hcju/proofwikiqa huggingface dataset:
==== Inverse of Natural Logarithm implies Solution of Differential Equation ==== Let $\exp x$ be the real function defined as the inverse of the natural logarithm: :$y = \exp x \iff x = \ln y$ Then: {{begin-eqn}} {{eqn | l = x       | r = \ln y }} {{eqn | r = \int_{t \mathop = 1}^{t \mathop = y} \frac 1 t \rd t       | c = {{Defof|Natural Logarithm/Positive Real|index = 1|Natural Logarithm}} }} {{eqn | ll= \leadsto       | l = \map {D_y} x       | r = D_y \int_{t \mathop = 1}^{t \mathop = y} \frac 1 t \rd t       | c = Differentiation {{WRT|Differentiation}} $y$ }} {{eqn | l = \frac {\d x} {\d y}       | r = \frac 1 y       | c = Fundamental Theorem of Calculus }} {{eqn | ll= \iff       | l = \frac {\d y} {\d x}       | r = y       | c = Derivative of Inverse Function }} {{end-eqn}} This proves that $y$ is a solution of the differential equation. It remains to be proven that $y$ fulfils the initial condition:  {{begin-eqn}} {{eqn | l = \map f 0        | r = 1 }} {{eqn | ll= \iff       | l = \map {f^{-1} } 1       | r = 0       | c = Image of Element under Inverse Mapping }} {{eqn | l = \bigintlimits {\ln y} {y \mathop = 1} {}       | r = \int_{t \mathop = 1}^{t \mathop = 1} \frac 1 t \rd t }} {{eqn | r = 0       | c = Integral on Zero Interval }} {{end-eqn}} That is: $\exp x$ is the particular solution of the differential equation: :$\dfrac {\d y} {\d x} = y$ satisfying the initial condition $\map f 0 = 1$. {{qed|lemma}} ==== Solution of Differential Equation implies Inverse of Natural Logarithm ==== Let $\exp x$ be the real function defined as the particular solution of the differential equation: :$\dfrac {\d y} {\d x} = y$ satisfying the initial condition $\map f  = 1$. Thus: {{begin-eqn}} {{eqn | l = \frac {\d y} {\d x}       | r = y }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d y}       | r = \frac 1 y       | c = Derivative of Inverse Function }} {{eqn | ll= \leadsto       | l = \int \rd x       | r = \int \frac 1 y \rd y       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = x + C       | r = \int_{t \mathop = 1}^{t \mathop = y} \frac 1 t \rd t       | c = Fundamental Theorem of Calculus }} {{eqn | r = \ln y       | c = {{Defof|Natural Logarithm/Positive Real|index = 1|Natural Logarithm}} }} {{end-eqn}} To solve for $C$, put $\tuple {x_0, y_0} = \tuple {0, 1}$: {{begin-eqn}} {{eqn | l = 0 + C       | r = \int_{t \mathop = 1}^{t \mathop = 1}\frac 1 t \rd t }} {{eqn | ll= \leadsto       | l = C       | r = 0       | c = Integral on Zero Interval }} {{end-eqn}} That is: :$y = \exp x \iff x = \ln y$ {{qed}}
==== Sum of Series implies Solution of Differential Equation ==== Let $\exp x$ be the real function defined as the sum of the power series: :$\exp x := \displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ Let $y = \displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$. Then: {{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d x}       | r = \dfrac \d {\d x} \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} }} {{eqn | r = \map {\dfrac \d {\d x} } {\frac {x^0} {0!} + \sum_{n \mathop = 1}^\infty \frac {x^n} {n!} }       | c = extracting the zeroth term }} {{eqn | r = 0 + \sum_{n \mathop = 1}^\infty \frac {n x^{n - 1} } {n!