Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 1 / 9D / i / Solution

Let $M$ bound all $|a_n{w}^n|$ and put $q=|z|/|w|<1$. ThenThe geometric series ${\sum }_{n=0}^{\infty }M{q}^n$ converges, so the comparison test for series proves absolute convergence of $\sum a_n{z}^n$. This is the bounded power-series terms force interior absolute convergence principle.

Define the radius of convergence byallowing $\rho =0$ and $\rho =\infty$. If $|z|<\rho$, the supremum supplies a point of convergence $w$ with $|w|>|z|$; convergence makes its terms bounded, so the argument above gives absolute convergence at $z$. If $|z|>\rho$, convergence is impossible by definition. At $|z|=\rho$, no general conclusion follows. Thus the convergence region has a circular interior centered at zero, with boundary behavior depending on the power series.