Question

(i-1)(3i-1)/(2-i)^2 ХЕЛП

Answer 1

(i-1)(3i-1)/((2-i)²=

(3i²-i-3i+1)/(4-4i+i²)=

(-3-4i+1)/(4-4i-1)=
(-2-4i)/(3-4i)=-(2+4i)(3+4i)/(3-4i)(3+4i)=
(-6-8i-12i-16i²)(9-16i²)=

(-20i+10)/25=10/25 -20i/25= 2/5-4i/5