If v = z/y, the factor 1/(y-z), in the above integral, becomes 1/y times 1/(1-v). Now begin to expand 1/(1-v) as a geometric series. The nth approximation is as follows.
1/(1-v) = 1 + v + v2 + v3 + … + vn-1 + vn/(1-v)
Multiply through by f(y) over 2πiy, then evaluate the contour integral around our circle of radius r, for the left and right sides of the equation. The left integrand has become f(y) over 2πiy×(1-v), or f(y) over 2πi×(y-z), and the resulting integral is simply f(z). The right side is a bit more complicated.
Look at the kth term on the right. Replace vk with zk/yk. The contour integral runs with respect to y, and z is a constant. Pull zk out of the integral, and concentrate on f(y) over 2πiyk+1.
At this point we must refer to an earlier theorem on analytic functions, which describes the kth derivative of f at 0. It is equal to k!/2πi times the integral of f(y)/yk+1. This looks a lot like our expression for the kth term on the right. After a little algebra, the kth term becomes the kth derivative of f at 0, times zk, over k factorial. This holds up to k = n, giving the following formula, which looks like a taylor approximation with an error term.
f(z) = f(0) + f′(0)z + f′′(0)z2/2 + f′′′(0)z3/6 + … (up to n-1) +
∫ f(y)vn over 2πi×(y-z)
Let m be the maximum of |f| on the disk of radius r, and use this to bound the error term. Replace f(y) with m. With z strictly inside the circle, the norm of the denominator is bounded above 0, and the norm of v is bounded below 1. Thus the error term approaches 0 as n approaches infinity, and the power series of f converges to f.
Let s be the radius of a smaller circle inside our circle of radius r. For any z inside or on s, v is bounded below s/r, which is bounded below 1. The power series converges uniformly to f on the closed disk of radius s.