfrac {z_3 - z_4} {z_3 - z_2} }       | r = \pi       | c =  }} {{eqn | ll= \leadsto       | l = \arg \paren {\paren {\dfrac {z_1 - z_2} {z_1 - z_4} } \paren {\dfrac {z_3 - z_4} {z_3 - z_2} } }       | r = \pi       | c = Argument of Product equals Sum of Arguments }} {{eqn | ll= \leadsto       | l = \operatorname {Im} \paren {\dfrac {\paren {z_1 - z_2} \paren {z_3 - z_4} } {\paren {z_1 - z_4} \paren {z_3 - z_2} } }       | r = 0       | c = Argument of Negative Real Number is Pi }} {{end-eqn}} Hence the result. {{qed}}
From Geometrical Interpretation of Complex Subtraction, the four sides of $UABC$ can be defined as: {{begin-eqn}} {{eqn | l = UA       | r = a - u       | c =  }} {{eqn | ll= \leadsto       | l = AU       | r = u - a       | c = (easier in this form) }} {{eqn | l = AB       | r = b - a       | c =  }} {{eqn | l = UV       | r = v - u       | c =  }} {{eqn | l = BV       | r = v - b       | c =  }} {{end-eqn}} Thus:  {{begin-eqn}} {{eqn | l = UA       | r = \lambda \paren {AB}       | c =  }} {{eqn | ll= \leadsto       | l = \cmod {a - u}       | r = \lambda \cmod {a - u}       | c =  }} {{eqn | l = BAU       | r = \alpha       | c =  }} {{eqn | ll= \leadsto       | l = \alpha + \arg \paren {b - a}       | r = \arg \paren {u - a}       | c =  }} {{eqn | ll= \leadsto       | l = u - a       | r = \lambda e^{i \alpha} \paren {b - a}       | c = Product of Complex Numbers in Exponential Form }} {{eqn | ll= \leadsto       | l = u       | r = a + q \paren {b - a}       | c = where $q = \lambda e^{i \alpha}$ }} {{eqn | r = \paren {1 - q} a + q b        | c =  }} {{end-eqn}} {{qed|lemma}} Then we have: {{begin-eqn}} {{eqn | l = v - u       | r = b - a       | c = Opposite Sides and Angles of Parallelogram are Equal }} {{eqn | ll= \leadsto       | l = v       | r = u + b - a       | c =  }} {{eqn | r = \paren {1 - q} a + q b + b - a       | c =  }} {{eqn | r = -q a + \paren {1 + q} b       | c =  }} {{end-eqn}} {{qed}}
By the geometry, the locus described by this equation is a circle of Apollonius. {{finish|etc.}}
First we note that the $p$-adic norm is a norm. Let $\nu_p$ denote the $p$-adic valuation on the rational numbers. Recall the definition of the $p$-adic norm: :$\forall q \in \Q: \norm q_p := \begin{cases}   0 & : q = 0 \\   p^{-\map {\nu_p} q} & : q \ne 0 \end{cases}$ We must show the following holds for all $x, y \in \Q$: :$\norm {x + y}_p \le \max \set {\norm x_p, \norm y_p}$ If $x = 0$ or $y = 0$, or $x + y = 0$, the result is trivial, as follows: Let $x = 0$. Then: {{begin-eqn}} {{eqn | l = x       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \norm x_p       | r = 0       | c = {{Defof|P-adic Norm|$p$-adic Norm}} }} {{eqn | ll= \leadsto       | l = \max \set {\norm x_p, \norm y_p}       | r = \norm y_p       | c = as $\norm y_p \ge 0 = \norm x_p$ from Norm Axioms: Axiom $(\text N 1)$ }} {{eqn | r = \norm {x + y}_p       | c =  }} {{end-eqn}} and so $\norm {x + y}_p \le \max \left( \norm x_p, \norm y_p \right)$ The same argument holds for $y = 0$. Let $x + y = 0$. {{begin-eqn}} {{eqn | l = x + y       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \norm {x + y}_p       | r = 0       | c = {{Defof|P-adic Norm|$p$-adic Norm}} }} {{eqn | o = \le       | r = \max \set {\norm x_p, \norm y_p}       | c = as $\norm x_p \ge 0$ and $\norm y_p \ge 0$ from Norm Axioms: Axiom $(\text N 1)$ }} {{end-eqn}} Let $x, y, x + y \in \Q_{\ne 0}$. From $p$-adic Valuation is Valuation: :$\map {\nu_p} {x + y} \ge \min \set {\map {\nu_p} x, \map {\nu_p} y}$ Then: {{begin-eqn}} {{eqn | l = \norm {x + y}_p       | r = p^{-\map {\nu_p} {x + y} }       | c = {{Defof|P-adic Norm|$p$-adic Norm}} }} {{eqn | o = \le       | r = \max \set {p^{- \map {\nu_p} x}, p^{-\map {\nu_p} y} } }} {{eqn | r = \max \set {\norm x_p, \norm y_p}       | c = {{Defof|P-adic Norm|$p$-adic Norm}} }} {{end-eqn}} {{qed}}
Let $A = \paren {x_a, y_a}, B = \paren {x_b, y_b}$. Let $X = \paren {x, y}$. We have: {{begin-eqn}} {{eqn | l = XA       | r = \lambda \paren {XB}       | c =  }} {{eqn | ll= \leadsto       | l = \sqrt {\paren {x - x_a}^2 + \paren {y - y_a}^2}       | r = \lambda \sqrt {\paren {x - x_b}^2 + \paren {y - y_b}^2}       | c = Pythagoras's Theorem }} {{eqn | ll= \leadsto       | l = \paren {x - x_a}^2 + \paren {y - y_a}^2       | r = \lambda^2 \paren {\paren {x - x_b}^2 + \paren {y - y_b}^2}       | c =  }} {{eqn | ll= \leadsto       | l = x^2 - 2 x x_a + {x_a}^2 + y^2 - 2 y y_a + {y_a}^2       | r = \lambda^2 \paren {x^2 - 2 x x_b + {x_b}^2 + y^2 - 2 y y_b + {y_b}^2}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\lambda^2 - 1} \paren {x^2 + y^2} + 2 x \paren {\lambda^2 x_b + x_a} + 2 y \paren {\lambda^2 y_b + y_a} + {x_a}^2 + {y_a}^2 - \lambda^2 \paren { {x_b}^2 + {y_b}^2}       | r = 0       | c =  }} {{end-eqn}} From Equation of Circle (Cartesian): Corollary 1, this is an equation in the form: :$A \left({x^2 + y^2}\right) + B x + C y + D = 0$ with radius $R$ and center $\left({a, b}\right)$, where: :$R = \dfrac 1 {2 A} \sqrt {B^2 + C^2 - 4 A D}$ :$\left({a, b}\right) = \left({\dfrac B {2 A}, \dfrac C {2 A} }\right)$ Here, we have: {{begin-eqn}} {{eqn | l = A       | r = \lambda^2 - 1       | c =  }} {{eqn | l = B       | r = 2 \paren {\lambda^2 x_b + x_a}       | c =  }} {{eqn | l = C       | r = 2 \paren {\lambda^2 y_b + y_a}       | c =  }} {{eqn | l = D       | r = {x_a}^2 + {y_a}^2 - \lambda^2 \paren { {x_b}^2 + {y_b}^2}       | c =  }} {{end-eqn}} Hence the center $\left({a, b}\right)$ can be evaluated: {{begin-eqn}} {{eqn | l = a       | r = \dfrac B {2 A}       | c =  }} {{eqn | r = \dfrac {2 \paren {\lambda^2 x_b + x_a} } {2 \paren {\lambda^2 - 1} }       | c =  }} {{eqn | r = \dfrac {\lambda^2 x_b + x_a} {\lambda^2 - 1}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = b       | r = \dfrac C {2 A}       | c =  }} {{eqn | r = \dfrac {2 \paren {\lambda^2 y_b + y_a} } {2 \paren {\lambda^2 - 1} }       | c =  }} {{eqn | r = \dfrac {\lambda^2 y_b + y_a} {\lambda^2 - 1}       | c =  }} {{end-eqn}} Then the radius can be evaluated: {{begin-eqn}} {{eqn | l = R       | r = \dfrac 1 {2 A} \sqrt {B^2 + C^2 - 4 A D}       | c =  }} {{eqn | r = \dfrac 1 {2 \paren {\lambda^2 - 1} } \sqrt {4 \paren {\lambda^2 x_b + x_a}^2 + 4 \paren {\lambda^2 y_b + y_a}^2 - 4 \paren {\lambda^2 - 1} \paren { {x_a}^2 + {y_a}^2 - \lambda^2 \paren { {x_b}^2 + {y_b}^2} } }       | c =  }} {{end-eqn}} {{finish|Good luck with that.}} Category:Circle of Apollonius jt9h2bpcth5z3yw0km1x9k4dvads2t2
