Annuli in the complex plane

Let $a\in \mathbb{C}$ and let $0\le r<R\le \mathrm{\infty }$. The annulus about $a$ with inner radius $r$ and outer radius $R$ is

$A(a;r,R):=\{z\in \mathbb{C}:r<\mathrm{\mid }z-a\mathrm{\mid }<R\}\mathrm{.}$

When $R=\mathrm{\infty }$, the condition $\mathrm{\mid }z-a\mathrm{\mid }<R$ is omitted, so $A(a;r,\mathrm{\infty })=\{z:\mathrm{\mid }z-a\mathrm{\mid }>r\}$. When $r=0$ and $R<\mathrm{\infty }$, the annulus is the punctured disc $0<\mathrm{\mid }z-a\mathrm{\mid }<R$.

The boundary circles $\mathrm{\mid }z-a\mathrm{\mid }=r$ and $\mathrm{\mid }z-a\mathrm{\mid }=R$ are not part of the annulus. In particular $A(a;0,R)$ is not the open disc $\mathrm{\mid }z-a\mathrm{\mid }<R$, because the centre $a$ is missing.

The finite annulus $A(a;r,R)$ with $0<r<R<\mathrm{\infty }$, the punctured disc $A(a;0,R)$, and the exterior domain $A(a;r,\mathrm{\infty })$ are treated by the same notation because Laurent expansions on all three have the same local form.