Una desigualdad que involucra dos números complejos

Question

Una desigualdad que involucra dos números complejos

desigualdad
Matemáticas
números complejos
álgebra-precálculo

usuario350331

Dejar $z_1, z_2 \in \mathbb C$ y $a,b \in \mathbb{R} \setminus \{0\}$ . Pruebalo

$| z_{1} |^{2} + | z_{2} |^{2} - | z_{1}^{2} + z_{2}^{2} | \leq 2 \frac{| a z_{1} + b z_{2} |^{2}}{a^{2} + b^{2}} \leq | z_{1} |^{2} + | z_{2} |^{2} + | z_{1}^{2} + z_{2}^{2} |$ $|z_1|^2+|z_2|^2-|z_1^2+z_2^2|\le 2\dfrac{|az_1+bz_2|^2}{a^2+b^2}\le |z_1|^2+|z_2|^2+|z_1^2+z_2^2|$

Intento de solución: Dejar $z=a+ib$ . Entonces, $2\dfrac{|az_1+bz_2|^2}{a^2+b^2}$ se puede simplificar en lo siguiente

2 \frac{| a z_{1} + b z_{2} |^{2}}{a^{2} + b^{2}} ⟹ 2 {| \frac{(\frac{z + \bar{z}}{2}) z_{1} + (\frac{z - \bar{z}}{2 i}) z_{2}}{z} |}^{2} ⟹ \frac{{| ℜ (z (z_{1} + i z_{2})) |}^{2}}{| z |^{2}}

$\begin{equation} 2\dfrac{|az_1+bz_2|^2}{a^2+b^2} \implies 2\left|\dfrac{\left(\dfrac{z+\bar{z}}{2}\right)z_1+\left(\dfrac{z-\bar{z}}{2i}\right)z_2}{z}\right|^2 \implies\dfrac{\left|\Re(z(z_1+iz_2))\right|^2}{|z|^2} \end{equation}$

Traté de sustituir $z=a+ib; z_1=x_1+iy_1; z_2=x_2+iy_2$ pero simplemente se convirtió en un lío que creo que no puedo reorganizar para que sea algo útil para probar la desigualdad. Si es posible, proporcione también el significado geométrico de esta desigualdad.

laboratorio bhattacharjee

Dejar

z_{r} = x_{r} + i y_{r}

$z_r=x_r+iy_r$ para

r = 1, 2

$r=1,2$

2 | a z_{1} + b z_{2} |^{2} - (a^{2} + b^{2}) (| z_{1} |^{2} + | z_{2} |^{2}) = \dots

$2|az_1+bz_2|^2-(a^2+b^2)(|z_1|^2+|z_2|^2)=\cdots$

= (a X_{1} + b X_{2})^{2} + (a y_{1} + b y_{2})^{2} - {(a X_{2} - b X_{1}) + (a y_{2} - b y_{1})^{2}}

$=(ax_1+bx_2)^2+(ay_1+by_2)^2-\{(ax_2-bx_1)+(ay_2-by_1)^2\}$

| z_{1}^{2} + z_{2}^{2} | = \sqrt{(a_{1}^{2} X_{1}^{2} - b_{1}^{2} y_{1}^{2} + a_{2}^{2} X_{2}^{2} - b_{2}^{2} y_{2}^{2})^{2} + (2 a_{1} b_{1} X_{1} y_{1} + 2 a_{2} b_{2} X_{2} y_{2})^{2}}

$|z_1^2+z_2^2|=\sqrt{(a_1^2x_1^2-b_1^2y_1^2+a_2^2x_2^2-b_2^2y_2^2)^2+(2a_1b_1x_1y_1+2a_2b_2x_2y_2)^2}$

Respuestas (1)

