Not so. Roundoff errors happen because when you add a big number and a small number in finite precision, the result loses information that was in the small numb

Not so. Roundoff errors happen because when you add a big number and a small number in finite precision, the result loses information that was in the small number. The magic of complex step differentiation is keeping the small part (the imaginary part) separate from the (potentially) large part (the value of the function evaluated at a real argument). You can use a very small step size (and so get very high accuracy) with traditional finite difference differentiation at x=0 for functions with f(0) = 0. Complex step differentiation takes advantage of the fact that real valued functions of a real argument have 0 imaginary part everywhere on the real line. > most problems won't be [well conditioned] and the complex step method won't help much then either Try coming up with an example of this. You might be able to find one if there is a pole or branch cut very near the real line, but otherwise, complex step differentiation is very robust.