Isolated Singularities and Laurent Series

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.

On the circle ${\gamma }_{\rho }(t)=a+\rho \exp (it)$, the Laurent series for $f$ converges uniformly by [L1], so [L2] gives ${\int }_{{\gamma }_{\rho }}\frac{f(\zeta )}{(\zeta -a{)}^{n+1}}d\zeta ={\lim }_{N\to \mathrm{\infty }}{\int }_{{\gamma }_{\rho }}{\sum }_{k=-N}^N{c}_k(\zeta -a{)}^{k-n-1}d\zeta \mathrm{.}$

By [L3] each finite integral in step 1.1 is ${\sum }_{k=-N}^N{c}_k{\int }_{{\gamma }_{\rho }}(\zeta -a{)}^{k-n-1}d\zeta$, and [L4] kills every summand except $k=n$, for which the integral is $2\pi i$; therefore every finite sum equals $2\pi i{c}_n$.

Letting $N\to \mathrm{\infty }$ in step 2.1 proves the contour formula for $c_n$.

If also $f(z)={\sum }_{n\in \mathbb{Z}}{d}_n(z-a{)}^n$ on the same annulus, the same contour formula gives $d_n=(2\pi i{)}^{-1}{\int }_{{\gamma }_{\rho }}f(\zeta )(\zeta -a{)}^{-n-1}d\zeta =c_n$ for every integer $n$, so the coefficients are unique.

By [L1], the integral over $\mathrm{\mid }\zeta -a\mathrm{\mid }={\rho }_1$ equals the coefficient $c_n$, and so does the integral over $\mathrm{\mid }\zeta -a\mathrm{\mid }={\rho }_2$.

Therefore the two integrals are equal to each other.

Let $0<R_0<R$ and choose $\sigma$ with $R_0<\sigma <R$; the coefficient formula gives $\mathrm{\mid }{c}_n\mathrm{\mid }\le M_{\sigma }\mathrm{/}{\sigma }^n$ for $n\ge 0$, where $M_{\sigma }={\max }_{\mathrm{\mid }\zeta -a\mathrm{\mid }=\sigma }\mathrm{\mid }f(\zeta )\mathrm{\mid }$, so $\mathrm{\mid }{c}_n(z-a{)}^n\mathrm{\mid }\le M_{\sigma }(R_0\mathrm{/}\sigma {)}^n$ for $\mathrm{\mid }z-a\mathrm{\mid }\le R_0$.

Let $\rho >r$ and choose ${\rho }_0$ with $r<{\rho }_0<\rho$; the coefficient formula gives $\mathrm{\mid }{c}_{-m}\mathrm{\mid }\le M_{{\rho }_0}{\rho }_0^m$ for $m\ge 1$, where $M_{{\rho }_0}={\max }_{\mathrm{\mid }\zeta -a\mathrm{\mid }={\rho }_0}\mathrm{\mid }f(\zeta )\mathrm{\mid }$, so $\mathrm{\mid }{c}_{-m}(z-a{)}^{-m}\mathrm{\mid }\le M_{{\rho }_0}({\rho }_0\mathrm{/}\rho {)}^m$ for $\mathrm{\mid }z-a\mathrm{\mid }\ge \rho$.

On the original annulus, [L1] gives that the Laurent series converges to $f$, and by definition that series is the sum of its nonnegative-power and negative-power subseries. So $f=f_{\mathrm{r}\mathrm{e}\mathrm{g}}+f_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}}$ there.

Uniqueness of the Laurent coefficients from [L2] makes both subseries unique term by term.

The geometric majorant in step 1.1 converges, so the regular part converges uniformly on $\mathrm{\mid }z-a\mathrm{\mid }\le R_0$; since $R_0<R$ was arbitrary, the convergence is locally uniform on the disc of radius $R$, and when $R=\mathrm{\infty }$ it is locally uniform on every bounded disc.

The geometric majorant in step 1.2 converges, so the principal part converges uniformly on $\mathrm{\mid }z-a\mathrm{\mid }\ge \rho$; since $\rho >r$ was arbitrary, the convergence is locally uniform on the exterior region $\mathrm{\mid }z-a\mathrm{\mid }>r$.

Steps 2.1, 2.2, 1.3, and 1.4 are exactly the claimed decomposition.

If $a$ is removable, let $F$ be a holomorphic extension to $\mathrm{\mid }z-a\mathrm{\mid }<\epsilon$; by [L3], $F$ is continuous at $a$, so $F$ is bounded on some smaller disc, and hence $f$ is bounded on the corresponding punctured disc.

If $f$ has a finite limit at $a$, then $f$ is bounded on some punctured neighbourhood of $a$.

