Задание 9. Z²+(3i–2)z+5–3i=0
z2+(3i–2)z+(5–3i)=0 D=(3i–2)2–4·(5–3i)=9i2–12i+4–20+12i=–25 √D=5i z1=(–3i+2–5i)/2=1–4i или z2=(–3i+2+5i)/2=1+i О т в е т. 1–4i; 1+i