{{begin-eqn}} {{eqn | l = \cos 6 \theta + i \sin 6 \theta       | r = \paren {\cos \theta + i \sin \theta}^6       | c = De Moivre's Formula }} {{eqn | r = \paren {\cos \theta}^6 + \binom 6 1 \paren {\cos \theta}^5 \paren {i \sin \theta} + \binom 6 2 \paren {\cos \theta}^4 \paren {i \sin \theta}^2       | c = Binomial Theorem }} {{eqn | o =        | ro=+       | r = \binom 6 3 \paren {\cos \theta}^3 \paren {i \sin \theta}^3 + \binom 6 4 \paren {\cos \theta}^2 \paren {i \sin \theta}^4 + \binom 6 5 \paren {\cos \theta} \paren {i \sin \theta}^5 + \paren {i \sin \theta}^6 }} {{eqn | r = \cos^6 \theta + 6 i \cos^5 \theta \sin \theta - 15 \cos^4 \sin^2 \theta       | c = substituting for binomial coefficients }} {{eqn | o =        | ro=-       | r = 20 i \cos^3 \theta \sin^3 \theta + 15 \cos^2 \theta \sin^4 \theta + 6 i \cos \theta \sin^5 \theta - \sin^6 \theta       | c = and using $i^2 = -1$ }} {{eqn | n = 1       | r = \cos^6 \theta - 15 \cos^4 \sin^2 \theta + 15 \cos^2 \theta \sin^4 \theta - \sin^6 \theta }} {{eqn | o =        | ro=+       | r = i \paren {6 \cos^5 \theta \sin \theta - 20 \cos^3 \theta \sin^3 \theta + 6 \cos \theta \sin^5 \theta}       | c = rearranging }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \cos 6 \theta       | r = \cos^6 \theta - 15 \cos^4 \sin^2 \theta + 15 \cos^2 \theta \sin^4 \theta - \sin^6 \theta       | c = equating real parts in $(1)$ }} {{eqn | r = \cos^6 \theta - 15 \cos^4 \paren {1 - \cos^2 \theta} + 15 \cos^2 \theta \paren {1 - \cos^2 \theta}^2 \theta -  \paren {1 - \cos^2 \theta}^3       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 32 \cos^6 \theta - 48 \cos^4 \theta + 18 \cos^2 \theta - 1       | c = multiplying out and gathering terms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos 6 \theta + i \sin 6 \theta       | r = \paren {\cos \theta + i \sin \theta}^6       | c = De Moivre's Formula }} {{eqn | r = \paren {\cos \theta}^6 + \binom 6 1 \paren {\cos \theta}^5 \paren {i \sin \theta} + \binom 6 2 \paren {\cos \theta}^4 \paren {i \sin \theta}^2       | c = Binomial Theorem }} {{eqn | o =        | ro=+       | r = \binom 6 3 \paren {\cos \theta}^3 \paren {i \sin \theta}^3 + \binom 6 4 \paren {\cos \theta}^2 \paren {i \sin \theta}^4 + \binom 6 5 \paren {\cos \theta} \paren {i \sin \theta}^5 + \paren {i \sin \theta}^6 }} {{eqn | r = \cos^6 \theta + 6 i \cos^5 \theta \sin \theta - 15 \cos^4 \sin^2 \theta       | c = substituting for binomial coefficients }} {{eqn | o =        | ro=-       | r = 20 i \cos^3 \theta \sin^3 \theta + 15 \cos^2 \theta \sin^4 \theta + 6 i \cos \theta \sin^5 \theta - \sin^6 \theta       | c = and using $i^2 = -1$ }} {{eqn | n = 1       | r = \cos^6 \theta - 15 \cos^4 \sin^2 \theta + 15 \cos^2 \theta \sin^4 \theta - \sin^6 \theta }} {{eqn | o =        | ro=+       | r = i \paren {6 \cos^5 \theta \sin \theta - 20 \cos^3 \theta \sin^3 \theta  + 6 \cos \theta \sin^5 \theta}       | c = rearranging }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \map \sin {6 \theta}       | r = 6 \cos^5 \theta \sin \theta - 20 \cos^3 \theta \sin^\theta + 6 \cos \theta \sin^5 \theta       | c = equating imaginary parts in $(1)$ }} {{eqn | ll= \leadsto       | l = \dfrac {\map \sin {6 \theta} } {\sin \theta}       | r = 6 \cos^5 \theta - 20 \cos^3 \theta \sin^2 \theta + 6 \cos \theta \sin^4 \theta       | c =  }} {{eqn | r = 6 \cos^5 \theta - 20 \cos^3 \theta \paren {1 - \cos^2 \theta} + 6 \cos \theta \paren {1 - \cos^2 \theta}^2       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 32 \cos^5 \theta - 32 \cos^3 \theta + 6 \cos \theta       | c = multiplying out and gathering terms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \cos {\theta + n \pi}       | r = \cos \theta \cos n \pi - \sin \theta \sin n \pi       | c = Cosine of Sum }} {{eqn | r = \cos \theta \cos n \pi       | c = Sine of Integer Multiple of Pi }} {{eqn | r = \paren {-1}^n \cos \theta       | c = Cosine of Integer Multiple of Pi }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \sin {\theta + n \pi}       | r = \sin \theta \cos n \pi + \cos \theta \sin n \pi       | c = Sine of Sum }} {{eqn | r = \sin \theta \cos n \pi       | c = Sine of Integer Multiple of Pi }} {{eqn | r = \paren {-1}^n \sin \theta       | c = Cosine of Integer Multiple of Pi }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac {1 + \sin \theta + i \cos \theta} {1 + \sin \theta - i \cos \theta}       | r = \frac {\paren {1 + \sin \theta + i \cos \theta}^2} {\paren {1 + \sin \theta - i \cos \theta} \paren {1 + \sin \theta + i \cos \theta} }       | c =  }} {{eqn | r = \frac {\paren {1 + \sin \theta}^2 + 2 i \cos \theta \paren {1 + \sin \theta} - \cos^2 \theta} {\paren {1 + \sin \theta}^2 + \cos^2 \theta}       | c =  }} {{eqn | r = \frac {1 + 2 \sin \theta + \sin^2 \theta + 2 i \cos \theta \paren {1 + \sin \theta} - \cos^2 \theta} {1 + 2 \sin \theta + \sin^2 \theta + \cos^2 \theta}       | c =  }} {{eqn | r = \frac {1 + 2 \sin \theta + 2 \sin^2 \theta + 2 i \cos \theta \paren {1 + \sin \theta} - \paren {\cos^2 \theta + \sin^2 \theta} } {1 + 2 \sin \theta + \sin^2 \theta + \cos^2 \theta}       | c =  }} {{eqn | r = \frac {2 \sin \theta \paren {1 + \sin \theta} + 2 i \cos \theta \paren {1 + \sin \theta} } {2 + 2 \sin \theta}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \sin \theta + i \cos \theta       | c = simplification }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | r = \paren {1 + \sin \dfrac \pi 5 + i \cos \dfrac \pi 5}^5 + i \paren {1 + \sin \dfrac \pi 5 - i \cos \dfrac \pi 5}^5       | o =        | c =  }} {{eqn | r = \frac {\paren {1 + \sin \dfrac \pi 5 + i \cos \dfrac \pi 5}^5} {\paren {1 + \sin \dfrac \pi 5 - i \cos \dfrac \pi 5}^5} \paren {1 + \sin \dfrac \pi 5 - i \cos \dfrac \pi 5}^5 + i \paren {1 + \sin \dfrac \pi 5 - i \cos \dfrac \pi 5}^5       | c =  }} {{eqn | r = \paren {\sin \dfrac \pi 5 + i \cos  \dfrac \pi 5}^5 \paren {1 + \sin \dfrac \pi 5 - i \cos \dfrac \pi 5}^5 + i \paren {1 + \sin \dfrac \pi 5 - i \cos \dfrac \pi 5}^5       | c = Complex Division Examples: $\dfrac {1 + \sin \theta + i \cos \theta} {1 + \sin \theta - i \cos \theta}$ }} {{eqn | r = i^5 \paren {-i \sin \dfrac \pi 5 + \cos  \dfrac \pi 5}^5 \paren {1 + \sin \dfrac \pi 5 - i \cos \dfrac \pi 5}^5 + i \paren {1 + \sin \dfrac \pi 5 - i \cos \dfrac \pi 5}^5       | c = multiplying left hand term by $i^5 \times -i = 1$ }} {{eqn | r = i \paren {\cos \pi - i \sin \pi} \paren {1 + \sin \dfrac \pi 5 - i \cos \dfrac \pi 5}^5 + i \paren {1 + \sin \dfrac \pi 5 - i \cos \dfrac \pi 5}^5       | c = De Moivre's Formula }} {{eqn | r = i \paren {-1} \paren {1 + \sin \dfrac \pi 5 - i \cos \dfrac \pi 5}^5 + i \paren {1 + \sin \dfrac \pi 5 - i \cos \dfrac \pi 5}^5       | c = Sine of Straight Angle and Cosine of Straight Angle }} {{eqn | r = -i \paren {1 + \sin \dfrac \pi 5 - i \cos \dfrac \pi 5}^5 + i \paren {1 + \sin \dfrac \pi 5 - i \cos \dfrac \pi 5}^5       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}}
By definition of integer power: :$1^n = 1$ for all $n$. Hence the result. {{qed}}
We have that $1$ is a positive real number. The result follows from Existence and Uniqueness of Positive Root of Positive Real Number. {{qed}}
From Positive Real Complex Root of Unity, we have that $1$ is the only positive real number in $U_n$. {{AimForCont}} $z \in \R$ such that $z \in U_n$ and $z < 0$. From Odd Power of Negative Real Number is Negative, $z^n < 0$. But this contradicts the fact that $z_n = 1 > 0$. Hence by Proof by Contradiction it follows there is no negative real number $z$ such that $z^n = 1$. Thus $1$ is the only real number in $U_n$. {{qed}}
From Positive Real Complex Root of Unity, we have that $1$ is the only element of $U_n$ which is a positive real number. We note that $\paren {-1}^n = 1$ as $n$ is even. Thus $-1$ is also an element of $U_n$. Now let $z \in U_n$ such that $\cmod z \ne 1$. Let $z > 0$. From Positive Power Function on Non-negative Reals is Strictly Increasing it follows that: :$z < 1 \implies z^n < 1$ and: :$z > 1 \implies z^n > 1$ Let $z < 0$. From Positive Power Function on Negative Reals is Strictly Decreasing it follows that: :$z < -1 \implies z^n > 1$ and: :$z > -1 \implies z^n < 1$ That is, in all cases where $\cmod z \ne 1$ we have that $z^n \ne 1$. {{qed}}
{{begin-eqn}} {{eqn | l = z^n       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \cmod {z^n}       | r = \cmod 1       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \cmod z^n       | r = 1       | c = Power of Complex Modulus equals Complex Modulus of Power }} {{eqn | ll= \leadsto       | l = \cmod z       | r = 1       | c = Positive Real Complex Root of Unity }} {{end-eqn}} {{qed}}
From Condition for Complex Root of Unity to be Primitive: :$\gcd \set {n, k} = 1$ where: :$\alpha_k = \exp \paren {\dfrac {2 k \pi i} n}$ Here we have: :$k = 1$ and: :$\gcd \set {n, 1} = 1$ Hence the result. {{qed}}
