Как решить: 2+3i/5-2i i-1/i+1
=(2+3i)(5+2i)/(5-2i)(5+2i)= (10+4i+15i+6i²)/(25-4i²)= (4+19i)/29 2)=(i-1)(i-1)/(i+1)(i-1)=(i²-i-i+1)/(i²-1)=2i/2=i (i²=-1)