This is quite different from the real numbers. Integrate abs(x) to find a function that is everywhere differentiable, but is not twice differentiable at 0. Somehow, complex differentiation is far more powerful. You can differentiate again and again, forever.
Remember that analytic means differentiable in a neighborhood about p, so let r be the radius of a circle, centered at p, that lies entirely inside this neighborhood. Now f is analytic on and inside the circle of radius r. we'll be taking contour integrals around this circle.
Cauchy's integral formula says that f(p) = w times the integral of f(z)/d, around the circle of radius r, where w = 1/2πi and d = z-p. (The symbols w and d are notational conveniences.) We will show that f′(p) is w times the integral of f(z)/d2. Furthermore, the second derivative is w times the integral of 2f(z)/d3. In general, the nth derivative is n!w times the integral of f(z)/dn+1. This formula holds for all p in the analytic neighborhood. Let's begin with the first derivative.
Let h be a small complex number, smaller than ½r, and consider the difference quotient f(p+h)-f(p) over h. Replace f(p) with w times the contour integral of f(z)/d. Similarly, replace f(p+h) with w times the contour integral of f(z)/(d-h). Subtract these two integrands and cancel h in the top and bottom to get an integrand of f(z) over d×(d-h). In the limit, as h approaches 0, we have the following.
f′(p) = w × ∫f(z) over d×(d-h)
We would like to show that f(z)/d2 would do just as well. So subtract the two integrands and show that the difference, as an integral, approaches 0 as h approaches 0.
f(z) / d(d-h) - f(z) / d2 = hf(z) / (d-h)d2
Let's see if we can make the right hand side, the difference integral, go to zero. Since f is continuous it is bounded on the disk of radius r. And d has a fixed norm of r. As long as h is less than half of r, d-h has a norm at least half of r. So the entire integrand is bounded by some constant times h. As h approaches 0 the integrand is bounded near 0. The direction vector of the path has norm r, so the contour integral around the circle approaches 0. Therefore we can write the following.
f′(p) = w × ∫f(z)/d2
Now evaluate the second derivative at the point p. Consider the difference quotient f′(p+h)-f′(p) over h, as h approaches 0.
Replace each first derivative in the numerator with its contour integral. Then subtract:
∫f(z) / (d-h)2 - ∫f(z) / d2 = ∫(2dh-h2)×f(z) / (d-h)2d2
Remember, the difference quotient has an h in the denoninator. Divide through by h and the first factor becomes 2d-h. Since d±h approaches d for small h, we have 2df(z)/d4, or 2f(z)/d3.
f′′(p) = w ∫2f(z)/d3
A similar process computes the third derivative, and so on. By induction, the nth derivative exists, and is equal to w×n! times the contour integral of f(z)/dn+1.
To be analytic in a region is to be infinitely differentiable throughout that region.