hcju/proofwikiqa huggingface dataset:
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 = {{Defof|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 = {{Defof|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}}
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}}
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}}
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 $P \paren {\zeta_1} = P \paren {\zeta_2} = \ldots = P \paren {\zeta_n} = 0$, then: : $\displaystyle P \paren z = k \prod_{j \mathop = 1}^n \paren {z - \zeta_j}$ 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}}
{{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}}
From Power of Complex Number minus 1: Corollary: :$\displaystyle \sum_{k \mathop = 0}^{n - 1} z^k = \prod_{k \mathop = 1}^{n - 1} \paren {z - \alpha^k}$ The result follows by setting $z = 1$. {{qed}}
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 $s = \dfrac n 2$. and in fact $\alpha^{n / 2}$ {{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}}
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 = {{NormAxiom|3}} }} {{eqn | r = \norm {x_n \paren {y_n - y_m } } + \norm {\paren {x_n - x_m} y_m}       | c =  }} {{eqn | o = \le       | r = \norm {x_n} \cdot \norm {y_n - y_m} + \norm {x_n - x_m} \cdot \norm {y_m}       | c = {{NormAxiom|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}}
From Power of Complex Number minus 1: :$z^{2 n + 1} - 1 = \displaystyle \prod_{k \mathop = 0}^{2 n} \paren {z - \alpha^k}$ 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 \cos \dfrac {2 k \pi} {2 n + 1} z + 1       | c = simplification }} {{end-eqn}} Hence the result. {{qed}}
From Power of Complex Number minus 1: :$z^{2 n} - 1 = \displaystyle \prod_{k \mathop = 0}^{2 n - 1} \paren {z - \alpha^k}$ where: {{begin-eqn}} {{eqn | l = \alpha       | r = e^{2 i \pi / \paren {2 n} }       | c =  }} {{eqn | r = \cos \dfrac {2 \pi} {2 n} + i \sin \dfrac {2 \pi} {2 n}       | c =  }} {{eqn | r = \cos \dfrac \pi n + i \sin \dfrac \pi n       | c =  }} {{end-eqn}} From Complex Roots of Unity occur in Conjugate Pairs: :$U_{2 n} = \set {1, \tuple {\alpha, \alpha^{2 n - 1} }, \tuple {\alpha^2, \alpha^{2 n - 2} }, \ldots, \tuple {\alpha^k, \alpha^{2 n - k} }, \ldots, \tuple {\alpha^{n - 1}, \alpha^{n + 1} }, -1}$ where $U_{2 n}$ denotes the complex $2 n$th roots of unity: :$U_{2 n} = \set {z \in \C: z^{2 n} = 1}$ The case $k = 0$ is taken care of by setting $\alpha^0 = 1$, from whence we have the factor $z - 1$. The case $k = n$ is taken care of by setting $\alpha^k = -1$, from whence we have the factor $z + 1$. Taking the product of each of the remaining factors of $z^{2 n} - 1$: {{begin-eqn}} {{eqn | l = \paren {z - \alpha^k} \paren {z - \alpha^{2 n - k} }       | 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 {k \pi} n + i \sin \dfrac {k \pi} n + \cos \dfrac {k \pi} n - i \sin \dfrac {k \pi} n} + 1       | c = Definition of $\alpha$ }} {{eqn | r = z^2 - 2 \cos \dfrac {k \pi} n z + 1       | c = simplification }} {{end-eqn}} Hence the result. {{qed}}
