Question

решить уравнение над полем комплексных чисел x^2-(3-2i)x+5-5i=0

Answer 1

х²-(3-2i)x+5-5i=0
D=(3-2i)²-4(5-5i)=9-12i+4i²-20+20i=

8i-15
x=(3-2i±√(8i-15))/2

D=2*4*i-16-i²=-(i²-8x+16)=
-(i-4)²
x=(3-2i±|i-4|*i)/4