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1) Dados los números complejos $\\color{blue} {z_1=3-2i}$ y $\\color{red} {z_2=5+7i}$ ¿Cuál alternativa muestra el resultado de $z_1 \\cdot z_2$.
\nA) $29+11i$
\nB) $15+14i$
\nC) $29-11i$
\nD) $11+29i$
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\nPaso 1: Multiplicar término a término los binomios que conforman cada número complejo. | \n
\n $(3-2i) \\cdot (5+7i)$ \n$=$[[0]]$+$[[1]]$i$ $-$ [[2]]$i$ $-$[[3]]$i^2$ \n | \n
Paso 2: Recordar que $i^2=-1$, por lo tanto, al final se produce un cambio de signo en la expresión quedando: | \n
$=$[[0]]$+$[[1]]$i$ $-$[[2]]$i$ $+$[[3]] | \n
Paso 3: Se reduce la expresión del paso 2, juntando lo real con real (lo que no tiene $i$), y luego lo imaginario con imaginario (todos aquellos número con $i$) | \n
$=$[[4]][[6]][[5]]$i$ | \n
Por lo tanto la alternativa correcta es: [[7]]
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