hcju/proofwikiqa huggingface dataset:
We have that: From Newton-Girard Formulas: :$\displaystyle \sum_{a \mathop \le j_1 \mathop < \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\left({-S_2}\right)^{k_2} } {2^{k_2} k_2 !} \cdots \dfrac {\left({\left({-1}\right)^{m + 1} S_m}\right)^{k_m} } {m^{k_m} k_m !}$ where: :$S_r = \displaystyle \sum_{k \mathop = a}^b {x_k}^r$ for $r \in \Z_{\ge 0}$. Setting $m = 1$, and setting $x_i := x_{j_1}$: {{begin-eqn}} {{eqn | l = \sum_{a \mathop \le i \mathop \le b} x_{j_1}       | r = \sum_{\substack {k_1, k_2 \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop = 2} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !}       | c =  }} {{eqn | r = \dfrac { {S_1}^1} {1^1 \times 1!}        | c = as $k_1 = 1$ is the only $k_1$ to fulfil the criteria }} {{eqn | r = S_1       | c = simplifying }} {{end-eqn}} {{qed}}
=== Outline === The proof is divided into three cases: $k < n$, $k=n$ and $k>n$. The tools are Viete's Formulas, symmetric function recursion, telescoping sums and  homogeneous functions of degree $k$.
Let $G \left({z}\right)$ be the generating function for the sequence $\left\langle{h_m}\right\rangle$. {{begin-eqn}} {{eqn | l = \sum_{m \mathop \ge 1} \dfrac {S_m z^m} m       | r = \ln \left({G \left({z}\right)}\right)       | c = Summation of Products of n Numbers taken m at a time with Repetitions: Lemma 2 }} {{eqn | r = \ln \left({1 + h_1 z + h_2 z^2 + \cdots}\right)       | c =  }} {{eqn | r = \sum_{k \mathop \ge 1} \dfrac {\left({-1}\right)^{k - 1} } k \left({h_1 z + h_2 z^2 + \cdots}\right)^k       | c = Power Series Expansion for Logarithm of 1 + x }} {{eqn | ll= \leadsto       | l = S_m       | r = m \sum_{k \mathop \ge 1} \dfrac {\left({-1}\right)^{k - 1} } k \left({\sum_{j \mathop = 1} h_j z^j}\right)^k       | c =  }} {{end-eqn}} {{finish|Not sure where this is going}}
{{begin-eqn}} {{eqn | l = \map G {w, z}       | r = \sum_{m, \, n \mathop \ge 0} a_{m n} w^m z^n       | c = {{Defof|Generating Function for Doubly Subscripted Sequence}} }} {{eqn | r = \sum_{m, \, n \mathop \ge 0} \dbinom n m w^m z^n       | c =  }} {{eqn | r = \sum_{n \mathop \ge 0} \paren {1 + w}^n z^n       | c = Binomial Theorem }} {{eqn | r = \sum_{n \mathop \ge 0} \paren {\paren {1 + w} z}^n       | c =  }} {{eqn | r = \dfrac 1 {1 - \paren {1 + w} z}       | c = Sum of Infinite Geometric Sequence }} {{eqn | r = \dfrac 1 {1 - z - w z}       | c =  }} {{end-eqn}} {{qed}}
We note that: {{begin-eqn}} {{eqn | n = 1       | l = L \left\{ {f \left({s}\right)}\right\}       | r = \int_0^\infty e^{-s t} f \left({t}\right) \rd t       | c = {{Defof|Laplace Transform}} }} {{end-eqn}} and: {{begin-eqn}} {{eqn | n = 2       | l = G \left({z}\right)       | r = \sum_{n \mathop \ge 0} a_n z^n       | c = {{Defof|Generating Function}} }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int_n^{n + 1} e^{-s t} f \left({t}\right) \rd t       | r = \int_n^{n + 1} \left({a_0 + a_1 + \cdots + a_n}\right) e^{-s t} \rd t       | c = Definition of $f \left({x}\right)$ }} {{eqn | r = \sum_{k \mathop = 0}^n a_k \left[{\dfrac {-1} s e^{-s t} }\right]_n^{n + 1}       | c = Primitive of Exponential Function }} {{eqn | r = \sum_{k \mathop = 0}^n \dfrac {a_k} s \left({e^{-s n} - e^{-s \left({n + 1}\right)} }\right)       | c =  }} {{eqn | ll= \leadsto       | l = L \left\{ {f \left({s}\right)}\right\}       | r = \int_0^\infty e^{-s t} f \left({t}\right) \rd t       | c = from $(1)$ }} {{eqn | r = \sum_{n \mathop \ge 0} \sum_{k \mathop = 0}^n \dfrac {a_k} s \left({e^{-s n} - e^{-s \left({n + 1}\right)} }\right)       | c =  }} {{eqn | r = \dfrac {a_0} s \left({1 - e^{-s} }\right) + \dfrac {a_0 + a_1} s \left({e^{-s} - e^{-2 s} }\right) + \dfrac {a_0 + a_1 + a_2} s \left({e^{-2 s} - e^{-3 s} }\right) + \cdots       | c =  }} {{eqn | r = \dfrac 1 s \left({a_0 + a_1 e^{-s} + a_2 e^{-2 s} + \cdots}\right)       | c =  }} {{eqn | r = \dfrac 1 s \sum_{n \mathop \ge 0} a_n e^{-s n}       | c =  }} {{eqn | r = \dfrac 1 s G \left({e^{-s} }\right)       | c = from $(2)$ }} {{end-eqn}} {{qed}}
