hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \sin 3 x       | r = 3 \sin x - 4 \sin^3 x       | c = Triple Angle Formula for Sine }} {{eqn | ll = \implies       | l = 4 \sin^3 x       | r = 3 \sin x - \sin 3 x       | c = rearranging }} {{eqn | ll = \implies       | l = \sin^3 x       | r = \frac {3 \sin x - \sin 3 x} 4       | c = dividing both sides by $4$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sin^3 x       | r = \paren {\frac {e^{i x} - e^{-i x} } {2 i} }^3       | c = Sine Exponential Formulation }} {{eqn | r = \frac {\paren {e^{i x} - e^{-i x} }^3} {8 i^3}       | c = rearranging }} {{eqn | r = -\frac 1 {8 i} \paren {\paren {e^{i x} }^3 - 3 \paren {e^{i x} }^2 \paren {e^{-i x} } + 3 \paren {e^{i x} } \paren {e^{-i x} }^2 - \paren {e^{-i x} }^3}       | c = multiplying out }} {{eqn | r = -\frac 1 {8 i} \paren {e^{3 i x} - 3 e^{i x} + 3 e^{-i x} - e^{-3 i x} }       | c = simplifying }} {{eqn | r = \frac 3 4 \paren {\frac {e^{i x} - e^{-i x} } {2 i} } - \frac 1 4 \paren {\frac {e^{3 i x} - e^{-3 i x} } {2 i} }       | c = gathering terms }} {{eqn | r = \frac {3 \sin x - \sin 3 x} 4       | c = Sine Exponential Formulation }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos^4 x       | r = \left({\cos^2 x}\right)^2 }} {{eqn | r = \left({\frac {1 + \cos 2 x} 2}\right)^2       | c = Square of Cosine }} {{eqn | r = \frac {1 + 2 \cos 2 x + \cos^2 2 x} 4       | c = multiplying out }} {{eqn | r = \frac {1 + 2 \cos 2 x + \frac {1 + \cos 4 x} 2} 4       | c = Square of Cosine }} {{eqn | r = \frac {2 + 4 \cos 2 x + 1 + \cos 4 x} 8       | c = multiplying top and bottom by $2$ }} {{eqn | r = \frac {3 + 4 \cos 2 x + \cos 4 x} 8       | c = rearrangement }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos 4 x       | r = \paren {\frac {e^{i x} + e^{-i x} } 2}^4       | c = Cosine Exponential Formulation }} {{eqn | r = \frac {\paren {e^{i x} + e^{-i x} }^4} {16}       | c = rearranging }} {{eqn | r = \frac {\paren {e^{i x} }^4 + 4 \paren {e^{i x} }^3 \paren {e^{-i x} } + 6 \paren {e^{i x} }^2 \paren {e^{-i x} }^2 + 4 \paren {e^{i x} } \paren {e^{-i x} }^3 + \paren {e^{-i x} }^4} {16}       | c = multiplying out }} {{eqn | r = \frac {e^{4 i x} + 4 e^{2 i x} + 6 + 4 e^{-2 i x} + e^{-4 i x} } {16}       | c = multiplying out }} {{eqn | r = \frac {3 + 4 \paren {\dfrac {e^{2 i x} + e^{-2 i x} } 2} + \paren {\dfrac {e^{4 i x} + e^{-4 i x} } 2} } 8       | c = gathering terms }} {{eqn | r = \frac {3 + 4 \cos 2 x + \cos 4 x} 8       | c = Cosine Exponential Formulation }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sin 5 \theta       | r = \map \sin {3 \theta + 2 \theta} }} {{eqn | r = \sin 3 \theta \cos 2 \theta + \cos 3 \theta \sin 2 \theta       | c = Sine of Sum }} {{eqn | r = \paren {3 \sin \theta - 4 \sin^3 \theta} \cos 2 \theta + \paren {4 \cos^3 \theta - 3 \cos \theta} \sin 2 \theta       | c = Triple Angle Formulas }} {{eqn | r = \paren {3 \sin \theta - 4 \sin^3 \theta} \paren {\cos^2 \theta - \sin^2 \theta} + \paren {4 \cos^3 \theta - 3 \cos \theta} 2 \sin \theta \cos \theta       | c = Double Angle Formulas }} {{eqn | r = \paren {3 \sin \theta - 4 \sin^3 \theta} \paren {\cos^2 \theta - \sin^2 \theta} + 8 \cos^4 \theta \sin \theta - 6 \cos^2 \theta \sin \theta       | c = multiplying out }} {{eqn | r = \paren {3 \sin \theta - 4 \sin^3 \theta} \paren {1 - 2 \sin^2 \theta} + 8 \paren {1 - \sin^2 \theta}^2 \sin \theta - 6 \paren {1 - \sin^2 \theta} \sin \theta       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 5 \sin \theta - 20 \sin^3 \theta + 16 \sin ^5 \theta       | c = multiplying out and gathering terms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos 5 \theta       | r = \map \cos {4 \theta + \theta} }} {{eqn | r = \cos 4 \theta \cos \theta - \sin 4 \theta \sin \theta       | c = Cosine of Sum }} {{eqn | r = \paren {8 \cos^4 \theta - 8 \cos^2 \theta + 1} \cos \theta - \paren {4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta} \sin \theta       | c = Quadruple Angle Formulas }} {{eqn | r = 8 \cos^5 \theta - 8 \cos^3 \theta + \cos \theta - 4 \sin^2 \theta \cos \theta - 8 \sin^4 \theta \cos \theta       | c = multiplying out }} {{eqn | r = 8 \cos^5 \theta - 8 \cos^3 \theta + \cos \theta - 4 \paren {1 - \cos^2 \theta} \cos \theta - 8 \paren {1 - \cos^2 \theta}^2 \cos \theta       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta       | c = multiplying out and gathering terms }} {{end-eqn}} {{qed}}
We have: {{begin-eqn}} {{eqn | l = \cos 5 \theta + i \sin 5 \theta       | r = \paren {\cos \theta + i \sin \theta}^5       | c = De Moivre's Formula }} {{eqn | r = \paren {\cos \theta}^5 + \binom 5 1 \paren {\cos \theta}^4 \paren {i \sin \theta} + \binom 5 2 \paren {\cos \theta}^3 \paren {i \sin \theta}^2       | c = Binomial Theorem }} {{eqn | o =        | ro=+       | r = \binom 5 3 \paren {\cos \theta}^2 \paren {i \sin \theta}^3 + \binom 5 4 \paren {\cos \theta} \paren {i \sin \theta}^4 + \paren {i \sin \theta}^5 }} {{eqn | r = \cos^5 \theta + 5 i \cos^4 \theta \sin \theta - 10 \cos^3 \theta \sin^2 \theta       | c = substituting for binomial coefficients }} {{eqn | o =        | ro=-       | r = 10 i \cos^2 \theta \sin^3 \theta + 5 \cos \theta \sin^4 \theta + i \sin^5 \theta       | c = and using $i^2 = -1$ }} {{eqn | n = 1       | r = \cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta }} {{eqn | o =        | ro=+       | r = i \paren {5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta}       | c = rearranging }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \cos 5 \theta       | r = \cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta       | c = equating real parts in $(1)$ }} {{eqn | r = \cos^5 \theta - 10 \cos^3 \theta \paren {1 - \cos^2 \theta} + 5 \cos \theta \paren {1 - \cos^2 \theta}^2       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta       | c = multiplying out and gathering terms }} {{end-eqn}} {{qed}}
