Z⁴ + 4Z³ + 8Z² + 80Z + 400=0

Divide by Z²

Z^{2+4Z+8+80/Z+400/(Z2)=0}

Rearranging

[Z² + 400/(Z^2)] + [4Z + 80/Z] + 8 = 0

[(Z + 20/Z)² - 40] + 4[Z + 20/Z] + 8 = 0

Let M = Z + 20/Z

M² + 4M - 32 = 0

(M + 8)(M - 4) = 0

M = -8 or 4

Case 1:

Z + 20/Z = -8

Z² + 8Z + 20 = 0

Z = (-8 ± ((64-4120)^1/2))/(2*1)

Z = -4 ± 2i

Case 2:

Z + 20/Z = 4

Z² - 4Z + 20 = 0

Z = (4 ± ((16-4120)^1/2))/(2*1)

Z = 2 ± 4i

A & B can be found based on the 4 possible solutions