}       | c = Power Rule for Derivatives and Derivative of Constant }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac {x^{n - 1} } {\paren {n - 1}!}       | c = simplifying }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}       | c = Translation of Index Variable of Summation }} {{eqn | r = y }} {{end-eqn}} Setting $x = 0$ we find: {{begin-eqn}} {{eqn | l = \map y 0       | r = \sum_{n \mathop = 0}^\infty \frac {0^n} {n!} }} {{eqn | r = \frac {0^0} {0!} + \sum_{n \mathop = 1}^\infty \frac {0^n} {n!} }} {{eqn | r = \frac {0^0} {0!}       | c = as $0^n = 0$ for all $n > 0$ }} {{eqn | r = 1       | c = Definition of $0^0$ }} {{end-eqn}} {{qed}} {{MissingLinks|uniform convergence}} That is: $\exp x$ is the particular solution of the differential equation: :$\dfrac {\d y} {\d x} = y$ satisfying the initial condition $\map f 0 = 1$. {{qed|lemma}} ==== Solution of Differential Equation implies Sum of Series ==== Let $\exp x$ be the real function defined as the particular solution of the differential equation: :$\dfrac {\d y} {\d x} = y$ satisfying the initial condition $\map f 0 = 1$. We have Taylor Series Expansion for Exponential Function: {{:Taylor Series Expansion for Exponential Function}}
Let $\exp x$ be the real function defined as the limit of the sequence: :$\exp x := \displaystyle \lim_{n \mathop \to \infty} \paren {1 + \frac x n}^n$ From the General Binomial Theorem: {{begin-eqn}} {{eqn | l = \paren {1 + \frac x n}^n       | r = 1 + x + \frac {n \paren {n - 1} x^2} {2! \ n^2} + \frac {n \paren {n - 1} \paren {n - 2} x^3} {3! \ n^3} + \cdots }} {{eqn | r = \frac {x^0} {0!} + \frac {x^1} {1!} + \paren {\frac {n - 1} n} \frac {x^2} {2!} + \paren {\frac {\paren {n - 1} \paren {n - 2} } {n^2} } \frac {x^3} {3!} + \cdots }} {{eqn | ll= \leadsto       | l = 0       | r = \paren {1 + \frac x n}^n - \paren {\frac {x^0} {0!} + \frac {x^1} {1!} + \paren {\frac {n - 1} n} \frac {x^2} {2!} + \paren {\frac {\paren {n - 1} \paren {n - 2} } {n^2} } \frac {x^3} {3!} + \cdots} }} {{end-eqn}} From Power over Factorial, this converges to: :$\exp x - \paren {\dfrac {x^0} {0!} + \dfrac {x^1} {1!} + \dfrac {x^2} {2!} + \dfrac {x^3} {3!} + \cdots} = 0$ as $n \to +\infty$. {{begin-eqn}} {{eqn | ll= \leadsto       | l = \exp x       | r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} }} {{end-eqn}} {{qed}}
Let $\exp x$ be the real function defined as the inverse of the natural logarithm: :$y = \exp x \iff x = \ln y$ Let $\left \langle {x_n} \right \rangle$ be the sequence in $\R$ defined as: :$x_n = \paren {1 + \dfrac x n}^n$ First it needs to be noted that $\sequence {x_n}$ does indeed converge to a limit. From Equivalence of Definitions of Real Exponential Function: Limit of Sequence implies Sum of Series, we have: :$\displaystyle \lim_{n \mathop \to \infty} \paren {1 + \dfrac x n}^n = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ From Series of Power over Factorial Converges, the {{RHS}} is indeed shown to converge to a limit. It will next be shown that: :$\displaystyle \map \ln {\lim_{n \mathop \to \infty} \sequence {x_n} } = x$ We have: {{begin-eqn}} {{eqn | l = \map \ln {\paren {1 + \frac x n}^n}       | r = n \, \map \ln {1 + x n^{-1} }       | c = Logarithms of Powers }} {{eqn | r = x \frac {\map \ln {1 + x n^{-1} } } {x n^{-1} }       | c = multiplying by $1 = \dfrac {x n^{-1} } {x n^{-1} }$ }} {{end-eqn}} From Limit of Sequence is Limit of Real Function, we can consider the differentiable analogue of the sequence. From Derivative of Logarithm at One we have: :$\displaystyle \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x = 1$ But $x n^{-1} \to 0$ as $n \to \infty$ from Sequence of Powers of Reciprocals is Null Sequence. Thus: :$\displaystyle x \frac {\map \ln {1 + x n^{-1} } } {x n^{-1} } \to x$ as $n \to \infty$. From Exponential Function is Continuous: :$\paren {1 + \dfrac x n}^n = \exp \paren {n \, \map \ln {1 + \dfrac x n} } \to \exp x = e^x$ as $n \to \infty$. {{qed}}
Let the restriction of the exponential function to the rationals be defined as: :$\displaystyle \exp \restriction_\Q: x \mapsto \lim_{n \mathop \to +\infty}\left ({1 + \frac x n}\right)^n$ Thus, let $e$ be Euler's Number defined as: :$e = \displaystyle \lim_{n \mathop \to +\infty}\left ({1 + \frac 1 n}\right)^n$ For $x = 0$: {{begin-eqn}} {{eqn | ll= \exp \restriction_\Q \paren 0       | r = \lim_{n \mathop \to +\infty} \paren {1 + \frac 0 n}^n }} {{eqn | r = 1 }} {{eqn | r = e^0 }} {{end-eqn}} For $x \ne 0$: {{begin-eqn}} {{eqn | ll= \exp \restriction_\Q \paren x       | r = \lim_{n \mathop \to +\infty}\paren {1 + \frac x n}^n       | c =  }} {{eqn | r = \lim_{\paren {n/x} \mathop \to +\infty}\paren {\paren {1 + \frac 1 {\paren {n/x} } }^{\paren {n/x} } }^x       | c = Exponent Combination Laws }} {{eqn | r = e^x }} {{end-eqn}} {{LinkWanted|We need that $y \mapsto y^x$ is continuous to make the last step}} For $x \in \R \setminus \Q$, we invoke Power Function to Rational Power permits Unique Continuous Extension. {{qed}}
Let $\exp x$ be the real function defined as the extension of rational exponential. Then we have: {{begin-eqn}} {{eqn | l = \map {D_x} {\exp x}       | r = \lim_{h \mathop \to 0} \frac {\map \exp {x + h} - \exp x} h       | c = {{Defof|Derivative}} }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\exp x \cdot \exp h - \exp x} h       | c = Exponential of Sum }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\exp x \paren {\exp h - 1} } h }} {{eqn | r = \exp x \paren {\lim_{h \mathop \to 0} \frac {\exp h - 1} h}       | c = Multiple Rule for Limits of Functions, as $\exp x$ is constant }} {{eqn | r = \exp x       | c = Derivative of Exponential at Zero: Proof 2 }} {{end-eqn}} The application of Derivative of Exponential at Zero is not circular as the referenced proof does not depend on $D_x \exp x = \exp x$. {{qed|lemma}} {{begin-eqn}} {{eqn | l = \exp 0       | r = \lim_{n \mathop \to +\infty} \paren {1 + \frac 0 n}^n }} {{eqn | r = 1 }} {{end-eqn}} {{qed}}
By definition of the real exponential function as the limit of a sequence: :$(1): \quad \exp x := \ds \lim_{n \mathop \to \infty} \paren {1 + \frac x n}^n$ By definition of Euler's number: :$e = e^1 = \exp 1$ The result follows by setting $x = 1$ in $(1)$. {{qed}}
