Question

z²-(10+4i)z+21+38i=0​

Answer 1

Ответ: z1 = 5+2i+3i√(2i).

z2 = 5+2i-3i√(2i).

Решение:

z²-(10+4i)z+21+38i=0​

a = 1; b = -(10+4i); c = 21+38i

D = b²-4ac

D = (-(10+4i))² - 4(21+38i) = (100+80i+16i²) -

(84+152i) = 100+80i+16i²-84-152i = 100+80i-16-84-152i = 100 - 100 - 72i = -72i

i² = -1

z1,z2 = (-b±√D)/2a

z1,z2 = ((10+4i)±√(-72i))/2 = ((10+4i)±√(72i³))/2

z1 = (10+4i+6i√(2i))/2 = 5+2i+3i√(2i).

z2 = 5+2i-3i√(2i).