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\begin{document}
\title{Problema 9-4\\Theory of Equations\\Uspensky}
\author{Eduardo Virueña Silva}
\date{Noviembre, 2025}
\maketitle
\begin{Prob}
Probar que si $z_0, z_1, z_2$ son tres números
complejos distintos, entonces la igualdad:
\[
 |z_2 - z_1|^2 = |z_1 - z_0|^2  + |z_2 - z_0|^2
\]
se cumple si y sólo si, existe un número real $\lambda$ tal que:
\[
        z_2 - z_0 = i \lambda (z_1 - z_0)
\]
\end{Prob}
\noindent{\em Observación}\\
Notemos que $z_1 - z_0 \ne 0$, porque los tres puntos son distintos.

\begin{proof}\phantom{.}\\

\noindent {\bf Necesidad} 
\begin{align*}
\intertext{Supongamos primero que:}
|z_2 - z_1|^2 &= |z_1 - z_0|^2  + |z_2 - z_0|^2
\intertext{Podemos suponer que: }
	z_0&= x_0 + i y_0, \quad x_0, y_0 \in \mathbb{R}\\ 
	z_1&= x_1 + i y_1, \quad x_1, y_1 \in \mathbb{R}\\
	z_2&= x_2 + i y_2, \quad x_2, y_2 \in \mathbb{R} 
\intertext{Tenemos entonces:}
|z_2 - z_1|^2	&= |(x_2 + i y_2)  - (x_1 + i y_1)|\\
		&= |(x_2-x_1) + i (y_2-y_1)|\\
		&= (x_2-x_1)^2 + (y_2-y_1)^2\\
		&= x_2^2-2x_1x_2+x_1^2 + y_2^2-2y_1y_2+y_2^2\\
\intertext{De igual manera:}
|z_1 - z_0|^2	&= x_1^2-2x_0x_1+x_0^2 + y_1^2-2y_0y_1+y_0^2\\
|z_2 - z_0|^2	&= x_2^2-2x_0x_2+x_0^2 + y_2^2-2y_0y_2+y_0^2
\end{align*}
\begin{align*}
\intertext{Ahora, la igualdad:}
	|z_2 - z_1|^2 &= |z_1 - z_0|^2  + |z_2 - z_0|^2
\intertext{equivale a:}
	x_2^2 - 2x_1x_2 + x_1^2 + y_1^2-2y_1y_2
		&+y_2^2 = \\
	x_1^2-2x_0x_1 + x_0^2 + y_1^2-2y_0y_1
		&+y_0^2+x_2^2 -2x_0x_2+x_0^2+y_2^2-2y_0y_2+y_0^2 
\intertext{Simplifiquemos algunos términos y agrupemos términos semejantes}
\cancel{x_2^2}-2x_1x_2+\bcancel{x_1^2}+\bcancel{y_1^2}-2y_1y_2+\cancel{y_2^2}
	&=\\
\bcancel{x_1^2}-2x_0x_1 + 2x_0^2 + \bcancel{y_1^2}-2y_0y_1+ 
	& \cancel{x_2^2}-2x_0x_2 + \cancel{y_2^2}-2y_0y_2+2y_0^2  \\
\intertext{al dividir entre $2$, tenemos:}
-x_1x_2-y_1y_2= -x_0x_1 + x_0^2 
	& -y_0y_1-x_0x_2 -y_0y_2+y_0^2 
\end{align*}
\begin{align*}
\intertext{Reacomodemos:}
x_0x_2-x_1x_2 + y_0y_1-y_1y_2 &= -x_0x_1 + x_0^2 -y_0y_2+ y_0^2  \\
\intertext{Factoricemos:}
(x_0-x_1)x_2+(y_0-y_2)y_1&= x_0(x_0-x_1) +y_0(y_0- y_2)  \\
(x_0-x_1)x_2-x_0(x_0-x_1))&=y_0(y_0- y_2)-(y_0-y_2)y_1  \\
	(x_0-x_1)(x_2-x_0)&= (y_0-y_2)(y_0-y_1)  \\
	(x_1-x_0)(x_2-x_0)&= (y_0-y_2)(y_1-y_0)  \\
	(x_1-x_0)(x_2-x_0)&= -(y_2-y_0)(y_1-y_0)  
\end{align*}
Entonces, tenemos:
\begin{equation}
	(x_1-x_0)(x_2-x_0)+(y_2-y_0)(y_1-y_0) = 0 \label{partereal}
\end{equation}