==== Sum of Series implies Solution of Differential Equation ==== Let $\exp z$ be the complex function defined as the sum of the power series: :$\exp z := \displaystyle \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$ Let $y = \displaystyle \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$. Then: {{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d z}       | r = \dfrac \d {\d z} \sum_{n \mathop = 0}^\infty \frac {z^n} {n!} }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac {z^{n - 1} } {\paren {n - 1}!}       | c = Derivative of Complex Power Series }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}       | c = Translation of Index Variable of Summation }} {{eqn | r = y }} {{end-eqn}} We show that $\displaystyle \sum_{n \mathop = 0}^\infty \dfrac {z^n} {n!}$ satisfies the initial condition: :$\exp \paren 0 = 1$. Setting $z = 0$ we find: {{begin-eqn}} {{eqn | l = y \paren 0       | r = \sum_{n \mathop = 0}^\infty \frac {0^n} {n!} }} {{eqn | r = \frac {0^0} {0!} + \sum_{n \mathop = 1}^\infty \frac {0^n} {n!} }} {{eqn | r = \frac {0^0} {0!}       | c = as $0^n = 0$ for all $n > 0$ }} {{eqn | r = 1       | c = Definition of $0^0$ }} {{end-eqn}} That is: $\exp z$ is the particular solution of the differential equation: :$\dfrac {\d y} {\d z} = y$ satisfying the initial condition $y \left({0}\right) = 1$. {{qed|lemma}} ==== Solution of Differential Equation implies Sum of Series ==== Let $\exp z$ be the complex function defined as the particular solution of the differential equation: :$\dfrac {\d y} {\d z} = y$ satisfying the initial condition $y \paren 0 = 1$. Let $f: \C \to \C$ be a solution to the differential equation $\dfrac {\d f} {\d z} = f$ with $f \paren 0 = 1$. Then Holomorphic Function is Analytic shows that $f$ can be expressed as a power series: :$\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ about any $\xi \in \C$. When $\xi = 0$, we have for all $n \in \N_{\ge 1}$: {{begin-eqn}} {{eqn | l = a_n       | r = \dfrac {f^{\paren n} \paren 0} {n!}       | c = Power Series is Taylor Series }} {{eqn | r = \dfrac {f^{\paren {n - 1} } \paren 0} {n!}       | c = as $f^{\paren n} \paren 0 = f \paren 0 = f^{\paren {n - 1} } \paren 0$ }} {{eqn | r = \dfrac 1 n a_{n - 1}       | c = Power Series is Taylor Series }} {{end-eqn}} As $a_0 = \dfrac{f^{\paren 0} \paren 0} {0!} = 1$ by the initial condition, it follows inductively that: :$a_n = \dfrac 1 {n!}$ Hence: :$\displaystyle f \paren z = \sum_{n \mathop = 0}^\infty \dfrac 1 {n!} z^n$
We have the result: :Sum of Series equivalent to Solution of Differential Equation which gives that the definition of $\exp z$ as the sum of the power series :$\exp z := \displaystyle \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$ is equivalent to the definition of $\exp z$ as the solution of the differential equation: :$\dfrac {\d y} {\d z} = y$ satisfying the initial condition $y \paren 0 = 1$. Let: :$e: \R \to \R$ denote the real exponential function :$\sin: \R \to \R$ denote the real sine function :$\cos: \R \to \R$ denote the real cosine function. Then: {{begin-eqn}} {{eqn | l = \exp z       | r = \map \exp {x + i y}       | c = where $x, y \in \R$ }} {{eqn | r = \map \exp x \, \map \exp {i y}       | c = Exponential of Sum: Complex Numbers }} {{eqn | r = e^x \, \map \exp {i y}       | c = Definition (as Sum of Series) agrees with Definition of Real Exponential for all $x \in \R$  }} {{eqn | r = e^x \paren {\cos y + i \sin y}       | c = Euler's Formula, which can be proven using Definition as Sum of Series }} {{end-eqn}}
