Laurent's expansion confusing

Question

Following H.A.Priestley's Introduction to Complex Analysis, it looks like Laurent's expansion only defined at $A:=\{z:R<|z|<T\}$ where $R$ can be zero. However, I also seen questions where we are required to find the expansion for the unit disc. In which case the region cannot be put in the form $\{z:R<|z|<T\}$. So is the question equivalently asking to find the expansion in the region $\{z:0<|z|<1\}?$ I rather get the feeling two things are not equivalent.

Answer 1

Facts:

1. Each holomorphic function $f : D \to \mathbb C$ defined on an open disk $D = D(c;R) = \{ z \in \mathbb C \mid \lvert z - c \rvert < R \}$ has a unique Taylor expansion $f(z) = \sum_{n=0}^\infty a_n (z-c)^n$ on $D$.

2. Each holomorphic function $f : A \to \mathbb C$ defined on an open annulus $A = A(c;R,T) = \{ z \in \mathbb C \mid R < \lvert z - c \rvert < T \}$ has a unique Laurent expansion $f(z) = \sum_{n=-\infty}^\infty a_n (z-c)^n$ on $A$.

For a Laurent expansion it may of course happen that $a_n = 0$ for $n < 0$. In that case $f$ has a holomomorphic extension $F : D(c;T) \to \mathbb C$.

Thus, if you have a holomorphic function $f$ on the punctured unit disk $D^*$, it has a Laurent expansion $f(z) = \sum_{n=-\infty}^\infty a_n z^n$ on $D^*$. This series has $a_n = 0$ for $n < 0$ iff $f$ is bounded. In fact, precisely when $f$ is bounded the singularity at $0$ is removable.