Question

calcule(1+i).(2-i).(3-2i)

Answer 1

$(1+i)\cdot(2-i)\cdot(3-2i)= \\ \\ =(2-i+2i-i^2)\cdot(3-2i)= \\ \\ =(2+i-i^2)\cdot(3-2i)= \\ \\ =(6-4i+3i-2i^2-3i^2+2i^3)= \\ \\ =(6-i-5i^2+2i^3)= \\ \\ =(6-i-5(-1)+2(-i))= \\ \\ =(6-i+5-2i)= \\ \\ =11-3i$