First note that when $\theta = 0, \pm \pi, \pm 2 \pi \ldots$: :$\sin \theta = 0$ so $\dfrac {\sin 5 \theta} {\sin \theta}$ is undefined. Therefore for the rest of the proof it is assumed that $\theta \ne 0, \pm \pi, \pm 2 \pi \ldots$ {{begin-eqn}} {{eqn | l = \sin 5 \theta       | r = 5 \sin \theta - 20 \sin^3 \theta + 16 \sin^5 \theta       | c = Quintuple Angle Formula for Sine }} {{eqn | ll= \leadsto       | l = \dfrac {\sin 5 \theta} {\sin \theta}       | r = 5 - 20 \paren {1 - \cos^2 \theta} + 16 \paren {1 - \cos^2 \theta}^2       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 16 \cos^4 \theta - 12 \cos^2 \theta + 1       | c = multiplying out and gathering terms }} {{end-eqn}} {{qed}}
We have: {{begin-eqn}} {{eqn | l = \cos 5 \theta + i \sin 5 \theta       | r = \paren {\cos \theta + i \sin \theta}^5       | c = De Moivre's Formula }} {{eqn | r = \paren {\cos \theta}^5 + \binom 5 1 \paren {\cos \theta}^4 \paren {i \sin \theta} + \binom 5 2 \paren {\cos \theta}^3 \paren {i \sin \theta}^2 }} {{eqn | o =        | ro=+       | r = \binom 5 3 \paren {\cos \theta}^2 \paren {i \sin \theta}^3 + \binom 5 4 \paren {\cos \theta} \paren {i \sin \theta}^4 + \paren {i \sin \theta}^5       | c = Binomial Theorem }} {{eqn | r = \cos^5 \theta + 5 i \cos^4 \theta \sin \theta - 10 \cos^3 \theta \sin^2 \theta       | c = substituting for binomial coefficients }} {{eqn | o =        | ro=-       | r = 10 i \cos^2 \theta \sin^3 \theta + 5 \cos \theta \sin^4 \theta + i \sin^5 \theta       | c = and using $i^2 = -1$ }} {{eqn | n = 1       | r = \cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta }} {{eqn | o =        | ro=+       | r = i \paren {5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta}       | c = rearranging }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \sin 5 \theta       | r = 5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta       | c = equating imaginary parts in $(1)$ }} {{eqn | r = 5 \paren {1 - \sin^2 \theta}^2 \sin \theta - 10 \paren {1 - \sin^2 \theta} \sin^3 \theta + \sin^5 \theta       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 5 \sin \theta - 20 \sin^3 \theta + 16 \sin^5 \theta       | c = multiplying out and gathering terms }} {{end-eqn}} {{qed}}
:500px Let $z = r e^{i \alpha}$. By Product of Complex Numbers in Exponential Form: :$z = r_1 r_2 e^{i \left({\theta_1 + \theta_2}\right)}$ Adding $\theta_2$ to $\theta_1$ is equivalent to rotation about the origin of $\C$ by $\theta_2$ in the positive direction. Similarly, the modulus of $z$ is obtained by multiplying the modulus of $z_1$ by $r_2$. {{qed}}