Proof by Counterexample: By definition, Euler's number $e$ can be defined as: :$e = \displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {n!}$ Each of the terms of the sequence of partial sums is rational. However, from Euler's Number is Irrational, $e$ itself is not. {{qed}}
From Real Numbers form Ring, the set $\R$ of real numbers forms a ring. The result then follows from Product of Ring Negatives. {{qed}}
$a$ can be expressed as a wholly real complex number $a + 0 i$. Then we have: {{begin-eqn}} {{eqn | l = a \times \left({c + d i}\right)       | r = \left({a + 0 i}\right) \times \left({c + d i}\right)       | c =  }} {{eqn | r = \left({a c - 0 d}\right) + \left({a d + 0 c}\right) i       | c =  }} {{eqn | r = a c + i a d       | c =  }} {{end-eqn}} The result for $\left({c + d i}\right) \times a$ follows from Complex Multiplication is Commutative. {{qed}}
Let $z = x + i y$. Then: {{begin-eqn}} {{eqn | l = \overline z       | r = -z       | c =  }} {{eqn | ll= \leadsto       | l = x - i y       | r = -\left({x + i y}\right)       | c = {{Defof|Complex Conjugate}} }} {{eqn | ll= \leadsto       | l = +x       | r = -x       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = 0       | c =  }} {{end-eqn}} Hence by definition, $z$ is wholly imaginary. {{qed|lemma}} Now suppose $z$ is wholly imaginary. Then: {{begin-eqn}} {{eqn | l = \overline z       | r = 0 - i y       | c =  }} {{eqn | r = -i y       | c =  }} {{eqn | r = -\left({0 + i y}\right)       | c =  }} {{eqn | r = -z       | c =  }} {{end-eqn}} {{qed}}
A direct consequence of Product of Complex Conjugates: : $\overline {z_1  z_2} = \overline {z_1} \cdot \overline {z_2}$ for two  complex numbers $z_1, z_2 \in \C$. {{qed}}
=== Necessary Condition === {{begin-eqn}} {{eqn | l = z       | r = 0       | c =  }} {{eqn | r = 0 + 0 i       | c =  }} {{eqn | ll= \leadsto       | l = \cmod z       | r = \sqrt {0^2 + 0^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === {{begin-eqn}} {{eqn | l = \cmod z       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \cmod {a + b i}       | r = 0       | c = Definition of $z$ }} {{eqn | ll= \leadsto       | l = \sqrt {a^2 + b^2}       | r = 0       | c = {{Defof|Complex Modulus}} }} {{eqn | ll= \leadsto       | l = a^2 + b^2       | r = 0       | c = squaring both sides }} {{eqn | ll= \leadsto       | l = a       | r = 0       | c = Square of Real Number is Non-Negative }} {{eqn | l = b       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = z       | r = 0       | c = Definition of $z$ }} {{end-eqn}} {{qed}}
