Respuesta :

[tex] \frac{10}{1 + 3i} = \frac{10}{1 + 3i} * \frac{1-3i}{1-3i} = \frac{10-30i}{1^{2} -(3i)^{2}} = \frac{10-30i}{1 -9 i^{2} } = \frac{10-30i}{1 -9*(-1)} = \frac{10(1-3i)}{1+9}= \frac{10(1-3i)}{10} [/tex] [tex]= 1 -3i[/tex]