Question

комплексные числа (3+2i)(2-2i)/(2+3i)(1+i)

Answer 1

(3+2i)(2-2i)/(2+3i)(1+i)=(6-6i+4i-4i²)/(2+2i+3i+3i²)=

=(6-2i+4)/(2+5i-3)=(10-2i)/(-1+5i)=

=(10-2i)(-1-5i)/(-1+5i)(-1-5i)=(-10-50i+2i+10i²)/(1+5i-5i-25i²)=

=(-10-48i-10)/(1+25)=(-20-48i)/26=-10/13-1 11/13i