avewright/proofwiki-math huggingface dataset:
Theorem: Positive Real Complex Root of Unity

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $U_n = \set {z \in \C: z^n = 1}$ be the set of complex $n$th roots of unity.
The only $x \in U_n$ such that $x \in \R_{>0}$ is:
:$x = 1$
That is, $1$ is the only complex $n$th root of unity which is a positive real number.

Proof:
We have that $1$ is a positive real number.
The result follows from Existence and Uniqueness of Positive Root of Positive Real Number.
{{qed}}
Theorem: Real Complex Roots of Unity for Odd Index

Let $n \in \Z_{>0}$ be a (strictly) positive integer such that $n$ is odd.
Let $U_n = \set {z \in \C: z^n = 1}$ be the set of complex $n$th roots of unity.
The only $x \in U_n$ such that $x \in \R$ is:
:$x = 1$
That is, $1$ is the only complex $n$th root of unity which is a real number.

Proof:
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}}
Theorem: Real Complex Roots of Unity for Even Index

Let $n \in \Z_{>0}$ be a (strictly) positive integer such that $n$ is even.
Let $U_n = \set {z \in \C: z^n = 1}$ be the set of complex $n$th roots of unity.
The only $x \in U_n$ such that $x \in \R$ are:
:$x = 1$ or $x \in -1$
That is, $1$ and $-1$ are the only complex $n$th roots of unity which are real number.

Proof:
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}}
Theorem: Modulus of Complex Root of Unity equals 1

Let $n \in \Z_{>0}$ be a (strictly) positive integer such that $n$ is even.
Let $U_n = \set {z \in \C: z^n = 1}$ be the set of complex $n$th roots of unity.
Let $z \in U_n$.
Then:
:$\cmod z = 1$
where $\cmod z$ denotes the modulus of $z$.

Proof:
{{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}}
Theorem: First Complex Root of Unity is Primitive

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $U_n$ denote the complex $n$th roots of unity:
:$U_n = \set {z \in \C: z^n = 1}$
Let $\alpha_1 = \exp \paren {\dfrac {2 \pi i} n}$ denote the first complex root of unity.
Then $\alpha_1$ is a primitive complex root of unity.

Proof:
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}}
Theorem: Powers of Primitive Complex Root of Unity form Complete Set

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $U_n$ denote the complex $n$th roots of unity:
:$U_n = \set {z \in \C: z^n = 1}$
Let $\alpha_k = \exp \paren {\dfrac {2 k \pi i} n}$ denote the $k$th complex root of unity.
Let $\alpha_k$ be a primitive complex root of unity.
Let $V_k = \set { {\alpha_k}^r: r \in \set {0, 1, \ldots, n - 1} }$.
Then:
: $V_k = U_n$
That is, $V_k = \set { {\alpha_k}^r: r \in \set {0, 1, \ldots, n - 1} }$ forms the complete set of complex $n$th roots of unity.

Proof:
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}}
Theorem: Sum of Powers of Primitive Complex Roots of Unity

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $U_n$ denote the complex $n$th roots of unity:
:$U_n = \set {z \in \C: z^n = 1}$
Let $\alpha = \exp \paren {\dfrac {2 k \pi i} n}$ denote a primitive complex $n$th root of unity.
Let $s \in \Z_{>0}$ be a (strictly) positive integer.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^{n - 1} \alpha^{j s}
      | r = 1 + \alpha^s + \alpha^{2 s} + \cdots + \alpha^{\paren {n - 1} s}
      | c = 
}}
{{eqn | r = \begin {cases} n & : n \divides s \\ 0 & : n \nmid s \end {cases}
      | c = 
}}
{{end-eqn}}
where:
:$n \divides s$ denotes that $n$ is a divisor of $s$
:$n \nmid s$ denotes that $n$ is not a divisor of $s$.

