Laurent expansion on an annulus

Fix $z\in A(a;r,R)$ and choose radii $\rho ,\sigma$ with $r<\rho <\mathrm{\mid }z-a\mathrm{\mid }<\sigma <R$; let ${\gamma }_{\rho }(t)=a+\rho \exp (it)$, let ${\gamma }_{\sigma }(t)=a+\sigma \exp (it)$, and put $\mathrm{\Gamma }:={\gamma }_{\sigma }-{\gamma }_{\rho }$.

If $r<{\rho }_1<{\rho }_2<R$, define $c_{-m}({\rho }_j):=\frac{1}{2\pi i}{\int }_{\mathrm{\mid }\zeta -a\mathrm{\mid }={\rho }_j}f(\zeta )(\zeta -a{)}^{m-1}d\zeta (j=1,2)\mathrm{.}$ For every $m\ge 1$ the integrand is holomorphic on $A(a;r,R)$, and the difference of the two circles is a cycle there whose index vanishes outside that annulus. Thus it is null-homologous in $A(a;r,R)$, and [L4] gives $c_{-m}({\rho }_2)=c_{-m}({\rho }_1)$.

If $p\mathrm{\notin }A(a;r,R)$, then either $\mathrm{\mid }p-a\mathrm{\mid }\le r<\rho$ or $\mathrm{\mid }p-a\mathrm{\mid }\ge R>\sigma$; [L1] gives $n({\gamma }_{\sigma },p)=n({\gamma }_{\rho },p)$, so [L2] gives $n(\mathrm{\Gamma },p)=0$. Both circles lie inside $A(a;r,R)$, hence $\mathrm{\Gamma }$ is null-homologous in that original annulus. Because $\rho <\mathrm{\mid }z-a\mathrm{\mid }<\sigma$, the same facts give $n(\mathrm{\Gamma },z)=1$.

On ${\gamma }_{\sigma }$ one has $\mathrm{\mid }(z-a)\mathrm{/}(\zeta -a)\mathrm{\mid }<1$, so $\frac{1}{\zeta -z}=\frac{1}{\zeta -a}\cdot \frac{1}{1-(z-a)\mathrm{/}(\zeta -a)}={\sum }_{n\ge 0}\frac{(z-a{)}^n}{(\zeta -a{)}^{n+1}},$ and the geometric series converges uniformly on that circle.

On ${\gamma }_{\rho }$ one has $\mathrm{\mid }(\zeta -a)\mathrm{/}(z-a)\mathrm{\mid }<1$, so $\frac{1}{\zeta -z}=-\frac{1}{z-a}\cdot \frac{1}{1-(\zeta -a)\mathrm{/}(z-a)}=-{\sum }_{m\ge 1}(\zeta -a{)}^{m-1}(z-a{)}^{-m},$ and this geometric series converges uniformly on that circle as well.

Applying [L3] on the original annulus $A(a;r,R)$ yields $f(z)=\frac{1}{2\pi i}{\int }_{{\gamma }_{\sigma }}\frac{f(\zeta )}{\zeta -z}d\zeta -\frac{1}{2\pi i}{\int }_{{\gamma }_{\rho }}\frac{f(\zeta )}{\zeta -z}d\zeta \mathrm{.}$

If $r<{\sigma }_1<{\sigma }_2<R$, define $c_n({\sigma }_j):=\frac{1}{2\pi i}{\int }_{\mathrm{\mid }\zeta -a\mathrm{\mid }={\sigma }_j}\frac{f(\zeta )}{(\zeta -a{)}^{n+1}}d\zeta (j=1,2)\mathrm{.}$ For every $n\ge 0$ the integrand is holomorphic on $A(a;r,R)$, the difference of the two circles is null-homologous there by the argument of step 2.1, and [L4] gives $c_n({\sigma }_2)=c_n({\sigma }_1)$.

For $n\ge 0$ set $c_n(\sigma ):=\frac{1}{2\pi i}{\int }_{{\gamma }_{\sigma }}\frac{f(\zeta )}{(\zeta -a{)}^{n+1}}d\zeta ,$ and for $m\ge 1$ set $c_{-m}(\rho ):=\frac{1}{2\pi i}{\int }_{{\gamma }_{\rho }}f(\zeta )(\zeta -a{)}^{m-1}d\zeta \mathrm{.}$ Indeed, the minus sign in the inner-circle part of step 3.1 cancels the minus sign in the geometric expansion of step 2.3. Thus [L5] applied to the uniformly convergent series of steps 2.2 and 2.3 turns step 3.1 into $f(z)={\sum }_{n\ge 0}{c}_n(\sigma )(z-a{)}^n+{\sum }_{m\ge 1}{c}_{-m}(\rho )(z-a{)}^{-m}\mathrm{.}$

Let $K_{{\rho }_0,{\sigma }_0}\subseteq A(a;r,R)$ be a closed subannulus, and choose $\rho ,\sigma$ with $r<\rho <{\rho }_0\le {\sigma }_0<\sigma <R$; writing $M_{\rho }={\max }_{{\gamma }_{\rho }}\mathrm{\mid }f\mathrm{\mid }$ and $M_{\sigma }={\max }_{{\gamma }_{\sigma }}\mathrm{\mid }f\mathrm{\mid }$, the integral formulas of steps 3.2 and 1.2 give $\mathrm{\mid }{c}_n(z_0-a{)}^n\mathrm{\mid }\le M_{\sigma }({\sigma }_0\mathrm{/}\sigma {)}^n$ for $n\ge 0$ and $\mathrm{\mid }{c}_{-m}(z_0-a{)}^{-m}\mathrm{\mid }\le M_{\rho }(\rho \mathrm{/}{\rho }_0{)}^m$ for $m\ge 1$ and every $z_0\in K_{{\rho }_0,{\sigma }_0}$.

Steps 3.2 and 1.2 let us write $c_n$ for the common value of the outer-circle integral when $n\ge 0$ and of the inner-circle integral when $n<0$, and step 4.1 becomes $f(z)={\sum }_{n\in \mathbb{Z}}{c}_n(z-a{)}^n$.

The geometric majorants in step 4.2 converge, so both one-sided subseries converge uniformly on $K_{{\rho }_0,{\sigma }_0}$; since the closed subannulus was arbitrary, the Laurent series converges locally uniformly on $A(a;r,R)$ and represents $f$ there.