Suppose $\mathrm{\mid }f(z)\mathrm{\mid }\le M$ whenever $0<\mathrm{\mid }z-a\mathrm{\mid }<\epsilon$. For $m\ge 1$ and $0<\rho <\epsilon$, the coefficient formula gives $\mathrm{\mid }{c}_{-m}\mathrm{\mid }\le \frac{1}{2\pi }{\int }_{\mathrm{\mid }\zeta -a\mathrm{\mid }=\rho }\mathrm{\mid }f(\zeta )\mathrm{\mid }\mathrm{\mid }\zeta -a{\mathrm{\mid }}^{m-1}\mathrm{\mid }d\zeta \mathrm{\mid }\le M{\rho }^m\mathrm{.}$

If the principal part is $0$, then $f(z)={\sum }_{n\ge 0}{c}_n(z-a{)}^n$ on the punctured disc, and [L1] makes this regular part holomorphic on $\mathrm{\mid }z-a\mathrm{\mid }<R$; defining $F(a)=c_0$ therefore extends $f$ holomorphically across $a$, so the singularity is removable.

Since step 1.3 holds for every sufficiently small $\rho >0$, letting $\rho \to 0$ gives $c_{-m}=0$ for every $m\ge 1$; so the principal part is $0$.

The extension from step 1.4 is continuous at $a$ by [L3], so $f(z)\to F(a)=c_0$ and, multiplying by $z-a$, one gets $(z-a)f(z)\to 0$.

Suppose $(z-a)f(z)\to 0$, and put $g(z):=(z-a)f(z)$ on the punctured disc. Then $g$ is holomorphic there and bounded near $a$, so the argument of steps 1.3 and 2.1 applied to the Laurent expansion $g(z)={\sum }_{n\in \mathbb{Z}}{c}_n(z-a{)}^{n+1}$ gives $c_{-m}=0$ for every $m\ge 2$.

With the coefficients from step 3.1 gone, $g(z)=c_{-1}+{\sum }_{n\ge 0}{c}_n(z-a{)}^{n+1}$, and [L1] makes the tail a holomorphic function vanishing at $a$; the hypothesis $g(z)\to 0$ therefore forces $c_{-1}=0$. So the whole principal part of $f$ is $0$.

Step 1.1 proves $1\Rightarrow 3$, step 1.2 proves $4\Rightarrow 3$, steps 1.3 and 2.1 prove $3\Rightarrow 2$, step 1.4 proves $2\Rightarrow 1$, step 2.2 proves $2\Rightarrow 4$ and $2\Rightarrow 5$, and steps 3.1 and 4.1 prove $5\Rightarrow 2$; therefore all five conditions are equivalent, and the extension value is the finite limit from step 2.2.

Suppose $a$ is a pole of order $m$. Then [L4] gives a holomorphic extension $g$ of $(z-a{)}^{m}f(z)$ with $g(a)\mathrm{\ne }0$. The singularity of $g$ at $a$ is removable, so [L5] writes $g(z)={\sum }_{n\ge 0}{b}_n(z-a{)}^n$ near $a$ with $b_0=g(a)\mathrm{\ne }0$; dividing by $(z-a{)}^m$ gives $f(z)={\sum }_{n\ge 0}{b}_n(z-a{)}^{n-m}$, whose principal part is finite and nonzero and ends at $(z-a{)}^{-m}$.

Suppose the principal part is finite and nonzero, and let $m$ be the largest index with $c_{-m}\mathrm{\ne }0$. Then $g(z):=(z-a{)}^{m}f(z)=c_{-m}+{\sum }_{k\ge 1-m}{c}_k(z-a{)}^{k+m}$ has zero principal part, so [L1] makes $g$ holomorphic at $a$ with $g(a)=c_{-m}\mathrm{\ne }0$. Therefore $a$ is a pole of order $m$ by [L4].

Suppose $\mathrm{\mid }f(z)\mathrm{\mid }\to \mathrm{\infty }$ as $z\to a$. Then $f$ is nonzero on some punctured neighbourhood of $a$, so $h:=1\mathrm{/}f$ is holomorphic there by [L3], and $h(z)\to 0$. By [L1], $h$ extends holomorphically across $a$ with value $0$, proving condition 4.

Suppose condition 4 holds. By [L2], the extension of $1\mathrm{/}f$ factors as $(z-a{)}^{m}u(z)$ for some $m\ge 1$ and some holomorphic $u$ with $u(a)\mathrm{\ne }0$; shrinking the disc if needed, $u$ stays nonzero there, so $f(z)=(z-a{)}^{-m}u(z{)}^{-1}$ and $a$ is a pole of order $m$ by [L3] and [L4].

