the Cauchy integral as a route to power series
How do you prove that a holomorphic function equals a power series — that f(z) = sum a_n (z - z_0)^n on a disk? The cleanest road starts at Cauchy's integral formula and expands its kernel as a geometric series. This is the bridge from the integral formula to Taylor's theorem, the place where 'holomorphic' becomes 'analytic.'
The idea in steps. Fix z_0 and a circle gamma around it; write f(z) = (1 / (2 pi i)) times the integral over gamma of f(w) / (w - z) dw for z inside. Split the kernel: 1 / (w - z) = 1 / ((w - z_0) - (z - z_0)) = (1 / (w - z_0)) times 1 / (1 - (z - z_0)/(w - z_0)). Since |z - z_0| is less than |w - z_0| on gamma, the second factor is a convergent geometric series, sum over n of ((z - z_0)/(w - z_0))^n. Substitute, swap sum and integral (justified by uniform convergence), and out drops f(z) = sum a_n (z - z_0)^n with a_n = (1 / (2 pi i)) times the integral over gamma of f(w) / (w - z_0)^(n+1) dw — which is exactly f^(n)(z_0) / n! by the generalized formula.
So the Cauchy integral does not just compute one value; expanded as a geometric series it manufactures the entire Taylor series at once, and even tells you the coefficients are the derivative values. The radius of convergence reaches all the way to the nearest singularity, because that is as far as you can grow gamma. (This term previews the result; the full power-series development, including convergence details, lives in the power-series field.)
For f(z) = 1 / (1 - z) and z_0 = 0, the kernel expansion reproduces the geometric series f(z) = 1 + z + z^2 + ..., with radius of convergence 1 — exactly the distance to the singularity at z = 1.
Expanding Cauchy's kernel geometrically yields the Taylor series and its radius.
The series is built from values on a circle, so its radius of convergence equals the distance from z_0 to the nearest singularity — you cannot expand past where f first fails to be holomorphic.