From Roots of Unity under Multiplication form Cyclic Group, $\struct {U_n, \times}$ is a group. The result follows from Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order. {{qed}}
First we address the case where $n \divides s$. Then: {{begin-eqn}} {{eqn | l = s       | r = q n       | c = for some $q \in \Z_{>0}$ }} {{eqn | ll= \leadsto       | l = \alpha^{j s}       | r = \alpha^{j q n}       | c =  }} {{eqn | r = \paren {\alpha^n}^{j q}       | c =  }} {{eqn | r = 1^{j q}       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^{n - 1} \alpha^{j s}       | r = \sum_{j \mathop = 0}^{n - 1} 1       | c =  }} {{eqn | r = n       | c =  }} {{end-eqn}} Now let $n \nmid S$. Then: {{begin-eqn}} {{eqn | l = s       | r = q n + r       | c = for $0 < r < 1$       | cc= Division Theorem }} {{eqn | ll= \leadsto       | l = \alpha^s       | r = \alpha^{q n + r}       | c =  }} {{eqn | r = \paren {\alpha^n}^q \alpha^r       | c =  }} {{eqn | r = 1^q \alpha^r       | c =  }} {{eqn | r = \alpha^r       | c =  }} {{eqn | o = \ne       | r = 1       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^{n - 1} \alpha^{j s}       | r = \sum_{j \mathop = 0}^{n - 1} \alpha^{j r}       | c =  }} {{eqn | r = \dfrac {\alpha^{n r} - 1} {\alpha^r - 1}       | c = valid because $\alpha^r \ne 1$       | cc= Sum of Geometric Sequence }} {{eqn | r = \dfrac {\paren {\alpha^n}^r - 1} {\alpha^r - 1}       | c =  }} {{eqn | r = \dfrac {1^r - 1} {\alpha^r - 1}       | c =  }} {{eqn | r = \dfrac {1 - 1} {\alpha^r - 1}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{qed}}
From Difference of Two Cubes: :$x^3 - y^3 = \paren {x - y} \paren {x^2 + x y + y^2}$ The result follows directly by setting $y = 1$. {{qed}}
{{begin-eqn}} {{eqn | l = 1 + \omega + \omega^2       | r = 1 + \paren {-\dfrac 1 2 + \dfrac {\sqrt 3} 2} + \paren {-\dfrac 1 2 - \dfrac {\sqrt 3} 2}       | c = Cube Roots of Unity }} {{eqn | r = 1 - \frac 1 2 - \frac 1 2 + \dfrac {\sqrt 3} 2 - \dfrac {\sqrt 3} 2       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | r = \paren {x + \omega y + \omega^2 z} \paren {x + \omega^2 y + \omega z}       | o =        | c =  }} {{eqn | r = x^2 + \omega^2 x y + \omega x z + \omega x y + \omega^3 y^2 + \omega^2 y z + x \omega^2 z + \omega^4 y z + \omega^3 z^2       | c =  }} {{eqn | r = x^2 + y^2 + z^2 + \paren {\omega + \omega^2} x y + \paren {\omega + \omega^2} x z + \paren {\omega + \omega^2} y z       | c = $\omega^3 = 1$ by definition }} {{eqn | r = x^2 + y^2 + z^2 - x y - x z - y z       | c = Sum of Cube Roots of Unity: $\omega + \omega^2 = -1$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | o =        | r = \paren {x + y + z} \paren {x^2 + y^2 + z^2 + - x y - x z - y z}       | c =  }} {{eqn | r = x^3 + x y^2 + x z^2 - x^2 y - x^2 z - x y z       | c =  }} {{eqn | o =        | ro= +       | r = y x^2 + y^3 + y z^2 - x y^2 - x y z - y^2 z       | c =  }} {{eqn | o =        | ro= +       | r = z x^2 + z y^2 + z^3 - x y x - x z^2 - y z^2       | c =  }} {{eqn | r = x^3 + y^3 + z^3 - 3 x y z       | c =  }} {{end-eqn}} {{qed}}