The proof proceeds by induction. For all $m \in \N_{>0}$, let $\map P m$ be the proposition: :$\displaystyle \map \exp {\sum_{j \mathop = 1}^m z_j} = \prod_{j \mathop = 1}^m \paren {\exp z_j}$ $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \map \exp {\sum_{j \mathop = 1}^1 z_j}       | r = \exp z_j       | c =  }} {{eqn | r = \prod_{j \mathop = 1}^1 \paren {\exp z_j}       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. === Basis for the Induction === $\map P 2$ is the case: :$\map \exp {z_1 + z_2} = \paren {\exp z_1} \paren {\exp z_2}$ This is proved in Exponential of Sum: Complex Numbers. Thus $\map P 2$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \map \exp {\sum_{j \mathop = 1}^k z_j} = \prod_{j \mathop = 1}^k \paren {\exp z_j}$ from which it is to be shown that: :$\displaystyle \map \exp {\sum_{j \mathop = 1}^{k + 1} z_j} = \prod_{j \mathop = 1}^{k + 1} \paren {\exp z_j}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \map \exp {\sum_{j \mathop = 1}^{k + 1} z_j}       | r = \map \exp {\sum_{j \mathop = 1}^k z_j + z_{k + 1} }       | c =  }} {{eqn | r = \map \exp {\sum_{j \mathop = 1}^k z_j} \exp z_{k + 1}       | c = Basis for the Induction }} {{eqn | r = \prod_{j \mathop = 1}^k \paren {\exp z_j} \exp z_{k + 1}       | c = Induction Hypothesis }} {{eqn | r = prod_{j \mathop = 1}^{k + 1} \paren {\exp 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: :$\forall m \in \N_{>0}: \displaystyle \map \exp {\sum_{j \mathop = 1}^m z_j} = \prod_{j \mathop = 1}^m \paren {\exp z_j}$ {{qed}}
From Exponential of Sum: Complex Numbers: General Result: :$\displaystyle \exp \paren {\sum_{j \mathop = 1}^m z_j} = \prod_{j \mathop = 1}^m \paren {\exp z_j}$ for complex numberst $z_1, z_2, \ldots, z_m \in \C$. The result follows by setting $z = z_1 = z_2 = \cdots = z_m$. {{qed}}
{{begin-eqn}} {{eqn | l = \map \exp {z + 2 \pi i}       | r = \map \exp z \, \map \exp {2 \pi i}       | c = Exponential of Sum: Complex Numbers }} {{eqn | r = \map \exp z \times 1       | c = Euler's Formula Example: $e^{2 i \pi}$ }} {{eqn | r = \map \exp z }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \paren {1 + e^{i \theta} }^n       | r = \sum_{k \mathop = 0}^n \dbinom n k e^{i k \theta}       | c = Binomial Theorem }} {{eqn | r = \sum_{k \mathop = 0}^n \dbinom n k \paren {\cos k \theta + i \sin k \theta}       | c = Euler's Formula }} {{eqn | ll= \leadsto       | l = \map \Im {\paren {1 + e^{i \theta} }^n}       | r = \map \Im {\sum_{k \mathop = 0}^n \dbinom n k \paren {\cos k \theta + i \sin k \theta} }       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^n \dbinom n k \sin k \theta       | c = taking imaginary parts }} {{eqn | r = \sum_{k \mathop = 1}^n \dbinom n k \sin k \theta       | c = as the zeroth term vanishes: $\sin 0 = 0$ }} {{end-eqn}} At the same time: {{begin-eqn}} {{eqn | l = \map \Im {\paren {1 + e^{i \theta} }^n}       | r = \map \Im {e^{i n \theta / 2} \paren {e^{i \theta / 2} + e^{-i \theta / 2} }^n}       | c =  }} {{eqn | r = \map \Im {e^{i n \theta / 2} \paren {2 \cos \dfrac \theta 2}^n}       | c = Cosine Exponential Formulation }} {{eqn | r = \map \Im {\paren {\cos \dfrac {n \theta} 2 + i \sin \dfrac {n \theta} 2} \paren {2 \cos \dfrac \theta 2}^n}       | c = Euler's Formula }} {{eqn | r = \sin \dfrac {n \theta} 2 \paren {2 \cos \dfrac \theta 2}^n       | c =  }} {{end-eqn}} {{qed}}