\begin{align*}
\intertext{Recordemos que $z_1-z_0\ne 0$, entonces es posible plantear
la siguente división:}
\frac{z_2-z_0}{z_1-z_0} &= \frac{(z_2-z_0)\overline{(z_1-z_0)}}
				{(z_1-z_0)\overline{(z_1-z_0)}} \\
			&= \frac{(z_2-z_0){(\bar{z}_1-\bar{z}_0)}}
				{(z_1-z_0){(\bar{z}_1-\bar{z}_0})}\\
	&= \frac{((x_2+iy_2)-(x_0+iy_0)) ((x_1-iy_1)-(x_0-iy_0))}
		{|z_0-z_1|^2}\\
	&= \frac{((x_2-x_0)+i(y_2-y_0)) ((x_1-x_0) -i(y_1-y_0))}
		{|z_0-z_1|^2}\\
	&= \frac{((x_2-x_0)(x_1-x_0)+(y_2-y_0)(y_1-y_0)}{|z_0-z_1|^2}\\
	&+ i\frac{(y_2-y_0)(x_1-x_0)-(x_2-x_0)(y_1-y_0)}{|z_0-z_1|^2}\\
\intertext{por la ecuación (\ref{partereal}), la parte real se cancela:}
	&= i\frac{(y_2-y_0)(x_1-x_0)-(x_2-x_0)(y_1-y_0)}{|z_0-z_1|^2}\\
	&= i \lambda.
\intertext{Donde, $\lambda=\frac{(y_2-y_0)(x_1-x_0)-(x_2-x_0)(y_1-y_0)}
					{|z_0-z_1|^2}$}
\frac{z_2-z_0}{z_1-z_0} &= i \lambda
\intertext{de manera que:}
	z_2-z_0 &= i \lambda (z_1-z_0)
\end{align*}


\noindent{\bf Suficiencia}\\
Recíprocamente, ahora supongamos que existe $\lambda\in\mathbb{R}$ tal que:
\[
        z_2 - z_0 = i \lambda (z_1 - z_0)
\]
entonces:
\begin{align*}
	|z_2 - z_1|^2 
		&= |z_2 - z_0 + z_0 - z_1|^2\\
		&= |z_2 - z_0 + z_0 - z_1|^2\\
		&= |i\lambda(z_1 - z_0) + z_0 - z_1|^2\\
		&= |i\lambda(z_1 - z_0) - (z_1 - z_0)|^2\\
		&= |(i\lambda-1)(z_1 - z_0)|^2\\
		&= |1-i\lambda|^2(z_1 - z_0)|^2\\
\intertext{como $\lambda \in \mathbb{R}$ entonces:}
		&= (1^2+\lambda^2) |z_1 - z_0|^2\\
		&= |z_1 - z_0|^2 + \lambda^2 |z_1 - z_0|^2\\
\intertext{ sabemos que $|i|=1$, por lo tanto $1= |i|^2$}
		&= |z_1 - z_0|^2 + |i|^2\lambda^2 |z_1 - z_0|^2\\
		&= |z_1 - z_0|^2 + |i\lambda|^2 |z_1 - z_0|^2\\
		&= |z_1 - z_0|^2 + |i\lambda(z_1 - z_0)|^2\\
	|z_2 - z_1|^2 
		&= |z_1 - z_0|^2 + |z_2 - z_0|^2
\end{align*}
			
\end{proof}

La idea geométrica que relaciona los módulos de las diferencias con
el Teorema de Pitágoras tiene validez.  El factor $i$ significa que
los números complejos $z_1-z_0$ y $z_2-z_0$ forman un ángulo 
que mide $\frac{\pi}{2}$; $\lambda$, es un factor de escala entre sus módulos.


\end{document}