The extension $g$ of step 1.1 is continuous and nonzero at $a$, so $\mathrm{\mid }g(z)\mathrm{\mid }\ge \delta >0$ near $a$; therefore $\mathrm{\mid }f(z)\mathrm{\mid }=\mathrm{\mid }g(z)\mathrm{\mid }\mathrm{\mid }z-a{\mathrm{\mid }}^{-m}\ge \delta \mathrm{\mid }z-a{\mathrm{\mid }}^{-m}\to \mathrm{\infty }$.

Step 1.1 proves $1\Rightarrow 2$, step 2.1 proves $1\Rightarrow 3$, step 1.2 proves $2\Rightarrow 1$, step 1.3 proves $3\Rightarrow 4$, and step 1.4 proves $4\Rightarrow 1$; hence all four conditions are equivalent, and the pole order is the largest negative exponent present in the finite principal part.

By [L4], the Laurent expansion exists, and its set of negative coefficients is either empty, finite nonempty, or infinite.

If there are no negative coefficients, the principal part is zero, so [L1] makes the singularity removable.

If there are finitely many negative coefficients and at least one is nonzero, the principal part is finite and nonzero, so [L2] makes the singularity a pole.

If there are infinitely many negative coefficients, the singularity is neither removable nor a pole by steps 2.1 and 2.2, so [L3] makes it essential.

The three coefficient cases are mutually exclusive and exhaustive, and steps 2.1 through 3.1 identify them with the three singularity types.

Suppose, for contradiction, that $f(\{0<\mathrm{\mid }z-a\mathrm{\mid }<r\})$ is not dense in $\mathbb{C}$. Then some $w\in \mathbb{C}$ and some $\epsilon >0$ satisfy $\mathrm{\mid }f(z)-w\mathrm{\mid }\ge \epsilon$ for every $z$ with $0<\mathrm{\mid }z-a\mathrm{\mid }<r$.

The function $g(z):=1\mathrm{/}(f(z)-w)$ is therefore holomorphic on $0<\mathrm{\mid }z-a\mathrm{\mid }<r$ by [L4] and bounded there by $1\mathrm{/}\epsilon$.

By [L2], the bounded function $g$ extends holomorphically across $a$. If the extension satisfies $g(a)\mathrm{\ne }0$, then $1\mathrm{/}g$ is holomorphic near $a$ and $f=w+1\mathrm{/}g$ is removable there by [L4]. If instead $g(a)=0$, then $1\mathrm{/}g$ has a pole at $a$ by [L3], so $f=w+1\mathrm{/}g$ has a pole there as well.

Either outcome in step 3.1 contradicts [L1], because an essential singularity is neither removable nor a pole. Therefore the assumption of step 1.1 is false, and every punctured neighbourhood image is dense in $\mathbb{C}$.

Applying [L2] with $n=-1$ gives the coefficient formula $c_{-1}=\frac{1}{2\pi i}{\int }_{\mathrm{\mid }\zeta -a\mathrm{\mid }=r}f(\zeta )(\zeta -a{)}^{0}d\zeta =\frac{1}{2\pi i}{\int }_{\mathrm{\mid }\zeta -a\mathrm{\mid }=r}f(\zeta )d\zeta \mathrm{.}$

By [L1], the coefficient $c_{-1}$ is exactly $\mathrm{Res}(f,a)$.

By [L1] and [L2], $g(z):=(z-a)f(z)$ extends holomorphically across $a$; write the extension again as $g$, so $g(a)$ is defined and $g(z)\to g(a)$ by [L4].

The Laurent expansion of $f$ is $f(z)=c_{-1}(z-a{)}^{-1}+{\sum }_{n\ge 0}{c}_n(z-a{)}^n$, because a simple pole has no terms $(z-a{)}^{-m}$ with $m\ge 2$. Multiplying by $(z-a)$ gives $g(z)=c_{-1}+{\sum }_{n\ge 0}{c}_n(z-a{)}^{n+1}$, so $g(a)=c_{-1}=\mathrm{Res}(f,a)$ by [L3].

Combining steps 1.1 and 1.2 gives ${\lim }_{z\to a}(z-a)f(z)=g(a)=\mathrm{Res}(f,a)$.

Let $g$ be the holomorphic extension from [L1]. On a sufficiently small punctured circle one has $f(\zeta )=g(\zeta )(\zeta -a{)}^{-m}$, so [L2] gives $\mathrm{Res}(f,a)=\frac{1}{2\pi i}{\int }_{\mathrm{\mid }\zeta -a\mathrm{\mid }=\rho }\frac{g(\zeta )}{(\zeta -a{)}^m}d\zeta \mathrm{.}$