From Complex Modulus of Product of Complex Numbers: : $\left\vert{z_1 z_2}\right\vert = \left\vert{z_1}\right\vert \cdot \left\vert{z_2}\right\vert$ for $z_1, z_2 \in \C$. Set $z = z_1 = z_2$ and the result follows. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\left\vert{z^n}\right\vert = \left\vert{z}\right\vert^n$ $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \left\vert{z^0}\right\vert       | r = \left\vert{1}\right\vert       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | r = \left\vert{z}\right\vert^0       | c =  }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \left\vert{z^1}\right\vert       | r = \left\vert{z}\right\vert       | c =  }} {{eqn | r = \left\vert{z}\right\vert^1       | c =  }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Basis for the Induction === $P \left({2}\right)$ is the case: : $\left\vert{z^2}\right\vert = \left\vert{z}\right\vert^2$ which is demonstrated in Square of Complex Modulus equals Complex Modulus of Square. Thus $P \left({2}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\left\vert{z^k}\right\vert = \left\vert{z}\right\vert^k$ from which it is to be shown that: :$\left\vert{z^{k + 1} }\right\vert = \left\vert{z}\right\vert^{k + 1}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \left\vert{z^{k + 1} }\right\vert       | r = \left\vert{z^k \cdot z}\right\vert       | c =  }} {{eqn | r = \left\vert{z^k}\right\vert \cdot \left\vert{z}\right\vert       | c = Complex Modulus of Product of Complex Numbers }} {{eqn | r = \left\vert{z}\right\vert^k \cdot \left\vert{z}\right\vert       | c = Induction Hypothesis }} {{eqn | r = \left\vert{z}\right\vert^{k + 1}       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \ Z_{\ge 0}: \left\vert{z^n}\right\vert = \left\vert{z}\right\vert^n$ {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\overline {z^n} = \left({\overline z}\right)^n$ $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \overline {z^0}       | r = \overline 1       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | r = \left({\overline z}\right)^0       | c =  }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \overline {z^1}       | r = \overline z       | c =  }} {{eqn | r = \left({\overline z}\right)^1       | c =  }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Basis for the Induction === $P \left({2}\right)$ is the case: :$\overline {z^2} = \left({\overline z}\right)^2$ which is demonstrated in Square of Complex Conjugate is Complex Conjugate of Square. Thus $P \left({2}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\overline {z^k} = \left({\overline z}\right)^k$ from which it is to be shown that: :$\overline {z^{k + 1} } = \left({\overline z}\right)^{k + 1}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \overline {z^{k + 1} }       | r = \overline {z^k \cdot z}       | c =  }} {{eqn | r = \overline {z^k} \cdot \overline z       | c = Product of Complex Conjugates }} {{eqn | r = \left({\overline z}\right)^k \cdot \overline z       | c = Induction Hypothesis }} {{eqn | r = \left({\overline z}\right)^{k + 1}       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \ Z_{\ge 0}: \overline {z^n} = \left({\overline z}\right)^n$ {{qed}} Category:Complex Conjugates Category:Complex Powers 6iojfs6m7fc0mwi33xhuq198eby3mwu
By Power of Complex Conjugate is Complex Conjugate of Power: :$\overline {\alpha^k} = \left({\overline \alpha}\right)^k$ for all $k$ between $0$ and $n$. Then from Product of Complex Conjugates: :$\overline {a_k \alpha^k} = \overline {a_k} \cdot \overline {\alpha^k}$ But $a_k$ is real. So by Complex Number equals Conjugate iff Wholly Real: :$\overline {a_k} = a_k$. From Sum of Complex Conjugates, it follows that: :$\overline {f \left({\alpha}\right)} = a_n \left({\overline \alpha}\right)^n + a_{n - 1} \left({\overline \alpha}\right)^{n - 1} + \cdots + a_1 \overline \alpha + a_0$ Hence the result. {{qed}}