Proof:
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}}
Theorem: Difference of Two Powers/Examples/Difference of Two Cubes/Corollary

:$x^3 - 1 = \paren {x - 1} \paren {x^2 + x + 1}$

Proof:
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}}
Theorem: Sum of Cube Roots of Unity

Let $U_3 = \set {1, \omega, \omega^2}$ denote the Cube Roots of Unity.
Then:
:$1 + \omega + \omega^2 = 0$

Proof:
{{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}}
Theorem: Sum of Cube Roots of Unity

Let $U_3 = \set {1, \omega, \omega^2}$ denote the Cube Roots of Unity.
Then:
:$1 + \omega + \omega^2 = 0$

Proof:
From Sum of Powers of Primitive Complex Roots of Unity:
:$\ds \sum_{j \mathop = 0}^{n - 1} \alpha^{j s} = \begin {cases} n & : n \divides s \\ 0 & : n \nmid s \end {cases}$
Here we have that $n = 3$ and $s = 1$.
Thus $n$ is not a divisor of $s$.
Hence the result.
{{qed}}
Theorem: Sum of Cube Roots of Unity

Let $U_3 = \set {1, \omega, \omega^2}$ denote the Cube Roots of Unity.
Then:
:$1 + \omega + \omega^2 = 0$

Proof:
Observe:
{{begin-eqn}}
{{eqn | l = \paren {1 - \omega} \paren {1 + \omega + \omega^2}
      | r = 1 - \omega^3
      | c = Difference of Two Cubes
}}
{{eqn | r = 1 - 1
      | c = Cube Roots of Unity
}}
{{eqn | r = 0
}}
{{end-eqn}}
As $\omega \ne 1$, it follows:
:$1 + \omega + \omega^2 = 0$
{{qed}}
Theorem: Sum of Cubes of Three Indeterminates Minus 3 Times their Product

For indeterminates $x, y, z$:
:$x^3 + y^3 + z^3 - 3 x y z = \paren {x + y + z} \paren {x + \omega y + \omega^2 z} \paren {x + \omega^2 y + \omega z}$
where $\omega = -\dfrac 1 2 + \dfrac {\sqrt 3} 2$

Proof:
{{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}}
Theorem: Sum of Cubes of Three Indeterminates Minus 3 Times their Product

For indeterminates $x, y, z$:
:$x^3 + y^3 + z^3 - 3 x y z = \paren {x + y + z} \paren {x + \omega y + \omega^2 z} \paren {x + \omega^2 y + \omega z}$
where $\omega = -\dfrac 1 2 + \dfrac {\sqrt 3} 2$

Proof:
Consider the determinant:
:$\Delta = \begin {vmatrix} x & z & y \\ y & x & z \\ z & y & x \end {vmatrix}$
We have:
{{begin-eqn}}
{{eqn | l = \Delta
      | r = x \paren {x^2 - y z} - z \paren {y x - z^2} + y \paren {y^2 - x z}
      | c = Determinant of Order 3
}}
{{eqn | r = x^3 + y^3 + z^3 - 3 x y z
      | c = 
}}
{{end-eqn}}
Then we note that adding rows $2$ and $3$ to rows $1$ gives:
{{begin-eqn}}
{{eqn | l = \Delta
      | r = \begin {vmatrix} x + y + z & x + y + z & x + y + z \\ y & x & z \\ z & y & x \end {vmatrix}
      | c = Multiple of Row Added to Row of Determinant
}}
{{eqn | r = \paren {x + y + z} \paren {x^2 - y z} - \paren {x + y + z} \paren {y x - z^2} + \paren {x + y + z} \paren {y^2 - x z}
      | c = Determinant of Order 3
}}
{{eqn | r = \paren {x + y + z} \paren {\paren {x^2 - y z} - \paren {y x - z^2} + \paren {y^2 - x z} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x + y + z}
      | o = \divides
      | r = \Delta
      | c = 
}}
{{end-eqn}}
Let $\omega$ denote the complex cube root of unity:
:$\omega = -\dfrac 1 2 + \dfrac {\sqrt 3} 2$
Hence adding $\omega$ times row $2$ and $\omega^2$ times row $3$ to rows $1$:
{{begin-eqn}}
{{eqn | l = \Delta
      | r = \begin {vmatrix} x + \omega y + \omega^2 z & \omega x + \omega^2 y + z & \omega^2 x + y + \omega z \\ y & x & z \\ z & y & x \end {vmatrix}
      | c = Multiple of Row Added to Row of Determinant:
}}
{{eqn | r = \begin {vmatrix} x + \omega y + \omega^2 z & \omega \paren {x + \omega y + \omega^2 z} & \omega^2 \paren {x + \omega y + \omega^2 z} \\ y & x & z \\ z & y & x \end {vmatrix}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x + \omega y + \omega^2}
      | o = \divides
      | r = \Delta
      | c = expanding as above
}}
{{end-eqn}}
and adding $\omega^2$ times row $2$ and $\omega$ times row $3$ to rows $1$:
{{begin-eqn}}
{{eqn | l = \Delta
      | r = \begin {vmatrix} x + \omega^2 y + \omega z & \omega^2 x + \omega y + z & \omega x + y + \omega^2 z \\ y & x & z \\ z & y & x \end {vmatrix}
      | c = Multiple of Row Added to Row of Determinant
}}
{{eqn | r = \begin {vmatrix} x + \omega^2 y + \omega z & \omega^2 \paren {x + \omega^2 y + \omega z} & \omega^2 \paren {x + \omega^2 y + \omega z} \\ y & x & z \\ z & y & x \end {vmatrix}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x + \omega^2 y + \omega}
      | o = \divides
      | r = \Delta
      | c = 
}}
{{end-eqn}}
Thus we have $3$ divisors of $x^3 + y^3 + z^3 - 3 x y z$, which is a polynomial of degree $3$.
There can be no other divisors except for a constant.
By examining the coefficient of $x^3$, for example, the constant is seen to be $1$.
Hence the result.
{{qed}}
Theorem: Unit Vectors in Complex Plane which are Vertices of Equilateral Triangle

