Power series are holomorphic in the disc of convergence

Statement

Let $\sum_{n \ge 0} a_n (z - c)^n$ be a power series with radius of convergence $R > 0$, and let $D = \{z : |z - c| < R\}$ be its disc of convergence. Let $f(z) = \sum_{n=0}^{\infty} a_n (z - c)^n$ for $z \in D$. Then $f$ is holomorphic on $D$, and for each $z \in D$, $$f'(z) = \sum_{n=1}^{\infty} n a_n (z - c)^{n-1}.$$ The series for $f'$ has the same radius of convergence $R$.

Remarks

By induction, $f$ has complex derivatives of all orders on $D$, and $a_n = f^{(n)}(c) / n!$. So the power series is the Taylor series of $f$ at $c$. For example, the complex exponential has $R = \infty$, so it is entire. The converse is also true: holomorphic functions are analytic.

Explanations

L1 human

Inside its disc of convergence, a power series can be differentiated term by term, like a polynomial. The result is again a power series with the same disc, so you can do it again and again.