Let: :$\alpha = x_1 + i y_1$ :$\beta = x_2 + i y_2$ Then: {{begin-eqn}} {{eqn | o =        | r = \left\lvert{\alpha + \beta}\right\rvert^2 + \left\lvert{\alpha - \beta}\right\rvert^2       | c =  }} {{eqn | r = \left\lvert{\left({x_1 + i y_1}\right) + \left({x_2 + i y_2}\right)}\right\rvert^2 + \left\lvert{\left({x_1 + i y_1}\right) - \left({x_2 + i y_2}\right)}\right\rvert^2       | c = Definition of $\alpha$ and $\beta$ }} {{eqn | r = \left\lvert{\left({x_1 + x_2}\right) + i \left({y_1 + y_2}\right)}\right\rvert^2 + \left\lvert{\left({x_1 - x_2}\right) + i \left({y_1 - y_2}\right)}\right\rvert^2       | c = {{Defof|Complex Addition}}, {{Defof|Complex Subtraction}} }} {{eqn | r = \left({x_1 + x_2}\right)^2 + \left({y_1 + y_2}\right)^2 + \left({x_1 - x_2}\right)^2 + \left({y_1 - y_2}\right)^2       | c = {{Defof|Complex Modulus}} }} {{eqn | r = {x_1}^2 + 2 x_1 x_2 + {x_2}^2 + {y_1}^2 + 2 y_1 y_2 + {y_2}^2 + {x_1}^2 - 2 x_1 x_2 + {x_2}^2 + {y_1}^2 - 2 y_1 y_2 + {y_2}^2       | c = Square of Sum, Square of Difference }} {{eqn | r = 2 {x_1}^2 + 2 {x_2}^2 + 2 {y_1}^2 + 2 {y_2}^2       | c = simplifying }} {{eqn | r = 2 \left({ {x_1}^2 + {y_1}^2}\right) + 2 \left({ {x_2}^2 + {y_2}^2}\right)       | c = simplifying }} {{eqn | r = 2 \left\lvert{x_1 + i y_1}\right\rvert^2 + 2 \left\lvert{x_2 + i y_2}\right\rvert^2       | c = {{Defof|Complex Modulus}} }} {{eqn | r = 2 \left\lvert{\alpha}\right\rvert^2 + 2 \left\lvert{\beta}\right\rvert^2       | c = Definition of $\alpha$ and $\beta$ }} {{end-eqn}} {{qed}}
Let $z = x + i y$. Let $\alpha = a + i b$. Thus: {{begin-eqn}} {{eqn | l = \cmod {z - \alpha}       | r = r       | c =  }} {{eqn | ll= \leadsto       | l = \cmod {x + i y - a + i b}       | r = r       | c =  }} {{eqn | ll= \leadsto       | l = \cmod {\paren {x - a} + i \paren {y - b} }       | r = r       | c = {{Defof|Complex Subtraction}} }} {{eqn | ll= \leadsto       | l = \sqrt {\paren {x - a}^2 + \paren {y - b}^2}       | r = r       | c = {{Defof|Complex Modulus}} }} {{eqn | ll= \leadsto       | l = \paren {x - a}^2 + \paren {y - b}^2       | r = r^2       | c = squaring both sides }} {{end-eqn}} The result follows from Equation of Circle. {{qed}}
From Equation of Circle in Complex Plane, the circle $C$ itself is given by: :$\left\lvert{z - \alpha}\right\rvert = r$ {{ProofWanted|This needs to be put into the rigorous context of Jordan curves, so as to define what is actually meant by "interior".  At the moment, the understanding is intuitive.}}
From Equation of Circle in Complex Plane, the circle $C$ itself is given by: :$\left\lvert{z - \alpha}\right\rvert = r$ {{ProofWanted|This needs to be put into the rigorous context of Jordan curves, so as to define what is actually meant by "exterior".  At the moment, the understanding is intuitive.}}
{{begin-eqn}} {{eqn | l = \cmod {z - 1}       | r = \cmod {z + 1}       | c =  }} {{eqn | ll= \leadsto       | l = \cmod {z - 1}^2       | r = \cmod {z + 1}^2       | c =  }} {{eqn | ll= \leadsto       | l = \paren {z - 1} \paren {\overline {z - 1} }       | r = \paren {z + 1} \paren {\overline {z + 1} }       | c = Modulus in Terms of Conjugate }} {{eqn | ll= \leadsto       | l = z \overline z - z - \overline z + 1       | r = z \overline z + z + \overline z + 1       | c =  }} {{eqn | ll= \leadsto       | l = 2 \paren {z + \overline z}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = 4 \map \Re z       | r = 0       | c = Sum of Complex Number with Conjugate }} {{eqn | ll= \leadsto       | l = \map \Re z       | r = 0       | c =  }} {{end-eqn}} The result follows by definition of imaginary axis. {{qed}}