Let $\epsilon_1, \epsilon_2, \epsilon_3$ be complex numbers embedded in the complex plane such that:
:$\epsilon_1, \epsilon_2, \epsilon_3$ all have modulus $1$
:$\epsilon_1 + \epsilon_2 + \epsilon_3 = 0$
Then:
:$\paren {\dfrac {\epsilon_2} {\epsilon_1} }^3 = \paren {\dfrac {\epsilon_3} {\epsilon_2} }^2 = \paren {\dfrac {\epsilon_1} {\epsilon_3} }^2 = 1$

Proof:
We have that:
{{begin-eqn}}
{{eqn | l = \epsilon_1 + \epsilon_2 + \epsilon_3
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \epsilon_1 - \paren {-\epsilon_2}
      | r = -\epsilon_3
      | c = 
}}
{{end-eqn}}
Thus by Geometrical Interpretation of Complex Subtraction, $\epsilon_1$, $\epsilon_2$ and $\epsilon_3$ form the sides of a triangle.
As the modulus of each of $\epsilon_1$, $\epsilon_2$ and $\epsilon_3$ equals $1$, $\triangle \epsilon_1 \epsilon_2 \epsilon_3$ is equilateral.
:420px
By Complex Multiplication as Geometrical Transformation:
{{begin-eqn}}
{{eqn | l = \arg \epsilon_1
      | r = \arg \epsilon_2 \pm \dfrac {2 \pi} 3
      | c = Complex Multiplication as Geometrical Transformation
}}
{{eqn | ll= \leadsto
      | l = \arg \paren {\dfrac {\epsilon_2} {\epsilon_1} }
      | r = \pm \dfrac {2 \pi} 3
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\epsilon_2} {\epsilon_1}
      | r = \cos \dfrac {2 \pi} 3 \pm i \sin \dfrac {2 \pi} 3
      | c = 
}}
{{eqn | r = \omega \text { or } \overline \omega
      | c = where $\omega$, $\overline \omega$ are the complex cube roots of unity
}}
{{eqn | ll= \leadsto
      | l = \paren {\dfrac {\epsilon_2} {\epsilon_1} }^3
      | r = 1
      | c = 
}}
{{end-eqn}}
The same analysis can be done to the other two pairs of sides of $\triangle \epsilon_1 \epsilon_2 \epsilon_3$.
Hence the result.
{{qed}}
Theorem: Combination Theorem for Cauchy Sequences/Sum Rule

:$\sequence {x_n + y_n}$ is a Cauchy sequence.