Una desigualdad que involucra dos números complejos

Dejar $z_r=x_r+iy_r$ para $r=1,2$ $2 | a z_{1} + b z_{2} |^{2} - (a^{2} + b^{2}) (| z_{1} |^{2} + | z_{2} |^{2}) = \dots$ $2|az_1+bz_2|^2-(a^2+b^2)(|z_1|^2+|z_2|^2)=\cdots$ $= (a X_{1} + b X_{2})^{2} + (a y_{1} + b y_{2})^{2} - {(a X_{2} - b X_{1}) + (a y_{2} - b y_{1})^{2}}$ $=(ax_1+bx_2)^2+(ay_1+by_2)^2-\{(ax_2-bx_1)+(ay_2-by_1)^2\}$ $| z_{1}^{2} + z_{2}^{2} | = \sqrt{(a_{1}^{2} X_{1}^{2} - b_{1}^{2} y_{1}^{2} + a_{2}^{2} X_{2}^{2} - b_{2}^{2} y_{2}^{2})^{2} + (2 a_{1} b_{1} X_{1} y_{1} + 2 a_{2} b_{2} X_{2} y_{2})^{2}}$ $|z_1^2+z_2^2|=\sqrt{(a_1^2x_1^2-b_1^2y_1^2+a_2^2x_2^2-b_2^2y_2^2)^2+(2a_1b_1x_1y_1+2a_2b_2x_2y_2)^2}$

amor matematico · Answer 1

Dejar $z_1=x_1+iy_1,z_2=x_2+iy_2$ dónde $x_1,y_1,x_2,y_2\in\mathbb R$ .

Entonces,

| z_{1} |^{2} + | z_{2} |^{2} - | z_{1}^{2} + z_{2}^{2} | \leq 2 \frac{| a z_{1} + b z_{2} |^{2}}{a^{2} + b^{2}} \leq | z_{1} |^{2} + | z_{2} |^{2} + | z_{1}^{2} + z_{2}^{2} |

$|z_1|^2+|z_2|^2-|z_1^2+z_2^2|\le 2\dfrac{|az_1+bz_2|^2}{a^2+b^2}\le |z_1|^2+|z_2|^2+|z_1^2+z_2^2|$ es equivalente a

X_{1}^{2} + y_{1}^{2} + X_{2}^{2} + y_{2}^{2} - \sqrt{(X_{1}^{2} - y_{1}^{2} + X_{2}^{2} - y_{2}^{2})^{2} + (2 X_{1} y_{1} + 2 X_{2} y_{2})^{2}} \leq 2 \frac{(a X_{1} + b X_{2})^{2} + (a y_{1} + b y_{2})^{2}}{a^{2} + b^{2}} \leq X_{1}^{2} + y_{1}^{2} + X_{2}^{2} + y_{2}^{2} + \sqrt{(X_{1}^{2} - y_{1}^{2} + X_{2}^{2} - y_{2}^{2})^{2} + (2 X_{1} y_{1} + 2 X_{2} y_{2})^{2}}

$x_1^2+y_1^2+x_2^2+y_2^2-\sqrt{(x_1^2-y_1^2+x_2^2-y_2^2)^2+(2x_1y_1+2x_2y_2)^2}\le 2\dfrac{(ax_1+bx_2)^2+(ay_1+by_2)^2}{a^2+b^2}\le x_1^2+y_1^2+x_2^2+y_2^2+\sqrt{(x_1^2-y_1^2+x_2^2-y_2^2)^2+(2x_1y_1+2x_2y_2)^2}$ Entonces, es suficiente demostrar que

\sqrt{(X_{1}^{2} - y_{1}^{2} + X_{2}^{2} - y_{2}^{2})^{2} + (2 X_{1} y_{1} + 2 X_{2} y_{2})^{2}} \geq | X_{1}^{2} + y_{1}^{2} + X_{2}^{2} + y_{2}^{2} - 2 \frac{(a X_{1} + b X_{2})^{2} + (a y_{1} + b y_{2})^{2}}{a^{2} + b^{2}} |

