Terms go to 0, so there is some m beyond which the norm of anpn is always less than 1. Move from p to z, where z is a point on or inside the circle of radius r. Change pk to zk, and the norm has been multiplied by (r/t)k. The terms are bounded by a geometric series, and f converges absolutely at z. Since r/t does not depend on z, the series of functions converges uniformly on the disk of radius r.

When the individual functions are continuous, and convergence is uniform, the sum is also continuous. Thus f is well defined, and continuous, inside the circle of radius t.

If p is the farthest point then t defines the "circle of convergence". The power series converges, and is continuous, inside the circle of radius t. It may converge at all, some, or none of the boundary points where |z| = t. There is no convergence beyond t.

When f begins life as an analytic function at 0, the power series, defined by the derivitives of f at 0, converges to f on a small neighborhood about 0, i.e. the neighborhood where f is analytic. By the above, the series converges to f inside a circle of radius t, and does not converge beyond t. This is the circle of convergence for f. Of course t may be infinite, if the power series converges everywhere.

How far does t go? As far as f is defined. If f is analytic out to a radius of t then the laurent operator creates a power series that converges for any z up to t. Convergence only stops if f becomes undefined. Set f = 1/(z-2), and the power series converges inside the circle of radius 2, and nowhere outside this circle, because f is not defined at 2.

Now let f be the convergence of an arbitrary power series, inside a circle of radius t. consider a path p wholly contained within our circle of radius t. Consider the contour integral of f along p. Since p has definite endpoints, it is bounded away from our circle of radius t. Hence the power series is uniformly convergent on a disk that contains all of p, and the contour integral can be evaluated term by term. If the path runs from a to b, the integral of zk along this path is (bk-ak)/k. Do this for all the terms in the series and express the integral of f along p as an infinite sum. If p returns to its starting point, i.e. a closed curve, then every term in this infinite sum is 0, and the integral is 0. In other words, the contour integral of every simple closed curve is 0. By Morera's theorem, f is analytic. If a power series converges, it is analytic inside its circle of convergence.

How about a general laurent series? Let an arbitrary series s converge at points p and q, whose distances define an annulus about 0. (Let p be the outermost point.)

Say we are given one of two things: s is a forward or reverse laurent series, or, s converges absolutely at p and q. We might know the latter if s comes from the laurent operator applied to a slightly larger annulus. This would make s absolutely convergent at p and q, and every other point in the larger annulus.

If s is bidirectional, split s at the constant term, building a forward laurent series sf and a reverse laurent series sr. If this is the case, we know that s converges absolutely at p and q, hence sf converges at p, and sr converges at q. We can divide and conquer.

If s was already a forward or reverse laurent series, then set sf or sr equal to s, and set the "other" series equal to 0. We won't need the other series, nor will we need absolute convergence at p and q.

Select a point z inside the annulus. Step back from p to z, and show that sf converges absolutely at z, using a proof similar to the one presented at the top of this page. Note that sf might contain a few terms with negative exponents, which increase as we move from p to z, but we can always skip past these, as there are only finitely many of them, hence f becomes absolutely convergent at z. Similarly, move from q to z to show sr converges absolutely at z. If we had to split the series s, write s = sf + sr, the sum of two absolutely convergent series, and s converges absolutely at z.

Show that the series converges uniformly on any annulus that is bounded away from p and q.

If f was already a forward laurent series, proof of convergence applies for all z between 0 and p. In other words, we can shrink the inner circle all the way down to the origin, and the annulus becomes a punctured disk. Similarly, if s is a reverse laurent series, we have convergence out to infinity. The annulus expands to the exterior of a circle in the complex plane.

Assume the series s converges as above. Is it analytic? Let p be a simple closed curve inside our annulus of convergence, such that p does not enclose the origin. Thus every power of z, even the negative powers of z, are analytic on and inside p. Evaluate the integral of f around p by integrating term by term, which is valid, thanks to uniform convergence. The integral of each term drops to 0, and the integral of f around p is 0. Use Morera's theorem again, and f is analytic everywhere inside our annulus.

To find the integral of an analytic power series f, integrate f along a straight line from 0 to z. Since f is uniformly convergent, this can be done term by term, and we know how to integrate zk. Therefore the integration of f occurs term by term. This is what we would expect, since differentiation occurs term by term, and that is the inverse of integration.

If f comes from a general laurent series, and is analytic on a certain annulus, select a point q on the positive x axis, inside the annulus, and draw an arc from q to z. Don't let this arc cross the negative x axis. The contour integral of f, from q to z, can now be computed term by term. The result is well defined, independent of the arc connecting q and z, as long as arcs never cross the negative x axis. Of course we could have chosen some other fixed point q and some other forbiden radius. We confront this ambiguity head on when we integrate a-1z-1 in the series. All other terms have well defined integrals, but 1/z becomes log(z), which cannot be continuous all the way around the annulus.