Radius of convergence
Definition. A complex power series centered at \(a\) has the form \(\sum_{n=0}^\infty c_n(z-a)^n\). Its radius is \(R=(\limsup |c_n|^{1/n})^{-1}\), with the usual conventions. It converges absolutely for \(|z-a|<R\) and diverges for \(|z-a|>R\).
Uniform convergence on compact subdiscs
Proposition. For every \(0<r<R\), the series converges uniformly and absolutely on \(\overline{D(a,r)}\). This follows from the Weierstrass test after choosing \(\rho\) with \(r<\rho<R\) and comparing \(|c_n|r^n\) with a geometric tail.
Termwise differentiation
Theorem. Inside the disc of convergence, a power series defines a holomorphic function and may be differentiated term by term: \[ \frac{\mathrm{d}}{\mathrm{d}z}\sum_{n=0}^\infty c_n(z-a)^n=\sum_{n=1}^\infty n c_n(z-a)^{n-1}. \] The differentiated series has the same radius of convergence.
Identity of coefficients
Proposition. If two power series agree on a set with an accumulation point inside the common disc of convergence, then all their coefficients agree. In particular, if \(f(z)=\sum c_n(z-a)^n\), then \(c_n=f^{(n)}(a)/n!\).
Analytic continuation warning
Remark. A power series gives a local representation. The same holomorphic function may require different power series around different centers, and singularities determine how far each local expansion can reach.
Geometric series model
Example. The function \((1-z)^{-1}\) equals \(\sum_{n\ge0}z^n\) on \(|z|<1\). Around \(a\ne1\), it has a different expansion whose radius is exactly \(|1-a|\), the distance to the nearest singularity.
Cauchy--Hadamard radius
Theorem. For \(\sum_{n\ge0}a_n(z-z_0)^n\), let \[ \frac1R=\limsup_{n\to\infty}|a_n|^{1/n} \] with the usual conventions. The series converges absolutely for \(|z-z_0|<R\) and diverges for \(|z-z_0|>R\).
Proof. Apply the root test: \[ \limsup |a_n(z-z_0)^n|^{1/n} =|z-z_0|/R. \] The test is inconclusive on the boundary, where behavior can vary from point to point.
Uniform convergence and differentiation
Theorem. Inside its radius of convergence, a power series converges uniformly on every closed subdisc and may be differentiated term by term: \[ \left(\sum_{n\ge0}a_n(z-z_0)^n\right)' =\sum_{n\ge1}na_n(z-z_0)^{n-1}. \] The differentiated series has the same radius \(R\).
Proof. Choose \(r<\rho<R\). Since \(\sum|a_n|\rho^n<\infty\), \(|a_n(z-z_0)^n|\le|a_n|\rho^n\) on \(|z-z_0|\le r\), proving uniform convergence. Also \[ n|a_n|r^{n-1} \le\frac1r n(r/\rho)^n|a_n|\rho^n. \] The factor \(n(r/\rho)^n\) is bounded, so the derivative series converges uniformly on the smaller disc. The standard theorem on uniform convergence of derivatives identifies the derivative. Since \(n^{1/n}\to1\), Cauchy--Hadamard gives the same radius.
Algebra and composition
Power series with positive radius can be added and multiplied by the Cauchy product. If \(g(z_0)=w_0\) and the image of a small disc lies inside the convergence disc of \(\sum b_n(w-w_0)^n\), then substitution defines the composition. Absolute uniform convergence justifies rearrangement. If \(a_0\ne0\), recursive coefficient comparison constructs a reciprocal series, so analytic functions form a local algebra.
Uniqueness of coefficients
Proposition. If a power series represents \(f\) near \(z_0\), then \[ a_n=\frac{f^{(n)}(z_0)}{n!}. \] Consequently, two analytic functions agreeing on a neighborhood have identical Taylor coefficients.
Proof. Differentiate termwise \(n\) times and set \(z=z_0\); all terms vanish except \(n!a_n\).
Boundary behavior and continuation
The radius of a Taylor series is the distance to the nearest obstruction to holomorphic continuation in the complex plane, not to a singularity visible only on the real line. For example, \[ \frac1{1+z^2}=\sum_{n\ge0}(-1)^nz^{2n} \] has radius \(1\) because of poles at \(\pm i\), although the function is smooth for every real \(z\). Analytic continuation may extend beyond one convergence disc through overlapping expansions.