Let $z = x + i y$. Let $a = l - i m$. Then: {{begin-eqn}} {{eqn | l = \map \Re {a z}       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\paren {a z + \overline {a z} } } 2       | r = 1       | c = Sum of Complex Number with Conjugate }} {{eqn | ll= \leadsto       | l = a z + \overline a \cdot \overline z       | r = 2       | c = Complex Modulus of Product of Complex Numbers }} {{eqn | ll= \leadsto       | l = \paren {l - i m} \paren {x + i y} + \paren {l + i m} \paren {x - i y}       | r = 2       | c = {{Defof|Complex Conjugate}} }} {{eqn | ll= \leadsto       | l = \paren {\paren {l x + m y} + i \paren {l y - m x} } + \paren {\paren {l x + m y} - i \paren {l y - m x} }       | r = 2       | c = {{Defof|Complex Multiplication}} }} {{eqn | ll= \leadsto       | l = l x + m y       | r = 1       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac 1 z       | r = \dfrac {\overline z} {z \overline z}       | c = Inverse for Complex Multiplication }} {{eqn | r = \dfrac {r \left({\cos \theta - i \sin \theta}\right)} {r \left({\cos \theta + i \sin \theta}\right) r \left({\cos \theta - i \sin \theta}\right)}       | c = Polar Form of Complex Conjugate }} {{eqn | r = \dfrac {\left({\cos \theta - i \sin \theta}\right)} {r \left({\cos^2 \theta - i^2 \sin^2 \theta}\right)}       | c = simplifying }} {{eqn | r = \dfrac {\left({\cos \theta - i \sin \theta}\right)} {r \left({\cos^2 \theta + \sin^2 \theta}\right)}       | c = {{Defof|Imaginary Unit}} }} {{eqn | r = \dfrac {\cos \theta - i \sin \theta} r       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} {{qed}}
Let $z$ be expressed in polar form: :$z := r \left({\cos \theta + i \sin \theta}\right)$ Then: {{begin-eqn}} {{eqn | l = \overline z       | r = r \left({\cos \theta - i \sin \theta}\right)       | c = Polar Form of Complex Conjugate }} {{eqn | r = r \left({\cos \left({-\theta}\right) + i \sin \left({-\theta}\right)}\right)       | c = Cosine Function is Even, Sine Function is Odd }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \dfrac 1 z       | r = \dfrac 1 r \left({\cos \theta - i \sin \theta}\right)       | c = Polar Form of Reciprocal of Complex Number }} {{eqn | r = \dfrac 1 r \left({\cos \left({-\theta}\right) + i \sin \left({-\theta}\right)}\right)       | c = Cosine Function is Even, Sine Function is Odd }} {{end-eqn}} The result follows by definition of the argument of a complex number {{qed}}
:600px Let $z$ be expressed in polar form as: :$z = r \left({\cos \theta + i \sin \theta}\right)$ From Polar Form of Complex Number: $i$: :$i = \cos \dfrac \pi 2 + i \sin \dfrac \pi 2$ and so: : the modulus of $i$ is $1$ : the argument of $i$ is $\dfrac \pi 2$. By Product of Complex Numbers in Polar Form: : the modulus of $i z$ is $r$ : the argument of $i$ is $\theta + \dfrac \pi 2$. That is, the result of multiplying $z$ by $i$ is the same as rotating $z$ through $\dfrac \pi 2$, which is a right angle. {{qed}}
:600px By definition of the imaginary unit: :$-1 = i^2$ and so: :$-1 \times z = i \paren {i z}$ From Multiplication by Imaginary Unit is Equivalent to Rotation through Right Angle, multiplication by $i$ is equivalent to rotation through a right angle, in an anticlockwise direction. So multiplying by $i^2$ is equivalent to rotation through two right angles in an anticlockwise direction. {{qed}}
