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If f is holomorphic in a simply connected domain D and γ encloses z₀, then fn(z₀) = (n!/2πi) ∮_γ f(z)/(z - z₀)n+1 dz for any positive integer n. This shows every holomorphic function is infinitely differentiable and all derivatives are also holomorphic. It is the gateway to Taylor series.
From Cauchy's Integral Formula you know that the value of a holomorphic function at any interior point is completely determined by its values on the boundary: f(z₀) = (1/2πi) ∮_γ f(z)/(z - z₀) dz. The generalization to derivatives extends this result: not just f(z₀) but every derivative fn(z₀) is recoverable from the boundary integral, via fn(z₀) = (n!/2πi) ∮_γ f(z)/(z - z₀)n+1 dz. Each successive derivative introduces one higher power of (z - z₀)−1 in the denominator and one factor of n! in the numerator.
The derivation follows by differentiating Cauchy's Integral Formula with respect to z₀ under the integral sign. Starting from f(z₀) = (1/2πi) ∮ f(z)/(z - z₀) dz, differentiate the integrand: d/dz₀[1/(z - z₀)] = 1/(z - z₀)². Differentiating n times: dn/dz₀^n[1/(z - z₀)] = n!/(z - z₀)n+1. This is simply the power rule applied to the function of z₀. The formula for fn(z₀) follows directly, with the factor n! appearing because differentiating 1/(z - z₀) exactly n times produces n! in the numerator.
The conceptual significance is profound. In real analysis, differentiability once does not imply differentiability twice — a function can have exactly one derivative and no more. In complex analysis, holomorphicity (complex differentiability) implies infinite differentiability: since the integral formula exists for every n, fn(z₀) exists for every n, and each derivative is again holomorphic. This is one of the deepest asymmetries between real and complex analysis. Holomorphic functions form a far more rigid class than real-differentiable functions.
The practical payoff is in computing Taylor coefficients. If f has a Taylor series f(z) = Σ aₙ(z - z₀)ⁿ, the n-th coefficient is aₙ = fn(z₀)/n!. Substituting the derivative formula gives aₙ = (1/2πi) ∮_γ f(z)/(z - z₀)n+1 dz. This bridges the integral representation of a holomorphic function and its power series expansion, and it is the key step in proving that every holomorphic function equals its Taylor series on its disk of convergence — the result you will encounter next.