Let $z_1$ and $z_2$ be expressed in polar form. :$z_1 = \polar {r_1, \theta_1}$ :$z_2 = \polar {r_2, \theta_2}$ From Division of Complex Numbers in Polar Form: :$\dfrac {z_1} {z_2} = \dfrac {r_1} {r_2} \paren {\map \cos {\theta_1 - \theta_2} + i \, \map \sin {\theta_1 - \theta_2} }$ By the definition of argument: :$\map \arg {z_1} = \theta_1$ :$\map \arg {z_2} = \theta_2$ :$\map \arg {\dfrac {z_1} {z_2} } = \theta_1 - \theta_2$ There are $3$ possibilities for the size of $\theta_1 + \theta_2$: :$(1): \quad \theta_1 - \theta_2 > \pi$ Then: :$-\pi < \theta_1 - \theta_2 - 2 \pi \le \pi$ and we have: {{begin-eqn}} {{eqn | l = \map \cos {\theta_1 - \theta_2}       | r = \map \cos {\theta_1 - \theta_2 - 2 \pi}       | c = Cosine of Angle plus Full Angle }} {{eqn | l = \map \sin {\theta_1 - \theta_2}       | r = \map \sin {\theta_1 - \theta_2 - 2 \pi}       | c = Sine of Angle plus Full Angle }} {{end-eqn}} and so $\theta_1 + \theta_2 - 2 \pi$ is the argument of $\dfrac {z_1} {z_2}$ within its principal range. :$(2): \quad \theta_1 - \theta_2 \le -\pi$ Then: :$-\pi < \theta_1 - \theta_2 + 2 \pi \le \pi$ and we have: {{begin-eqn}} {{eqn | l = \map \cos {\theta_1 - \theta_2}       | r = \map \cos {\theta_1 - \theta_2 + 2 \pi}       | c = Cosine of Angle plus Full Angle }} {{eqn | l = \map \sin {\theta_1 - \theta_2}       | r = \map \sin {\theta_1 - \theta_2 + 2 \pi}       | c = Sine of Angle plus Full Angle }} {{end-eqn}} and so $\theta_1 - \theta_2 + 2 \pi$ is within the principal range of $\dfrac {z_1} {z_2}$. :$(3): \quad -\pi < \theta_1 + \theta_2 \le \pi$ Then $\theta_1 - \theta_2$ is already within the principal range of $\dfrac {z_1} {z_2}$. {{qed}}
By definition of the canonical form of a rational number, $p$ and $q$ are coprime. Substitute $\dfrac p q$ for $z$ in $P$ and multiply by $q^n$: :$(1): \quad a_n p^n + a_{n - 1} p^{n - 1} q + \cdots + a_1 p q^{n - 1} + a_0 q^n = 0$ Dividing $(1)$ by $p$ gives: :$(2): \quad a_n p^{n - 1} + a_{n - 1} p^{n - 2} q + \cdots + a_1 q^{n - 1} = -\dfrac {a_0 q^n} p$ The {{LHS}} of $(2)$ is an integer and therefore so is the {{RHS}}. We have that $p$ and $q$ are coprime. By Euclid's Lemma it follows that $p$ divides $a_0$. Similarly, dividing $(1)$ by $q$ gives: :$(3): \quad -\dfrac {a_n p^n} q = a_{n - 1} p^{n - 1} + \cdots + a_1 p q^{n - 2} + a_0 q^{n - 1}$ By Euclid's Lemma it follows that $q$ divides $a_n$. {{qed}}
By Isomorphism from R^n via n-Term Sequence, $V$ is isomorphic to the $K$-vector space $K^{\map \dim V}$. Thus: :$\size V = \size {K^{\map \dim V} }$ By Cardinality of Cartesian Space: :$\size {K^{\map \dim V} } = \size K^{\map \dim V}$ Thus: :$\size V = \size K^{\map \dim V}$ {{qed}} Category:Vector Spaces 9shqkcdbnuaf4de5rjyxlcmkv8y4r6g
Let the roots of $P$ be $z_1, z_2, \ldots, z_n$. Then $P$ can be written in factored form as: :$\displaystyle a_n \prod_{k \mathop = 1}^n \paren {z - z_k} = a_n \paren {z - z_1} \paren {z - z_2} \cdots \paren {z - z_n}$ Multiplying this out, $P$ can be expressed as: :$a_n \paren {z^n - \paren {z_1 + z_2 + \cdots + z_n} z^{n - 1} + \cdots + \paren {-1}^n z_1 z_2 \cdots z_n} = 0$ where the coefficients of $z^{n - 2}, z^{n - 3}, \ldots$ are more complicated and irrelevant. Equating powers of $z$, it follows that: :$-a_n \paren {z_1 + z_2 + \cdots + z_n} = a_{n - 1}$ from which: :$z_1 + z_2 + \cdots + z_n = - \dfrac {a_{n - 1} } {a_n}$ {{qed}}
