Question

(3-2i)÷(-2+i) Division de numeros complejos

Answer 1

i^2=-1
(3-2i)/(-2+i)=(3-2i)(-2-i)/(-2+i)(-2-i)=(-6+4i-3i+2i^2)/[(-2)^2-(i)^2]=
(-6+i-2)/(4+1)=(-8+i)/5=-8/5+1/5*i