the generalized formula for derivatives
If the boundary values of a holomorphic function determine its value inside, can they also determine its slope, its curvature, every higher derivative? Yes — and by the same kind of integral. The generalized Cauchy formula gives every derivative f^(n)(z_0) as a contour integral of f over a surrounding loop, with no differentiation of f actually carried out.
The statement: for f holomorphic on and inside a simple closed contour gamma and z_0 inside, f^(n)(z_0) = (n! / (2 pi i)) times the integral over gamma of f(z) / (z - z_0)^(n+1) dz, for every n = 0, 1, 2, and so on (n = 0 recovers the ordinary formula). You can believe it by differentiating Cauchy's formula under the integral sign with respect to z_0: differentiating the kernel 1 / (z - z_0) once gives 1 / (z - z_0)^2, again gives 2 / (z - z_0)^3, and after n steps you collect n! / (z - z_0)^(n+1). The derivatives of f are read off the same boundary data, just weighted by a higher power of the kernel.
The consequence is enormous and has no real-variable counterpart: because the right-hand side makes sense for every n whenever f is merely once complex-differentiable, f automatically has derivatives of all orders. One complex derivative secretly contains infinitely many. This is the step that turns 'holomorphic' into 'infinitely smooth' and, soon after, into 'equal to its own Taylor series.'
To get f''(z_0), use f''(z_0) = (2! / (2 pi i)) times the integral over gamma of f(z) / (z - z_0)^3 dz; with f(z) = z^2 and z_0 = 0, the integral picks out 2! times the coefficient pattern and returns f''(0) = 2.
The n-th derivative is a single contour integral with kernel 1 / (z - z_0)^(n+1) and a factor n!.
Differentiating under the integral sign is legitimate here because the kernel is smooth in z_0 away from gamma and gamma is bounded — but that justification is a theorem, not a license to swap limits and integrals carelessly in general.