From Viète's Formulas: :$\displaystyle a_{n - k} = \paren {-1}^k a_n \sum_{1 \mathop \le i_1 \mathop < \dotsb \mathop < i_k \mathop \le n} z_{i_1} \dotsm z_{i_k}$ for $k = 1, 2, \ldots, n$. The result follows for $k = 1$. {{qed}}
Let the roots of $P$ be $z_1, z_2, \ldots, z_n$. Then $P$ can be written in factored form as: :$\displaystyle a_n \prod_{k \mathop = 1}^n \paren {z - z_k} = a_0 \paren {z - z_1} \paren {z - z_2} \dotsm \paren {z - z_n}$ Multiplying this out, $P$ can be expressed as: :$a_n \paren {z^n - \paren {z_1 + z_2 + \dotsb + z_n} z^{n - 1} + \dotsb + \paren {-1}^n z_1 z_2 \dotsm z_n} = 0$ where the coefficients of $z^{n - 2}, z^{n - 3}, \ldots$ are more complicated and irrelevant. Equating powers of $z$, it follows that: :$a_n \paren {-1}^n z_1 z_2 \dotsm z_n = a_0$ from which: :$z_1 z_2 \dotsm z_n = \dfrac {\paren {-1}^n a_0} {a_n}$ {{qed}}
Let $\alpha = p + q i$. Let $p + q i$ be expressed in exponential form as $\alpha = r e^{i \theta}$. As $\alpha = r e^{i \theta}$ satisfies $\map f \alpha = 0$, it follows that: :$a_n r^n e^{n i \theta} + a_{n - 1} r^{n - 1} e^{\paren {n - 1} i \theta} + \dotsb + a_1 r e^{i \theta} + a_0 = 0$ Taking the conjugate of both sides: :$a_n r^n e^{-n i \theta} + a_{n - 1} r^{n - 1} e^{-\paren {n - 1} i \theta} + \dotsb + a_1 r e^{-i \theta} + a_0 = 0$ it follows that $\overline \alpha = p - q i$ is also a root of $f$. If any of the $a_k$ are complex, then the conjugate of $f$ is not the same polynomial as $f$. Therefore the result holds only for real coefficients. {{qed}}
Consider the equation: :$z^n - 1 = 0$ whose solutions are the complex roots of unity: :$1, e^{2 \pi i / n}, e^{4 \pi i / n}, e^{6 \pi i / n}, \ldots, e^{2 \paren {n - 1} \pi i / n}$ By Sum of Roots of Polynomial: :$1 + e^{2 \pi i / n} + e^{4 \pi i / n} + e^{6 \pi i / n} + \cdots + e^{2 \paren {n - 1} \pi i / n} = 0$ From Euler's Formula: :$e^{i \theta} = \cos \theta + i \sin \theta$ from which comes: :$\paren {1 + \cos \dfrac {2 \pi} n + \cos \dfrac {4 \pi} n + \cdots + \cos \dfrac {2 \paren {n - 1} \pi} n} + i \paren {\sin \dfrac {2 \pi} n + \sin \dfrac {4 \pi} n + \cdots + \sin \dfrac {2 \paren {n - 1} \pi} n} = 0$ Equating real parts: :$1 + \cos \dfrac {2 \pi} n + \cos \dfrac {4 \pi} n + \cdots + \cos \dfrac {2 \paren {n - 1} \pi} n = 0$ whence the result. {{qed}}