Proof:
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon 2 > 0$.
Since $\sequence {x_n}$ is a Cauchy sequence, we can find $N_1$ such that:
:$\forall n, m > N_1: \norm{x_n - x_m} < \dfrac \epsilon 2$
Similarly, $\sequence {y_n}$ is a Cauchy sequence, we can find $N_2$ such that:
: $\forall n, m > N_2: \norm{y_n - y_m} < \dfrac \epsilon 2$
Now let $N = \max \set {N_1, N_2}$.
Then if $n, m > N$, both the above inequalities will be true.
Thus $\forall n, m > N$:
{{begin-eqn}}
{{eqn | l = \norm {\paren {x_n + y_n} - \paren {x_m + y_m} }
      | r = \norm {\paren {x_n - x_m} + \paren {y_n - y_m} }
      | c = 
}}
{{eqn | o = \le
      | r = \norm {x_n - x_m} + \norm {y_n - y_m}
      | c = Axiom $\text N 3$ of norm: Triangle Inequality
}}
{{eqn | o = <
      | r = \frac \epsilon 2 + \frac \epsilon 2 = \epsilon
      | c = 
}}
{{end-eqn}}
Hence:
:$\sequence {x_n + y_n}$ is a Cauchy sequences in $R$.
{{qed}}
Theorem: Power of Complex Number minus 1

Let $z \in \C$ be a complex number.
Then:
:$z^n - 1 = \ds \prod_{k \mathop = 0}^{n - 1} \paren {z - \alpha^k}$
where $\alpha$ is a primitive complex $n$th root of unity.

Proof:
Follows directly from the corollary to the Polynomial Factor Theorem:
If $\zeta_1, \zeta_2, \ldots, \zeta_n \in \C$ such that all are different, and $\map P {\zeta_1} = \map P {\zeta_2} = \ldots = \map P {\zeta_n} = 0$, then:
:$\ds \map P z = k \prod_{j \mathop = 1}^n \paren {z - \zeta_j}$
where $k \in \C$.
In this context, each of $\alpha^k$ is a primitive complex $n$th root of unity:
:$\paren {\alpha^k}^n = 1$
and the result follows.
{{qed}}
Theorem: Power of Complex Number minus 1/Corollary

Let $z \in \C$ be a complex number.
Then:
:$\ds \sum_{k \mathop = 0}^{n - 1} z^k = \prod_{k \mathop = 1}^{n - 1} \paren {z - \alpha^k}$
where $\alpha$ is a primitive complex $n$th root of unity.

Proof:
{{begin-eqn}}
{{eqn | l = z^n - 1
      | r = \prod_{k \mathop = 0}^{n - 1} \paren {z - \alpha^k}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {z^n - 1} {z - 1}
      | r = \prod_{k \mathop = 1}^{n - 1} \paren {z - \alpha^k}
      | c = as $\alpha^k = 1$ when $k = 0$
}}
{{eqn | ll= \leadsto
      | l = \sum_{k \mathop = 0}^{n - 1} z^k
      | r = \prod_{k \mathop = 1}^{n - 1} \paren {z - \alpha^k}
      | c = Sum of Geometric Sequence
}}
{{end-eqn}}
{{qed}}
Theorem: Product of Differences between 1 and Complex Roots of Unity

Let $\alpha$ be a primitive complex $n$th root of unity.
Then:
:$\ds \prod_{k \mathop = 1}^{n - 1} \paren {1 - \alpha^k} = n$

Proof:
From {{Corollary|Power of Complex Number minus 1}}:
:$\ds \sum_{k \mathop = 0}^{n - 1} z^k = \prod_{k \mathop = 1}^{n - 1} \paren {z - \alpha^k}$
The result follows by setting $z = 1$.
{{qed}}
Theorem: Complex Roots of Unity occur in Conjugate Pairs

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $U_n$ denote the complex $n$th roots of unity:
:$U_n = \set {z \in \C: z^n = 1}$
Let $\alpha \in U_n$ be the first complex $n$th root of unity.
Then:
:$\forall k \in \Z_{>0}, k < \dfrac n 2: \overline {\alpha^k} = \alpha^{n - k}$
That is, each of the complex $n$th roots of unity occur in conjugate pairs:
:$\tuple {\alpha, \alpha^{n - 1} }; \tuple {\alpha^2, \alpha^{n - 2} }; \ldots; \tuple {\alpha^s, \alpha^{n - s} }$
where:
:$s = \dfrac {n - 1} 2$ for odd $n$
:$s = \dfrac {n - 2} 2$ for even $n$.

