Question

(-2+i)÷(4+6i) Division de numeros complejos

Answer 1

i^2=-1
(-2+i)/(4+6i)=(-2+i)*(4-6i)/(4+6i)(4-6i)=(-8+4i+12i-36i^2)/(16-36i^2)=
(-8+16i+36)/(16+36)=(28+16i)/52=28/52+16/52i=7/13+4/13i