Consider the equation: :$z^n - 1 = 0$ whose solutions are the complex roots of unity: :$1, e^{2 \pi i / n}, e^{4 \pi i / n}, e^{6 \pi i / n}, \ldots, e^{2 \paren {n - 1} \pi i / n}$ By Sum of Roots of Polynomial: :$1 + e^{2 \pi i / n} + e^{4 \pi i / n} + e^{6 \pi i / n} + \cdots + e^{2 \paren {n - 1} \pi i / n} = 0$ From Euler's Formula: :$e^{i \theta} = \cos \theta + i \sin \theta$ from which comes: :$\paren {1 + \cos \dfrac {2 \pi} n + \cos \dfrac {4 \pi} n + \cdots + \cos \dfrac {2 \paren {n - 1} \pi} n} + i \paren {\sin \dfrac {2 \pi} n + \sin \dfrac {4 \pi} n + \cdots + \sin \dfrac {2 \paren {n - 1} \pi} n} = 0$ Equating imaginary parts: :$\sin \dfrac {2 \pi} n + \sin \dfrac {4 \pi} n + \cdots + \sin \dfrac {2 \paren {n - 1} \pi} n = 0$ {{qed}}
:400px From Area of Parallelogram: :$\AA = \text{base} \times \text{height}$ In this context: :$\text {base} = \cmod {z_2}$ and: :$\text {height} = \cmod {z_1} \sin \theta$ The result follows by definition of complex cross product. {{qed}}
:400px Let $A$, $B$ and $C$ be defined as complex numbers in the complex plane. The vectors from $C$ to $A$ and from $C$ to $B$ are given by: :$z_1 = \paren {x_1 - x_3} + i \paren {y_1 - y_3}$ :$z_2 = \paren {x_2 - x_3} + i \paren {y_2 - y_3}$ From Area of Triangle in Terms of Side and Altitude, $\mathcal A$ is half that of a parallelogram contained by $z_1$ and $z_2$. Thus: {{begin-eqn}} {{eqn | l = \mathcal A       | r = \frac 1 2 z_1 \times z_2       | c = Area of Parallelogram in Complex Plane }} {{eqn | r = \frac 1 2 \size {\paren {\map \Im {\paren {x_1 - x_3} - i \paren {y_1 - y_3} } \paren {\paren {x_2 - x_3} - i \paren {y_2 - y_3} } } }       | c = {{Defof|Vector Cross Product|subdef = Complex|index = 3}} }} {{eqn | r = \frac 1 2 \size {\paren {x_1 - x_3} \paren {y_2 - y_3} - \paren {y_1 - y_3} \paren {x_2 - x_3} }       | c = {{Defof|Complex Multiplication}} }} {{eqn | r = \frac 1 2 \size {x_1 y_2 - y_1 x_2 + x_2 y_3 - y_2 x_3 + x_3 y_1 - y_3 x_1}       | c = multiplying out }} {{eqn | r = \frac 1 2 \size {\paren {\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix} } }       | c = {{Defof|Determinant of Order 3}} }} {{end-eqn}} {{qed}}
From Equation of Straight Line in Plane, the equation for a straight line is: :$A x + B y + C = 0$ Thus: {{begin-eqn}} {{eqn | l = A x + B y + C       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \frac A 2 \paren {z + \overline z} + B y + C       | r = 0       | c = Sum of Complex Number with Conjugate }} {{eqn | ll= \leadsto       | l = \frac A 2 \paren {z + \overline z} + \frac B {2 i} \paren {z - \overline z} + C       | r = 0       | c = Difference of Complex Number with Conjugate }} {{eqn | ll= \leadsto       | l = \paren {\frac A 2 + \frac B {2 i} } z + \paren {\frac A 2 - \frac B {2 i} } \overline z + C       | r = 0       | c = gathering terms }} {{end-eqn}} The result follows by setting $\beta := \dfrac A 2 + \dfrac B {2 i}$ and $\gamma := C$. {{qed}}
:400px === Sufficient Condition === {{:Vertices of Equilateral Triangle in Complex Plane/Sufficient Condition}}{{qed|lemma}} === Necessary Condition === {{:Vertices of Equilateral Triangle in Complex Plane/Necessary Condition}}{{qed}}