Proof:
Consider the polynomial equation:
:$(1): \quad z^n - 1 = 0$
The complex $n$th roots of unity are:
:$1, \alpha, \alpha^2, \ldots, \alpha^{n - 1}$
From Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs, the roots of $(1)$ occur in conjugate pairs.
Let $k \in \Z$ such that $1 \le k \le n$.
Then:
{{begin-eqn}}
{{eqn | l = \alpha^{n - k}
      | r = \cos \frac {2 \paren {n - k} \pi} n + i \sin \frac {2 \paren {n - k} \pi} n
      | c = {{Defof|Complex Roots of Unity}}
}}
{{eqn | r = \cos \paren {\frac {2 n \pi} n  - \frac {2 k \pi} n} + i \sin \paren {\frac {2 n \pi} n  - \frac {2 k \pi} n}
      | c = 
}}
{{eqn | r = \cos \paren {2 \pi  - \frac {2 k \pi} n} + i \sin \paren {2 \pi  - \frac {2 k \pi} n}
      | c = 
}}
{{eqn | r = \cos \paren {-\frac {2 k \pi} n} + i \sin \paren {-\frac {2 k \pi} n}
      | c = Cosine of Angle plus Full Angle, Sine of Angle plus Full Angle
}}
{{eqn | r = \cos \paren {\frac {2 k \pi} n} - i \sin \paren {\frac {2 k \pi} n}
      | c = Cosine Function is Even, Sine Function is Odd
}}
{{eqn | r = \overline {\alpha^k}
      | c = {{Defof|Complex Conjugate}}
}}
{{end-eqn}}
That is, the complex $n$th root of unity which is the other half of the conjugate pair with $\alpha^k$ is $\alpha^{n - k}$.
When $n$ is odd, these pair up as:
:$\tuple {\alpha, \alpha^{n - 1} }; \tuple {\alpha^2, \alpha^{n - 2} }; \ldots; \tuple {\alpha^s, \alpha^{n - s} }$
where $s$ is the largest integer less than $\dfrac n 2$; that last pair can be expressed:
:$\tuple {\alpha^s, \alpha^{s + 1} }$
When $n$ is even:
:$\alpha^s = \alpha^{n - s}$
when $s = \dfrac n 2$.
Indeed:
{{begin-eqn}}
{{eqn | l = \alpha^{n / 2}
      | r = \cos \frac {2 \paren {n / 2} \pi} n + i \sin \frac {2 \paren {n / 2} \pi} n
      | c = {{Defof|Complex Roots of Unity}}
}}
{{eqn | r = \cos \paren {\frac {n \pi} n} + i \sin \paren {\frac {n \pi} n}
      | c = 
}}
{{eqn | r = \cos \pi + i \sin \pi
      | c = 
}}
{{eqn | r = -1
      | c = Cosine of Straight Angle, Sine of Straight Angle
}}
{{end-eqn}}
and so is wholly real.
Thus from Complex Number equals Conjugate iff Wholly Real:
:$\alpha^{n / 2} = \alpha^{n - n / 2}$
Hence the complex $n$th roots of unity pair up as:
:$\tuple {\alpha, \alpha^{n - 1} }; \tuple {\alpha^2, \alpha^{n - 2} }; \ldots; \tuple {\alpha^s, \alpha^{n - s} }$
where $s$ is the largest integer less than $\dfrac n 2$; that last pair can be expressed:
:$\tuple {\alpha^s, \alpha^{s + 2} }$
The result follows.
{{qed}}
Theorem: Combination Theorem for Cauchy Sequences/Product Rule

:$\sequence {x_n y_n}$ is a Cauchy sequence.