From Geometrical Interpretation of Complex Subtraction, the lines $XA$ and $XB$ can be represented by the complex numbers $a - x$ and $b - x$. :400px From Multiplication by Imaginary Unit is Equivalent to Rotation through Right Angle $a - x$ and $b - x$ are perpendicular {{iff}} either: :$a - x = r i \paren {b - x}$ for some real numbers $r \in \R$. That is, {{iff}} $\dfrac {a - x} {b - x}$ are purely imaginary. Thus: {{begin-eqn}} {{eqn | l = \dfrac {a - x} {b - x}       | r = r i       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\paren {a - x} \paren {\overline b - x} } {\paren {b - x} \paren {\overline b - x} }       | r = r i       | c =  }} {{eqn | ll= \leadsto       | l = \paren {a_x + a_y i - x} \paren {b_x - b_y i - x}       | r = r' i       | c = for some $r' \in \R$: denominator is real }} {{eqn | ll= \leadsto       | l = a_x b_x - a_x b_y i - a_x x + a_y b_x i + a_y b_y - a_y x i - x b_x - x b_y i + x^2       | r = r' i       | c =  }} {{eqn | ll= \leadsto       | l = a_x b_x - a_x x + a_y b_y - x b_x + x^2       | r = 0       | c = equating real parts }} {{eqn | ll= \leadsto       | l = x^2 - \paren {a_x + b_x} x + a_x b_x + a_y b_y       | r = 0       | c = simplifying }} {{eqn | ll= \leadsto       | l = x       | r = \dfrac {a_x + b_x \pm \sqrt {\paren {a_x + b_x}^2 - 4 \paren {a_x b_x + a_y b_y} } } 2       | c = Quadratic Formula }} {{eqn | r = \dfrac {a_x + b_x \pm \sqrt {a_x^2 + b_x^2 + 2 a_x b_x - 4 a_x b_x - 4 a_y b_y} } 2       | c =  }} {{eqn | r = \dfrac {a_x + b_x \pm \sqrt {a_x^2 + b_x^2 - 2 a_x b_x - 4 a_y b_y} } 2       | c =  }} {{end-eqn}} {{qed}} {{Proofread|This did not work out as neatly as I expected. I may have made a mistake somewhere. The original example which inspired this page was a specific case where the numbers worked out neatly.}}
Let us label the points: {{begin-eqn}} {{eqn | l = A       | o = :=       | r = 2 + i       | c =  }} {{eqn | l = B       | o = :=       | r = 3 + 2 i       | c =  }} {{eqn | l = C       | o = :=       | r = 2 + 3 i       | c =  }} {{eqn | l = D       | o = :=       | r = 1 + 2 i       | c =  }} {{end-eqn}} From Geometrical Interpretation of Complex Subtraction, we have that the difference $p - q$ between two complex numbers $p, q$ represents the vector $\vec {q p}$. Let us take the differences of the complex numbers given: {{begin-eqn}} {{eqn | ll= B - A:       | l = \paren{3 + 2 i} - \paren{2 + i}       | r = 1 + i       | c =  }} {{eqn | ll= B - C:       | l = \paren{3 + 2 i} - \paren{2 + 3 i}       | r = 1 - i       | c =  }} {{eqn | ll= B - D:       | l = \paren{3 + 2 i} - \paren{1 + 2 i}       | r = 2 + 0 i       | c =  }} {{eqn | ll= C - A:       | l = \paren{2 + 3 i} - \paren{2 + i}       | r = 0 + 2 i       | c =  }} {{eqn | ll= C - D:       | l = \paren{2 + 3 i} - \paren{1 + 2 i}       | r = 1 + i       | c =  }} {{eqn | ll= A - D:       | l = \paren{2 + i} - \paren{1 + 2 i}       | r = 1 - i       | c =  }} {{end-eqn}} So: :$\vec {AB} = \vec {DC} = 1 + i$ :$\vec {DA} = \vec {CB} = 1 - i$ Thus by definition $ABCD$ forms a parallelogram. Next it is noted that: :$\paren {1 + i} i = i + i^2 = 1 - i$ and so $AB$ and $DC$ are perpendicular to $DA$ and $CB$. Thus by definition $ABCD$ is a rectangle. Finally note that: :$\cmod {1 + i} = \cmod {1 - i} = \sqrt {1^2 + 1^2} = \sqrt 2$ and so: :$\cmod {AB} = \cmod {DC} = \cmod {DC} = \cmod {DC}$ That is, all four sides of $ABCD$ are the same length. Thus by definition $ABCD$ is a square. {{qed}}
