Taylor Series in ℂ

We now have complex power series and their disks of convergence. The central theorem of complex analysis says these power series are everywhere: every holomorphic function is, locally, the sum of one.

If f is holomorphic on a domain containing the disk |z - z_0| < R, then on that whole disk f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}\,(z - z_0)^n.
The series converges on the largest disk about z_0 that fits inside the domain of holomorphy.
In ℂ, holomorphic and analytic mean the same thing — a function differentiable once on an open set automatically equals its Taylor series there.

That last point is the surprise. On the real line, smoothness does not imply being a power series. In ℂ it does, and the infinite differentiability we already proved is exactly what supplies the coefficients f^{(n)}(z_0).

Why it works: expand the Cauchy kernel

Step 1 — start from Cauchy's integral formula. For z inside a circle \gamma about z_0,

f(z) = \frac{1}{2\pi i}\oint_\gamma \frac{f(w)}{w - z}\,dw.

Step 2 — rewrite the kernel around z_0. Factor w - z = (w - z_0) - (z - z_0) and pull out w - z_0:

\frac{1}{w - z} = \frac{1}{(w - z_0)\left(1 - \frac{z - z_0}{w - z_0}\right)}.

Step 3 — expand as a geometric series. Since w is on \gamma and z is inside, \left|\frac{z - z_0}{w - z_0}\right| < 1, so the geometric series applies:

\frac{1}{w - z} = \sum_{n=0}^{\infty} \frac{(z - z_0)^n}{(w - z_0)^{n+1}}.

Step 4 — integrate term by term. Insert this and swap sum and integral. Each coefficient is a Cauchy integral, which the generalised formula identifies as a derivative:

f(z) = \sum_{n=0}^{\infty} (z - z_0)^n \cdot \frac{1}{2\pi i}\oint_\gamma \frac{f(w)}{(w - z_0)^{n+1}}\,dw = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}\,(z - z_0)^n.

Worked series and the radius rule

About z_0 = 0, the familiar series carry straight over to ℂ, now valid on a disk:

e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}, \qquad \sin z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!}, \qquad \frac{1}{1 - z} = \sum_{n=0}^{\infty} z^n.

The first two are entire (holomorphic everywhere), so R = \infty. The third has a pole at z = 1, and its radius is exactly R = 1 — the distance from 0 to that pole.

The radius of convergence of the Taylor series of f about z_0 equals the distance from z_0 to the nearest singularity of f.
The disk grows until it touches the closest point where f fails to be holomorphic — there it must stop.

On the real line, \frac{1}{1 + x^2} = 1 - x^2 + x^4 - \cdots converges only for |x| < 1, even though the function is smooth and finite for every real x. Nothing on the line explains the wall at 1. The complex plane does: the denominator 1 + z^2 vanishes at z = \pm i, each at distance 1 from the origin. The real series feels a singularity it cannot see — a pole hiding off the real axis. The radius of convergence is the distance to the nearest complex singularity, and that single fact explains every otherwise-mysterious real radius.

The disk grows until it hits a pole

Two fixed singularities sit at z = \pm i (the poles of 1/(1 + z^2)). Move the expansion centre z_0 with the sliders. The Taylor disk about z_0 grows until its boundary just touches the nearer of the two poles — so its radius R is precisely that distance, shown in the readout.