Taylor's theorem for complex functions

1) For real functions we only get an approximation (with error bounds) while for complex functions we actually have that the power series is equal to the original function.
2) We only require a complex function to be once complex differentiable, while a real function has to be several times differentiable to apply the theorem.

Indeed the complex Taylor theorem allows us to deduce that any analytic function is in fact infinitely complex differentiable, while on the other hand even if a real function is infinitely (real) differentiable this does not guarantee the existence of a power series that equals the function (the traditional counterexample being f(x) = exp(-x ^-2).
The proof is essentially an application of the Cauchy integral formula, together with some technical fiddling to ensure that everything converges properly. Unfortunately the HTML makes for very unpleasant reading.

Taylor's theorem for complex functions:
Let f : D → C be an analytic function on a domain D containing 0. Then if R > 0 is such that z ∈ D for all |z| ≤ R then for any |z| < R (sum from k = 0 to ∞)

f(z) = Σ z^ka_k

where (integrals round the circle of radius R centred at 0)

a_k = (2πi)^-1∫ f(w)/w^k dw

Proof:
Note the identity (this and subsequent sums from k = 0 to n)

(w-z)^-1 = w^-1(Σ (z/w)^k) + (z/w)ⁿ⁺¹/(z-w)

which holds for all n ∈ N when |z| < |w|. Hence by the Cauchy integral formula

2π*|f(z) - Σ z^ka_k| = |∫ (w-z)^-1f(w) dw - Σ (z/w)^kf(w) dw| = |∫ (z/w)ⁿ⁺¹(z-w)^-1f(w) dw| ≤ (|z|/R)ⁿ⁺¹|∫ (z-w)^-1f(w) dw| → 0

as n → ∞ for any |z| < R. So the series converges to f(z).

Corollary:
Any analytic function f : D → C is infinitely complex differentiable on D.

Proof:
By Taylor's theorem f can be written locally as a power series, and any power series is infinitely differentiable.