z2 = -1 + i = sqrt(2)*(-1/sqrt(2) + i*sqrt(2)) = sqrt(2)*(cos(3π/4) + i*sin(3π/4)) = sqrt(2)*e^(i*3π/4)
z3 = -i = 1*(cos(3π/2) + i*sin(3π/2)) = 1*e^(i*3π/2)
z4 = -4 = 4*(cos(π) + i*sin(π)) = 4*e^(i*π)
И в конце, на всякий случай, равенство Эйлера:
e^(i*π) = -1