Proof:
Because $\sequence {x_n} $ is a Cauchy sequence, it is bounded by Cauchy Sequence is Bounded.
Suppose $\norm {x_n} \le K_1$ for $n = 1, 2, 3, \ldots$.
Because $\sequence {y_n} $ is a is a Cauchy sequence, it is bounded by Cauchy Sequence is Bounded.
Suppose $\norm {y_n} \le K_2$ for $n = 1, 2, 3, \ldots$.
Let $K = \max \set {K_1, K_2}$. 
Then both sequences are bounded by $K$.
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon {2K} > 0$.
Since $\sequence {x_n}$ is a Cauchy sequence, we can find $N_1$ such that:
:$\forall n, m > N_1: \norm {x_n - x_m} < \dfrac \epsilon {2 K}$
Similarly, $\sequence {y_n} $ is a Cauchy sequence, we can find $N_2$ such that:
:$\forall n, m > N_2: \norm {y_n - y_m} < \dfrac \epsilon {2 K}$
Now let $N = \max \set {N_1, N_2}$.
Then if $n, m > N$, both the above inequalities will be true.
Thus $\forall n, m > N$:
{{begin-eqn}}
{{eqn | l = \norm {x_n y_n - x_m y_m}
      | r = \norm {x_n y_n - x_n y_m + x_n y_m - x_m y_m}
      | c = 
}}
{{eqn | o = \le
      | r = \norm {x_n y_n - x_n y_m} + \norm {x_n y_m - x_m y_m}
      | c = {{Norm-axiom-mult|3}}
}}
{{eqn | r = \norm {x_n \paren {y_n - y_m } } + \norm {\paren {x_n - x_m} y_m}
      | c = 
}}
{{eqn | r = \norm {x_n} \cdot \norm {y_n - y_m} + \norm {x_n - x_m} \cdot \norm {y_m}
      | c = {{Norm-axiom-mult|2}}
}}
{{eqn | o = \le
      | r = K \cdot \norm {y_n - y_m} + \norm {x_n - x_m} \cdot K
      | c = as both sequences are bounded by $K$
}}
{{eqn | o = \le
      | r = K \cdot \dfrac \epsilon {2K} + \dfrac \epsilon {2K} \cdot K
      | c = 
}}
{{eqn | r = \dfrac \epsilon 2 + \dfrac \epsilon 2
      | c = 
}}
{{eqn | r = \epsilon
      | c = 
}}
{{end-eqn}}
Hence:
:$\sequence {x_n y_n}$ is a Cauchy sequence in $R$.
{{qed}}
Theorem: Factorisation of x^(2n+1)-1 in Real Domain

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
:$\ds z^{2 n + 1} - 1 = \paren {z - 1} \prod_{k \mathop = 1}^n \paren {z^2 - 2 z \cos \dfrac {2 \pi k} {2 n + 1} + 1}$

Proof:
From Power of Complex Number minus 1:
:$\ds z^{2 n + 1} - 1 = \prod_{k \mathop = 0}^{2 n} \paren {z - \alpha^k}$
where:
{{begin-eqn}}
{{eqn | l = \alpha
      | r = e^{2 i \pi / \paren {2 n + 1} }
      | c = 
}}
{{eqn | r = \cos \dfrac {2 \pi} {2 n + 1} + i \sin \dfrac {2 \pi} {2 n + 1}
      | c = 
}}
{{end-eqn}}
From Complex Roots of Unity occur in Conjugate Pairs:
:$U_{2 n + 1} = \set {1, \tuple {\alpha, \alpha^{2 n} }, \tuple {\alpha^2, \alpha^{2 n - 1} }, \ldots, \tuple {\alpha^k, \alpha^{2 n - k + 1} }, \ldots, \tuple {\alpha^n, \alpha^{n + 1} } }$
where $U_{2 n + 1}$ denotes the complex $2 n + 1$th roots of unity:
:$U_{2 n + 1} = \set {z \in \C: z^{2 n + 1} = 1}$
The case $n = 0$ is taken care of by setting $\alpha^0 = 1$, from whence we have the factor $z - 1$.
Taking the product of each of the remaining factors of $z^{2 n + 1} - 1$ in pairs:
{{begin-eqn}}
{{eqn | l = \paren {z - \alpha^k} \paren {z - \alpha^{2 n - k + 1} }
      | r = \paren {z - \alpha^k} \paren {z - \overline {\alpha^k} }
      | c = Complex Roots of Unity occur in Conjugate Pairs
}}
{{eqn | r = z^2 - z \paren {\alpha^k + \overline {\alpha^k} } + \alpha^k \overline {\alpha^k}
      | c = 
}}
{{eqn | r = z^2 - z \paren {\alpha^k + \overline {\alpha^k} } + \cmod {\alpha^k}^2
      | c = Modulus in Terms of Conjugate
}}
{{eqn | r = z^2 - z \paren {\alpha^k + \overline {\alpha^k} } + 1
      | c = Modulus of Complex Root of Unity equals 1
}}
{{eqn | r = z^2 - z \paren {\cos \dfrac {2 k \pi} {2 n + 1} + i \sin \dfrac {2 k \pi} {2 n + 1} + \cos \dfrac {2 k \pi} {2 n + 1} - i \sin \dfrac {2 k \pi} {2 n + 1} } + 1
      | c = Definition of $\alpha$
}}
{{eqn | r = z^2 - 2 z \cos \dfrac {2 k \pi} {2 n + 1} + 1
      | c = simplification
}}
{{end-eqn}}
Hence the result.
{{qed}}