From $z^n - a = 0$ we have that: :$a = z^n$ Let $b = a^{1 / n}$, hence $b^n = a$, with $a, b \in \C$. From Roots of Complex Number: {{begin-eqn}} {{eqn | l = z^{1 / n}       | r = \set {a^{1 / n} e^{i \paren {\theta + 2 k \pi} / n}: k \in \set {0, 1, 2, \ldots, n - 1}, \theta = \arg a}       | c = }} {{eqn | r = \set {a^{1 / n} e^{2 i k \pi / n}: k \in \set {0, 1, 2, \ldots, n - 1} }       | c = $\theta = 0$ }} {{eqn | r = \set {b e^{2 i k \pi / n}: k \in \set {0, 1, 2, \ldots, n - 1} }       | c = $b = a^{1 / n}$ }} {{end-eqn}} and so each of $b e^{2 i k \pi / n}$ is a root of $z^n - a$. From First Complex Root of Unity is Primitive: For each $n \in \Z_{>0}$ there exists at least one primitive complex $n$th root of unity, $\alpha$, hence each of $b \alpha^k$ is a unique root of $z^n - a$. From the corollary to the Polynomial Factor Theorem: If $\zeta_1, \zeta_2, \ldots, \zeta_n \in \C$ such that all are different, and $\map P {\zeta_1} = \map P {\zeta_2} = \cdots = \map P {\zeta_n} = 0$, then: :$\displaystyle \map P z = k \prod_{j \mathop = 1}^n \paren {z - \zeta_j}$ where $k \in \C$. $z^n - a$ is a monic polynomial, hence $k = 1$ in the above product. Choose $\zeta_j = b \alpha^{j - 1}$ and we have the desired result. {{qed}}
{{begin-eqn}} {{eqn | l = z^6 + z^3 + 1       | r = \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}       | c = Complex Algebra Examples: $z^6 + z^3 + 1$ }} {{eqn | ll= \leadsto       | l = z^3 + z^0 + z^{-3}       | r = \paren {z - 2 \cos \dfrac {2 \pi} 9 + z^{-1} } \paren {z - 2 \cos \dfrac {4 \pi} 9 + z^{-1} } \paren {z - 2 \cos \dfrac {8 \pi} 9 + z^{-1} }       | c =  }} {{end-eqn}} Setting $z = e^{i \theta}$: {{begin-eqn}} {{eqn | l = e^{3 i \theta} + 1 + e^{-3 i \theta}       | r = \paren {e^{i \theta} - 2 \cos \dfrac {2 \pi} 9 + e^{-i \theta} } \paren {e^{i \theta} - 2 \cos \dfrac {4 \pi} 9 + e^{-i \theta} } \paren {e^{i \theta} - 2 \cos \dfrac {8 \pi} 9 + e^{-i \theta} }       | c =  }} {{eqn | ll= \leadsto       | l = 2 \dfrac {e^{3 i \theta} + e^{-3 i \theta} } 2 + 1       | r = \paren {2 \dfrac {e^{i \theta} + e^{-i \theta} } 2 - 2 \cos \dfrac {2 \pi} 9} \paren {2 \dfrac {e^{i \theta} + e^{-i \theta} } 2 - 2 \cos \dfrac {4 \pi} 9} \paren {2 \dfrac {e^{i \theta} + e^{-i \theta} } 2 - 2 \cos \dfrac {8 \pi} 9}       | c = rearranging }} {{eqn | ll= \leadsto       | l = 2 \cos 3 \theta + 1       | r = \paren {2 \cos \theta - 2 \cos \dfrac {2 \pi} 9} \paren {2  \cos \theta - 2 \cos \dfrac {4 \pi} 9} \paren {2  \cos \theta - 2 \cos \dfrac {8 \pi} 9}       | c = Cosine Exponential Formulation }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos 144 \degrees       | r = \cos \paren {180 \degrees - 36 \degrees}       | c =  }} {{eqn | r = -\cos 36 \degrees       | c = Cosine of Supplementary Angle }} {{eqn | r = -\dfrac {1 + \sqrt 5} 4       | c = Cosine of $36 \degrees$ }} {{end-eqn}} {{qed}} Category:Cosine Function l55vy4qa8yqo65r0so44d63cb14qtz0
{{begin-eqn}} {{eqn | l = \sin 144 \degrees       | r = \sin \paren {180 \degrees - 36 \degrees}       | c =  }} {{eqn | r = \sin 36 \degrees       | c = Sine of Supplementary Angle }} {{eqn | r = \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8}       | c = Sine of $36 \degrees$ }} {{end-eqn}} {{qed}} Category:Sine Function 4fz8zck3sk6pp1v6vules5zy4kohf9u