By definition of complex modulus: : $\cmod {z - \alpha}$ is the distance from $z$ to $\alpha$ : $\cmod {z - \beta}$ is the distance from $z$ to $\beta$. Thus $\cmod {z - \alpha} + \cmod {z - \beta}$ is the sum of the distance from $z$ to $\alpha$ and from $z$ to $\beta$. This is precisely the equidistance property of the ellipse. From Equidistance of Ellipse equals Major Axis, the constant distance $d$ is equal to the major axis of $E$. {{qed}}
=== Sufficient Condition === Let $z_1, z_2, z_3$ be collinear. Then by Condition for Collinearity of Points in Complex Plane: Formulation 1 there exists a real number $b$ such that: :$z_2 - z_1 = b \paren {z_3 - z_1}$ Then: {{begin-eqn}} {{eqn | l = z_2 - z_1       | r = b \paren {z_3 - z_1}       | c =  }} {{eqn | ll= \leadsto       | l = z_2 - z_1 - b z_3 + b z_1       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {b - 1} z_1 + z_2 - b z_3       | r = 0       | c =  }} {{end-eqn}} Setting $\alpha = b - 1, \beta = 1, \gamma = -b$ fits the bill, as $\paren {b - 1} + 1 + \paren {-b} = 0$. {{qed|lemma}} === Necessary Condition === Let $\alpha + \beta + \gamma = 0$ such that: :$\alpha z_1 + \beta z_2 + \gamma z_3 = 0$ :at least one of $\alpha, \beta, \gamma$ is not zero. {{WLOG}} let $\alpha \ne 0$. Then it follows that as $\alpha + \beta + \gamma = 0$, at least one of $\beta$ and $\gamma$ is also non-zero. {{WLOG}} let $\beta \ne 0$. In the following it is immaterial whether $\gamma = 0$ or not. We have: {{begin-eqn}} {{eqn | l = \alpha + \beta + \gamma       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \gamma       | r = -\paren {\alpha + \beta}       | c =  }} {{eqn | ll= \leadsto       | l = \alpha z_1 + \beta z_2       | r = \paren {\alpha + \beta} z_3       | c = from $\alpha z_1 + \beta z_2 + \gamma z_3 = 0$ }} {{eqn | ll= \leadsto       | l = z_1 + \frac \beta \alpha z_2       | r = \frac {\alpha + \beta} \alpha z_3       | c = which can be done because $\alpha \ne 0$ }} {{eqn | ll= \leadsto       | l = z_1 + \frac \beta \alpha z_2 - \paren {1 + \frac \beta \alpha} z_1       | r = \frac {\alpha + \beta} \alpha z_3 - \paren {1 + \frac \beta \alpha} z_1       | c =  }} {{eqn | ll= \leadsto       | l = \frac \beta \alpha \paren {z_2 - z_1}       | r = \frac {\alpha + \beta} \alpha \paren {z_3 - z_1}       | c = simplifying }} {{eqn | ll= \leadsto       | l = z_2 - z_1       | r = \frac \alpha \beta \frac {\alpha + \beta} \alpha \paren {z_3 - z_1}       | c = multiplying both sides by $\dfrac \alpha \beta$, which can be done because $\beta \ne 0$ }} {{eqn | ll= \leadsto       | l = z_2 - z_1       | r = \frac {\alpha + \beta} \beta \paren {z_3 - z_1}       | c =  }} {{end-eqn}} Thus it is seen that: :$z_2 - z_1 = b \paren {z_3 - z_1}$ for some $b \in \R$. Hence by Condition for Collinearity of Points in Complex Plane: Formulation 1, $z_1$, $z_2$ and $z_3$ are collinear. {{qed}}