Resposta :

[tex]\boxed{i^2=-1}\\\\ \frac{8+i}{1+2i} = \frac{1-2i}{1-2i} = \frac{8-16i+i-2i^2}{1-2i+2i-4i^2} = \frac{10-15i}{5} = \boxed{2-3i}[/tex]

Alternativa c) 2-3i

Outras perguntas