Consider the equation: :$z^m - 1 = 0$ whose solutions are the complex roots of unity: :$1, e^{2 \pi i / m}, e^{4 \pi i / m}, e^{6 \pi i / m}, \ldots, e^{2 \paren {m - 1} \pi i / m}$ Then: {{begin-eqn}} {{eqn | l = z^m - 1       | r = \paren {z - 1} \paren {z - e^{2 \pi i / m} } \paren {z - e^{4 \pi i / m} } \dotsm \paren {z - e^{2 \paren {m - 1} \pi i / m} }       | c = product of all the roots }} {{eqn | ll= \leadsto       | l = \frac {z^m - 1} {z - 1}       | r = \prod_{k \mathop = 1}^{m - 1} \paren {z - e^{2 k \pi i / m} }       | c = dividing by $z - 1$ }} {{eqn | ll= \leadsto       | l = 1 + z + \dotsb + z^{m-1}       | r = \prod_{k \mathop = 1}^{m - 1} \paren {z - e^{2 k \pi i / m} }       | c = Sum of Geometric Sequence }} {{eqn | n = 1       | ll= \leadsto       | l = m       | r = \prod_{k \mathop = 1}^{m - 1} \paren {1 - e^{2 k \pi i / m} }       | c = setting $z = 1$ }} {{eqn | n = 2       | ll= \leadsto       | l = m       | r = \prod_{k \mathop = 1}^{m - 1} \paren {1 - e^{-2 k \pi i / m} }       | c = taking complex conjugate of both sides }} {{eqn | ll= \leadsto       | l = m^2       | r = \prod_{k \mathop = 1}^{m - 1} \paren {\paren {1 - e^{2 k \pi i / m} } \paren {1 - e^{-2 k \pi i / m} } }       | c = multiplying $(1)$ by $(2)$ }} {{eqn | ll= \leadsto       | l = m^2       | r = \prod_{k \mathop = 1}^{m - 1} \paren {1 - e^{2 k \pi i / m} - e^{-2 k \pi i / m} + 1}       | c = multiplying out }} {{eqn | ll= \leadsto       | l = m^2       | r = \prod_{k \mathop = 1}^{m - 1} \paren {2 - 2 \frac {e^{2 k \pi i / m} + e^{-2 k \pi i / m} } 2}       | c = rearranging }} {{eqn | ll= \leadsto       | l = m^2       | r = \prod_{k \mathop = 1}^{m-1} \paren {2 - 2 \cos \frac {2 k \pi} m}       | c = Cosine Exponential Formulation }} {{eqn | ll= \leadsto       | l = m^2       | r = 2^{m - 1} \prod_{k \mathop = 1}^{m - 1} \paren {1 - \cos \frac {2 k \pi} m}       | c = factoring out $m - 1$ instances of $2$ }} {{eqn | ll= \leadsto       | l = m^2       | r = 2^{m - 1} \prod_{k \mathop = 1}^{m - 1} \paren {2 \sin^2 \frac {k \pi} m}       | c = Double Angle Formula for Cosine: Corollary 2 }} {{eqn | ll= \leadsto       | l = \frac {m^2} {2^{2 m - 2} }       | r = \prod_{k \mathop = 1}^{m - 1} \paren {\sin^2 \frac {k \pi} m}       | c = factoring out another $m - 1$ instances of $2$ }} {{eqn | ll= \leadsto       | l = \frac m {2^{m - 1} }       | r = \prod_{k \mathop = 1}^{m - 1} \paren {\sin \frac {k \pi} m}       | c = taking square root of both sides }} {{end-eqn}} {{qed}}
