z^2+3+4i=0
z = (a+bi)
z^2 = (a+bi)^2 = –3 –4i
a^2 + 2abi – b^2 = –3 –4i
(a^2 – b^2) + 2abi= –3 –4i
Система:
a^2 – b^2 = –3
2ab = -4 стремится к b = -2/a
a^2 – (-2/a)^2 = –3
a^2 – 4/(a^2) = –3
(a^4 + 3a^2 – 4) / (a^2) = 0
a^4 + 3a^2 – 4 = 0 (a≠0)
(a^2 + 4)(a^2 – 1)=0
Итого:
a = 1, b = -2: z_1 = 1-2i
b = -1, b = 2: z_2 = -1+2i