Power series in complex analysis

Question

We derived from cauchy's integral formula that a holomorphic function converges locally in a power series. Now we had the Identity theorem and I wanted to know whether I can conclude from this that the power series of a function is uniquely determined and converges on the whole domain of the function(which should be assumed to be connted and open) and not just locally anymore?

Answer 1

Consider the function $f(z)=\frac{1}{z}$. It is analytic in $\mathbb C - \{0\}$. This function can't be expressed as a power series on its entire domain of definition, for the following reason. Suppose $f$ is given as a power series expansion at $z=0$. This is impossible, since such a power series can be extended to $z=0$, while the function $f(z)$ has an essential discontinuity at $z=0$. So, suppose $f$ is given as a power series expansion about $z=z_0\ne 0$. The assumption is thus that the power series converges for every $w\ne 0$, and that it gives $f(w)$. But then the radius of convergence of the power series is $\infty $, and thus, again, the function $f(z)$ should be extendable to $z=0$, which is false.

Answer 2

Most certainly not. Consider the function $1/(1-z)$, analytic on $\mathbb C\setminus\{1\}$. Its power-series expansion at $0$ is $1+z+z^2+\cdots\,$, which isn’t convergent on the unit circle at all, nor, of course, beyond that.