Proof by induction: For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \overline {\prod_{j \mathop = 1}^n z_j} = \prod_{j \mathop = 1}^n \overline {z_j}$ $\map P 1$ is trivially true, as this just says $\overline {z_1} = \overline {z_1}$. === Basis for the Induction === $\map P 2$ is the case: :$\overline {z_1 z_2} = \overline {z_1} \cdot \overline {z_2}$ which has been proved in Product of Complex Conjugates. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \overline {\prod_{j \mathop = 1}^k z_j} = \prod_{j \mathop = 1}^k \overline {z_j}$ Then we need to show: :$\displaystyle \overline {\prod_{j \mathop = 1}^{k + 1} z_j} = \prod_{j \mathop = 1}^{k + 1} \overline {z_j}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \overline {\prod_{j \mathop = 1}^{k + 1} z_j}       | r = \overline {\paren {\prod_{j \mathop = 1}^k z_j} z_{k + 1} }       | c =  }} {{eqn | r = \overline {\paren {\prod_{j \mathop = 1}^k z_j} } \cdot \overline {z_{k + 1} }       | c = Basis for the Induction }} {{eqn | r = \paren {\prod_{j \mathop = 1}^k \overline {z_j} } \cdot \overline {z_{k + 1} }       | c = Induction Hypothesis }} {{eqn | r = \prod_{j \mathop = 1}^{k + 1} \overline {z_j}       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \N: \overline {\prod_{j \mathop = 1}^n z_j} = \prod_{j \mathop = 1}^n \overline {z_j}$ {{qed}}
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$, where $x_1, y_1, x_2, y_2 \in \R$. Then: {{begin-eqn}} {{eqn | l = \overline {\paren {\frac {z_1} {z_2} } }       | r = \overline {\paren {\frac {x_1 x_2 + y_1 y_2} { {x_2}^2 + {y_2}^2} + i \frac {x_2 y_1 - x_1 y_2} { {x_2}^2 + {y_2}^2} } }        | c = Division of Complex Numbers }} {{eqn | r = \frac {x_1 x_2 + y_1 y_2} { {x_2}^2 + {y_2}^2} - i \frac {x_2 y_1 - x_1 y_2} { {x_2}^2 + {y_2}^2}       | c = {{Defof|Complex Conjugate}} }} {{eqn | r = \frac {x_1 x_2 + \paren {-y_1} \paren {-y_2} } { {x_2}^2 + \paren {-y_2}^2} + i \frac {x_2 \paren {-y_1} - x_1 \paren {-y_2} } { {x_2}^2 + \paren {-y_2}^2}       | c =  }} {{eqn | r = \frac {x_1 - i y_1} {x_2 - i y_2}       | c = Division of Complex Numbers }} {{eqn | r = \frac {\paren {\overline {z_1} } } {\paren {\overline {z_2} } }       | c = {{Defof|Complex Conjugate}} }} {{end-eqn}} {{qed}}
:400px $ABCD$ is a parallelogram {{iff}}: :$\vec {AB} = \vec {DC}$ Then: {{begin-eqn}} {{eqn | l = \vec {AB}       | r = \vec {DC}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = z_2 - z_1       | r = z_3 - z_4       | c =  }} {{eqn | ll= \leadstoandfrom       | l = z_1 - z_2 + z_3 - z_4       | r = 0       | c =  }} {{end-eqn}} {{qed}}
The diagonals of $ABCD$ bisect each other if the position vectors of the midpoints of the diagonals are the same point. Let $z_1, z_2, z_3, z_4$ be the position vectors of the vertices of $ABCD$. Thus: {{begin-eqn}} {{eqn | l = z_1 + \frac {z_3 - z_1} 2       | r = z_2 + \frac {z_4 - z_2} 2       | c = condition for bisection }} {{eqn | ll= \leadsto       | l = \frac {z_1 + z_3} 2       | r = \frac {z_2 + z_4} 2       | c =  }} {{eqn | ll= \leadsto       | l = z_1 + z_3       | r = z_2 + z_4       | c =  }} {{eqn | ll= \leadsto       | l = z_1 - z_2 + z_3 - z_4       | r = 0       | c =  }} {{end-eqn}} The result follows from Condition for Points in Complex Plane to form Parallelogram. {{qed}}