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April 20

complex number differentaition

f(Z) is complex function as its first derivative exists,how can i prove that its higher derivative exists.why it is not true in real functions — Preceding unsigned comment added by 182.187.31.17 (talk) 09:08, 20 April 2013 (UTC)[reply]

For the former, presumably you want to use Cauchy's integral formula (assuming its first derivative exists in a suitable open set). As to why it's not true in real functions, consider a function that's x^2 for x<0 and -x^2 for x>0. Does that help? Straightontillmorning (talk) 11:13, 20 April 2013 (UTC)[reply]

If a complex function is continuously differentiable on an open set, then it is in fact complex analytic, meaning that it has smooth derivatives of all orders. This is pretty amazing, and very different from the case of real functions. For intuition as to why, consider that continuity of a complex derivative is far more restrictive for complex functions than for real functions. Real functions must only have limits from the left and right, but the complex function must have convergent limits for any sequence z_n->z*. SemanticMantis (talk) 15:05, 20 April 2013 (UTC)[reply]