:400px {{WLOG}}, let $ABCD$ be embedded in the complex plane so that vertex $A$ coincides with the origin $0 + 0 i$. Let $AB$ and $AD$ be represented by the complex numbers $a$ and $b$ respectively, expressed as vectors $\mathbf a$ and $\mathbf b$ respectively. By Geometrical Interpretation of Complex Addition, the diagonal $AC$ is represented by the complex number $a + b$, expressed as a vector $\mathbf a + \mathbf b$. By Geometrical Interpretation of Complex Subtraction, the diagonal $BD$ is represented by the complex number $a - b$, expressed as a vector $\mathbf a - \mathbf b$. Let $a - b = w \paren {a + b}$ for some complex number $w \in \C$. For $A \perp BD$, it is necessary and sufficient  from Complex Multiplication as Geometrical Transformation for the argument of $w$ to be $\dfrac \pi 2$. That is, for $w$ to be wholly imaginary. From Complex Number equals Negative of Conjugate iff Wholly Imaginary, that is for $w + \overline w = 0$. It remains to demonstrate this. So: {{begin-eqn}} {{eqn | l = w + \overline w       | r = \dfrac {a - b} {a + b} + \dfrac {\overline a - \overline b} {\overline a + \overline b}       | c =  }} {{eqn | r = \dfrac {2 a \overline a - 2 b \overline b} {\left({a + b}\right) \left({\overline a + \overline b}\right)}       | c =  }} {{eqn | r = \dfrac {2 \cmod a^2 - 2 \cmod b^2} {\left({a + b}\right) \left({\overline a + \overline b}\right)}       | c = Modulus in Terms of Conjugate }} {{eqn | r = 0       | c = as $\cmod a = \cmod b$ }} {{end-eqn}} The result follows. {{qed}}
:400px Let $a = a_x + i a_y$ and $b = b_x + i b_y$. Then by definition of complex addition: :$a + b = \paren {a_x + b_x} + i \paren {a_y + b_y}$ Thus $\mathbf a + \mathbf b$ is the vector whose components are $a_x + b_x$ and $a_y + b_y$. Similarly, we have: :$b + a = \paren {b_x + a_x} + i \paren {b_y + a_y}$ Thus $\mathbf b + \mathbf a$ is the vector whose components are $b_x + a_x$ and $b_y + a_y$. It follows that both $\mathbf a + \mathbf b$ and $\mathbf b + \mathbf a$ both correspond to the diagonal $OC$ of $OACB$. {{qed}}
:400px By definition of vector addition: :$OB + BA = OA$ That is: :$\mathbf b + \vec {BA} = \mathbf a$ which leads directly to: :$\vec {BA} = \mathbf a - \mathbf b$ {{qed}}
By Geometrical Interpretation of Complex Subtraction: :$z_1 - z_3$ can be represented as the line segment from $z_3$ to $z_1$ :$z_3 - z_2$ can be represented as the line segment from $z_2$ to $z_3$. Thus we have that $z_1$, $z_2$ and $z_3$ are collinear {{iff}} $z_1 - z_3$ is parallel to $z_3 - z_2$, when expressed as line segments. Let $\dfrac {z_1 - z_3} {z_3 - z_2} = \lambda$ for $\lambda \in \C$. That is: :$z_1 - z_3 = \lambda \paren {z_3 - z_2}$ By Complex Multiplication as Geometrical Transformation: :$\map \arg {z_1 - z_3} = \map \arg {z_3 - z_2} \iff \map \arg {z_3 - z_2} \arg \lambda \in \R_{>0}$ and: :$\map \arg {z_1 - z_3} = -\map \arg {z_3 - z_2} \iff \map \arg {z_3 - z_2} \arg \lambda \in \R_{<0}$ where $\arg$ denotes the argument of a complex number. Also by Complex Multiplication as Geometrical Transformation, $z_1 - z_3$ is $\lambda$ the length of $z_3 - z_2$. The result follows. {{Qed}}
:420px By Geometrical Interpretation of Complex Subtraction: :$z - a$ represents the line from $A$ to $Z$ :$z - b$ represents the line from $B$ to $Z$ {{begin-eqn}} {{eqn | l = \arg \dfrac {z - b} {z - a}       | r = \lambda       | c =  }} {{eqn | ll= \leadsto       | l = \map \arg {z - b} - \map \arg {z - a}       | r = \lambda       | c = Argument of Quotient equals Difference of Arguments }} {{end-eqn}} Thus: :$\arg \dfrac {z - b} {z - a} = \lambda$ represents the statement that the angle between $AZ$ and $BZ$ is constant: :$\angle AZB = \lambda$ That is, the angle subtended by $AB$ at $Z$ is $\lambda$. The result follows from the Inscribed Angle Theorem. {{qed}}