$\sqrt{(x_1^2-y_1^2+x_2^2-y_2^2)^2+(2x_1y_1+2x_2y_2)^2}\ge \left|x_1^2+y_1^2+x_2^2+y_2^2-2\dfrac{(ax_1+bx_2)^2+(ay_1+by_2)^2}{a^2+b^2}\right|$

Cuadrando ambos lados,

(X_{1}^{2} - y_{1}^{2} + X_{2}^{2} - y_{2}^{2})^{2} + (2 X_{1} y_{1} + 2 X_{2} y_{2})^{2} \geq {(X_{1}^{2} + y_{1}^{2} + X_{2}^{2} + y_{2}^{2} - 2 \frac{(a X_{1} + b X_{2})^{2} + (a y_{1} + b y_{2})^{2}}{a^{2} + b^{2}})}^{2}

$(x_1^2-y_1^2+x_2^2-y_2^2)^2+(2x_1y_1+2x_2y_2)^2\ge \left(x_1^2+y_1^2+x_2^2+y_2^2-2\dfrac{(ax_1+bx_2)^2+(ay_1+by_2)^2}{a^2+b^2}\right)^2$ que es equivalente a

(X_{1}^{2} - y_{1}^{2} + X_{2}^{2} - y_{2}^{2})^{2} + (2 X_{1} y_{1} + 2 X_{2} y_{2})^{2} - (X_{1}^{2} + y_{1}^{2} + X_{2}^{2} + y_{2}^{2})^{2} \geq - 4 (X_{1}^{2} + y_{1}^{2} + X_{2}^{2} + y_{2}^{2}) \frac{(a X_{1} + b X_{2})^{2} + (a y_{1} + b y_{2})^{2}}{a^{2} + b^{2}} + 4 {(\frac{(a X_{1} + b X_{2})^{2} + (a y_{1} + b y_{2})^{2}}{a^{2} + b^{2}})}^{2}

$(x_1^2-y_1^2+x_2^2-y_2^2)^2+(2x_1y_1+2x_2y_2)^2-(x_1^2+y_1^2+x_2^2+y_2^2)^2\ge -4(x_1^2+y_1^2+x_2^2+y_2^2)\dfrac{(ax_1+bx_2)^2+(ay_1+by_2)^2}{a^2+b^2}+4\left(\dfrac{(ax_1+bx_2)^2+(ay_1+by_2)^2}{a^2+b^2}\right)^2$ que es equivalente a

- (X_{1} y_{2} - y_{1} X_{2})^{2} \geq \frac{(a X_{1} + b X_{2})^{2} + (a y_{1} + b y_{2})^{2}}{a^{2} + b^{2}} \cdot \frac{- (a X_{2} - b X_{1})^{2} - (a y_{2} - b y_{1})^{2}}{a^{2} + b^{2}}

$-(x_1y_2-y_1x_2)^2\ge \dfrac{(ax_1+bx_2)^2+(ay_1+by_2)^2}{a^2+b^2}\cdot\frac{-(ax_2-bx_1)^2-(ay_2-by_1)^2}{a^2+b^2}$ Multiplicando ambos lados por

- (a^{2} + b^{2})^{2}

$-(a^2+b^2)^2$ ,

(a^{2} + b^{2})^{2} (X_{1} y_{2} - y_{1} X_{2})^{2} \leq ((a X_{1} + b X_{2})^{2} + (a y_{1} + b y_{2})^{2}) ((a y_{2} - b y_{1})^{2} + (b X_{1} - a X_{2})^{2})

$(a^2+b^2)^2(x_1y_2-y_1x_2)^2\color{red}{\le} ((ax_1+bx_2)^2+(ay_1+by_2)^2)((ay_2-by_1)^2+(bx_1-ax_2)^2)$ Ahora, esta desigualdad se cumple por la desigualdad de Cauchy-Schwarz.

Una desigualdad que involucra dos números complejos

usuario350331

laboratorio bhattacharjee

Respuestas (1)

amor matematico

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