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The Laurent theorem states that a holomorphic function on the annulus $R<|z-a|<S$ has a unique Laurent seriesThe integration circle is counterclockwise. The series converges absolutely at every point of the annulus and uniformly on every compact subannulus. The coefficient integrals are independent of the choice of $\rho$ by the Cauchy integral theorem applied between two such circles. These statements include $R=0$ or $S=\infty$ when the indicated annulus has those endpoints.

On an annulus $r<|z-z_0|<R$, the Laurent theorem represents an analytic function uniquely asFor any positively oriented simple closed contour $C$ in the annulus winding once around $z_0$, the coefficients in the Laurent series areThe formula follows from the Cauchy integral formula applied term by term, and it is independent of the chosen contour by the Cauchy integral theorem.

The Laurent theorem says that if $f$ is analytic on an annulusthen it has a unique Laurent seriesconverging locally uniformly on that annulus, wherefor any positively oriented circle $C$ in the annulus around $a$.

An isolated singularity at $a$ is a point at which $f$ is not analytic although it is analytic on some punctured neighbourhood. It is removable when every $c_n$ with $n<0$ vanishes; it is a pole of order $m$ when $c_{-m}\ne 0$ and $c_n=0$ for $n<-m$; and it is essential when infinitely many negative-index coefficients are nonzero.

For $0<|z|<1$,For $|z|>1$,The coefficients are unique after the annulus is fixed; these expansions differ because they represent the function on different annuli. At zero the first expansion has principal part $-z^{-1}$, so zero is a simple pole with residue $-1$.

The Liouville theorem states that every bounded entire function is constant.

By the Laurent theorem, $f$ has a unique Laurent series $f(z)={\sum }_{n=-\infty }^{\infty }{c}_n{z}^n$ on $R<|z|<S$. Split it intoHere $g$ is holomorphic on $D_S$ and $h$ is holomorphic on $A_{R,\infty }$. To check their extended regions of convergence explicitly, fix $r\in (R,S)$ and put $M_r={\max }_{|z|=r}|f(z)|$. The Laurent coefficient formula givesThe first bound ensures convergence uniformly on every closed disk $|z|\le r^{\prime }<r$; letting $r$ range up to $S$ covers $D_S$. The second gives convergence uniformly on $|z|\ge r^{\prime \prime }>r$; letting $r$ range down to $R$ covers $A_{R,\infty }$. Locally uniform convergence of these power series, in $z$ and $1/z$ respectively, proves the claimed holomorphy. On the original annulus their sum is $f$.

The second estimate also proves $h(z)\to 0$ as $|z|\to \infty$: for any fixed $r\in (R,S)$ and $|z|>r$,

For uniqueness under this condition, suppose $f=g_1+h_1=g_2+h_2$. Define $F=g_1-g_2$ on $D_S$ and $F=h_2-h_1$ on $A_{R,\infty }$. The two definitions agree on the overlap, and their domains cover $\mathbb{C}$, so they define an entire function. Moreover, $F(z)\to 0$ at infinity. It is bounded outside a large disk by this limit, and inside by continuity and the extreme value theorem. The Liouville theorem makes $F$ constant, and its limit makes that constant zero. Hence $g_1=g_2$ and $h_1=h_2$. Without the limiting condition, adding a constant to $g$ and subtracting it from $h$ would preserve the decomposition.