Let $z = x + i y$. From $\Re \paren z = -\dfrac 1 2$: :$x = -\dfrac 1 2$ by definition of the real part of $z$. Then: {{begin-eqn}} {{eqn | l = \cmod z       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = x^2 + y^2       | r = 1       | c = {{Defof|Complex Modulus}} }} {{eqn | ll= \leadsto       | l = \paren {-\dfrac 1 2}^2 + y^2       | r = 1       | c = substituting for $x$ }} {{eqn | ll= \leadsto       | l = y       | r = \sqrt {1 - \dfrac 1 4}       | c = simplifying }} {{eqn | r = \pm \dfrac {\sqrt 3} 2       | c =  }} {{end-eqn}} Thus: :$z = -\dfrac 1 2 \pm \dfrac {\sqrt 3} 2$ and the result follows from Cube Roots of Unity. {{qed}}
From Sum of Cubes of Three Indeterminates Minus 3 Times their Product: :$x^3 + y^3 + z^3 - 3 x y z = \paren {x + y + z} \paren {x + \omega y + \omega^2 z} \paren {x + \omega^2 y + \omega z}$ Setting $x \gets 0, y \gets a, z \gets b$: :$0^3 + a^3 + b^3 - 3 \times 0 \times a b = \paren {0 + a + b} \paren {0 + \omega a + \omega^2 b} \paren {0 + \omega^2 a + \omega b}$ The result follows. {{qed}}
Multiplying out: {{begin-eqn}} {{eqn | l = \paren {a + b + c}^3       | r = \paren {a + b + c} \paren {a^2 + b^2 + c^2 + 2 a b + 2 a c + 2 b c}       | c =  }} {{eqn | n = 1       | r = a^3 + b^3 + c^3 + 3 a^2 b + 3 a^2 c + 3 a b^2 + 3 b^2 c + 3 a c^2 + 3 b c^2 + 6 a b c       | c =  }} {{end-eqn}} Replacing $b$ with $b \omega$ and $c$ with $c \omega^2$ in $(1)$: {{begin-eqn}} {{eqn | l = \paren {a + b \omega + c \omega^2}^3       | r = a^3 + b^3 \omega^3 + c^3 \omega^6 + 3 a^2 b \omega + 3 a^2 c \omega^2 + 3 a b^2 \omega^2 + 3 b^2 c \omega^4 + 3 a c^2 \omega^4 + 3 b c^2 \omega^5 + 6 a b c \omega^3       | c =  }} {{eqn | n = 2       | r = a^3 + b^3 + c^3 + 3 a^2 b \omega + 3 a^2 c \omega^2 + 3 a b^2 \omega^2 + 3 b^2 c \omega + 3 a c^2 \omega + 3 b c^2 \omega^2 + 6 a b c       | c = as $\omega^3 = 1$ }} {{end-eqn}} Replacing $b$ with $b \omega^2$ and $c$ with $c \omega$ in $(1)$: {{begin-eqn}} {{eqn | l = \paren {a + b \omega^2 + c \omega}^3       | r = a^3 + b^3 \omega^6 + c^3 \omega^3 + 3 a^2 b \omega^2 + 3 a^2 c \omega + 3 a b^2 \omega^4 + 3 b^2 c \omega^5 + 3 a c^2 \omega^2 + 3 b c^2 \omega^4 + 6 a b c \omega^3       | c =  }} {{eqn | n = 3       | r = a^3 + b^3 + c^3 + 3 a^2 b \omega^2 + 3 a^2 c \omega + 3 a b^2 \omega + 3 b^2 c \omega^2 + 3 a c^2 \omega^2 + 3 b c^2 \omega + 6 a b c       | c = as $\omega^3 = 1$ }} {{end-eqn}} Adding together $(1)$, $(2)$ and $(3)$: {{begin-eqn}} {{eqn | r = \paren {a + b + c}^3 + \paren {a + b \omega + c \omega^2}^3 + \paren {a + b \omega^2 + c \omega}^3       | o =        | c =  }} {{eqn | r = a^3 + b^3 + c^3 + 3 a^2 b + 3 a^2 c + 3 a b^2 + 3 b^2 c + 3 a c^2 + 3 b c^2 + 6 a b c       | c =  }} {{eqn | o =       | ro= +       | r = a^3 + b^3 + c^3 + 3 a^2 b \omega + 3 a^2 c \omega^2 + 3 a b^2 \omega^2 + 3 b^2 c \omega + 3 a c^2 \omega + 3 b c^2 \omega^2 + 6 a b c       | c =  }} {{eqn | o =       | ro= +       | r = a^3 + b^3 + c^3 + 3 a^2 b \omega^2 + 3 a^2 c \omega + 3 a b^2 \omega + 3 b^2 c \omega^2 + 3 a c^2 \omega^2 + 3 b c^2 \omega + 6 a b c       | c =  }} {{eqn | r = 3 \paren {a^3 + b^3 + c^3 + 6 a b c}       | c =  }} {{eqn | o =       | ro= +       | r = \paren {3 a^2 b + 3 a^2 c + 3 a b^2 + 3 b^2 c + 3 a c^2 + 3 b c^2} \paren {1 + \omega + \omega^2}       | c =  }} {{eqn | r = 3 \paren {a^3 + b^3 + c^3 + 6 a b c}       | c = Sum of Cube Roots of Unity: $1 + \omega + \omega^2 = 0$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z^8 + 1       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = z       | r = \paren {-1}^{1/8}       | c =  }} {{end-eqn}} From Euler's Identity: :$-1 = e^{i \pi}$ Let $b$ be defined as: {{begin-eqn}} {{eqn | l = b       | r = \sqrt [8] 1 \paren {\cos \dfrac \pi 8 + i \sin \dfrac \pi 8}       | c =  }} {{eqn | r = \cos \dfrac \pi 8 + i \sin \dfrac \pi 8       | c =  }} {{end-eqn}} From Roots of Complex Number: Corollary: :$\paren {-1}^{1/8} = \cos \paren {\dfrac \pi 8 + \dfrac {2 k \pi} 8} + i \sin \paren {\dfrac \pi 8 + \dfrac {2 k \pi} 8}$ for $k = 0$ to $7$. {{begin-eqn}} {{eqn | l = \paren {-1}^{1/8}       | r = \set {\cos \paren {\dfrac \pi 8 + \dfrac {2 k \pi} 8} + i \sin \paren {\dfrac \pi 8 + \dfrac {2 k \pi} 8}: k \in \set {0, 1, \ldots, 7} }       | c =  }} {{eqn | r = \set {\cos \dfrac {\paren {2 k + 1} \pi} 8 + i \sin \dfrac {\paren {2 k + 1} \pi} 8: k \in \set {0, 1, \ldots, 7} }       | c =  }} {{eqn | r = \set {\exp \dfrac {\paren {2 k + 1} i \pi} 8: k \in \set {0, 1, \ldots, 7} }       | c =  }} {{end-eqn}} {{qed}} === Illustration === {{:Roots of Complex Number/Examples/z^8 + 1 = 0/Illustration}}
{{begin-eqn}} {{eqn | l = z^8 + 1       | r = \prod_{k \mathop = 0}^3 \paren {z^2 - 2 z \cos \dfrac {\paren {2 k + 1} \pi} 8 + 1}       | c = Complex Algebra Examples: $z^8 + 1$ }} {{eqn | ll= \leadsto       | l = z^4 + z^{-4}       | r = \prod_{k \mathop = 0}^3 \paren {z - 2 \cos \dfrac {\paren {2 k + 1} \pi} 8 + z^{-1} }       | c = dividing both sides by $z^4$ }} {{end-eqn}} Setting $z = e^{i \theta}$: {{begin-eqn}} {{eqn | l = e^{4 i \theta} + e^{-4 i \theta}       | r = \prod_{k \mathop = 0}^3 \paren {e^{i \theta} - 2 \cos \dfrac {\paren {2 k + 1} i \pi} 8 + e^{-i \theta} }       | c =  }} {{eqn | ll= \leadsto       | l = 2 \dfrac {e^{4 i \theta} + e^{-4 i \theta} } 2       | r = \prod_{k \mathop = 0}^3 \paren {2 \frac {e^{i \theta} + e^{-i \theta} } 2 - 2 \cos \dfrac {\paren {2 k + 1} i \pi} 8}       | c = rearranging }} {{eqn | ll= \leadsto       | l = 2 \cos 4 \theta       | r = \prod_{k \mathop = 0}^3 \paren {2 \cos \theta - 2 \cos \dfrac {\paren {2 k + 1} i \pi} 8}       | c = Cosine Exponential Formulation }} {{eqn | ll= \leadsto       | l = \cos 4 \theta       | r = \prod_{k \mathop = 0}^3 \paren {\cos \theta - \cos \dfrac {\paren {2 k + 1} i \pi} 8}       | c = simplifying }} {{end-eqn}} {{qed}}
From Factorisation of $z^n - a$, setting $a = -1$: :$z^n + 1 = \displaystyle \prod_{k \mathop = 0}^{n - 1} \paren {z - \alpha^k b}$ :$\alpha$ is a primitive complex $n$th root of unity :$b$ is any complex number such that $b^n = a$. From Euler's Identity: :$-1 = e^{i \pi}$ From Exponential of Product: :$\paren {\exp \dfrac {i \pi} n}^n = e^{i \pi}$ and so: :$b = \exp \dfrac {i \pi} n$ We also have by definition of the first complex $n$th root of unity, and from First Complex Root of Unity is Primitive: :$\alpha = \exp \dfrac {2 i \pi} n$ Hence: {{begin-eqn}} {{eqn | l = z^n + 1       | r = \prod_{k \mathop = 0}^{n - 1} \paren {z - \paren {\exp \dfrac {2 i \pi} n}^k \exp \dfrac {i \pi} n}       | c =  }} {{eqn | r = \prod_{k \mathop = 0}^{n - 1} \paren {z - \exp \dfrac {2 k i \pi} n \exp \dfrac {i \pi} n}       | c = Exponential of Product }} {{eqn | r = \prod_{k \mathop = 0}^{n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} n}       | c = Exponential of Sum }} {{end-eqn}} {{qed}} Category:Algebra Category:Complex Roots 7cb9fm6tks4g9bdjzddoblnczho52mj