The Radius Reaches the Nearest Singularity

A puzzle calculus left unanswered

Back in real calculus you may have met the function 1/(1 + x^2). It is perfectly smooth on the whole real line — no division by zero, no corners, no jumps anywhere. Yet its Taylor series at 0, which is 1 - x^2 + x^4 - x^6 + ..., converges only for |x| < 1 and blows up the moment |x| reaches 1. Nothing on the real line explains that wall at x = 1. The series simply refuses to work there, and real analysis can only shrug.

The mystery dissolves the instant you let the variable be complex. Write the same function as 1/(1 + z^2) and ask where it breaks: the denominator vanishes at z = i and z = -i, two points sitting off the real axis at distance exactly 1 from the origin. By Taylor's theorem, the series of a holomorphic function lives on a disk, and that disk can only grow until it bumps into trouble. The nearest trouble is i and -i, a distance 1 away — and that is the radius. The real line was hiding the explanation in a direction it could not see.

Stating the theorem honestly

Here is the precise claim, the one named the radius reaches the nearest singularity. Suppose f is holomorphic on some open region containing the point z_0, and let f = sum a_n (z - z_0)^n be its Taylor series there. Then the radius of convergence R equals the distance from z_0 to the nearest singular point of f — the nearest point you cannot reach while keeping f holomorphic. The disk of convergence is the largest open disk centred at z_0 on which f is honestly holomorphic, and not one bit larger.

Two halves make this true, and it is worth seeing both. First, R can be no larger than that distance: if the series converged on a bigger disk, its sum would be holomorphic there (a convergent power series is holomorphic inside its disk), and that would extend f holomorphically across the singular point — contradicting that the point is genuinely singular. Second, R is no smaller: Taylor's theorem guarantees the series converges on every disk where f is holomorphic, so the disk reaches all the way out to the first obstruction. Squeeze from both sides and R lands exactly on the nearest singularity.

R = dist(z_0, nearest singularity of f)

  |z - z_0| < R   ->  series converges, f holomorphic
  |z - z_0| = R   ->  boundary circle, at least one singularity sits on it
  |z - z_0| > R   ->  series diverges (the disk cannot grow past the obstruction)

What counts as the nearest trouble

The nearest singularity does not have to be a dramatic blow-up. Any isolated point where f cannot be holomorphic stops the disk — most commonly a pole, where f shoots off to infinity like 1/(z - i) does at z = i. But the obstruction can also be a branch point, where a multivalued function like a logarithm or a square root cannot be made single-valued, or an essential singularity, or even an accumulation of singularities along a curve. Whatever the nearest obstacle is, the radius reaches it and halts.

A clean worked example. Take f(z) = 1/((z - 1)(z - 3)) and expand it around z_0 = 0. Its only singularities are simple poles at z = 1 and z = 3, at distances 1 and 3 from the centre. The nearest is the pole at 1, so the Taylor series at 0 has radius exactly R = 1 — the disk fills up until it touches z = 1 and can go no further, even though z = 3 is sitting calmly just beyond. If instead you re-centred at z_0 = 2, the distances become 1 (to z = 1) and 1 (to z = 3), so the radius would be 1 again, now with two singularities on the boundary at once.

Reading a radius, step by step

For functions built from familiar pieces, you usually do not need to sum anything at all. You can locate the radius purely by hunting singularities. Here is the routine.

1. Identify your expansion centre z_0 — the point you are building the Taylor series around.
2. Find every point where f fails to be holomorphic: zeros of any denominator (poles), branch points of any log or fractional power, and any other singular point.
3. Measure the distance from z_0 to each such point using the modulus |singularity - z_0|.
4. Take the smallest of those distances. That minimum is R, the radius of convergence.
5. If there are no singularities at all, every distance is effectively infinite, so R = infinity and the series converges on the whole plane.

Try it on the real-calculus puzzle f(x) = 1/(1 + x^2) we opened with, re-centred at the everyday point x = 1, that is z_0 = 1. The singularities are still at i and -i. The distance from 1 to i is |i - 1| = sqrt(2), and the distance from 1 to -i is also sqrt(2). The smallest is sqrt(2), so the Taylor series of 1/(1 + x^2) about x = 1 has radius sqrt(2) — a value that would look utterly arbitrary on the real line but is read straight off the geometry once you step into the plane.

Subtleties to keep honest

Two honest cautions. First, the theorem tells you that at least one singularity sits on the boundary circle |z - z_0| = R, but it says nothing about behaviour on that circle itself. The series might converge at some boundary points and diverge at others; the boundary is delicate territory, the home of Abel's theorem and of subtle examples like sum z^n / n^2, which converges everywhere on |z| = 1 even though R = 1. The clean dichotomy (converge inside, diverge outside) holds strictly off the circle.

Second, the series diverging past R does not mean the function ends there. The function 1/(1 + z^2) is happily holomorphic far beyond the unit disk, all over the plane except at i and -i; it is only this particular series, anchored at this particular centre, that quits at R = 1. To follow the function outward you re-expand around a new centre and reach further — this is the doorway to analytic continuation, where overlapping disks are stitched together to extend a function as far as its singularities allow. The radius is a property of the series, not a fence around the function.

And one beautiful extreme to keep in view: some functions are walled in completely. A series like sum z^(2^n) has a natural boundary on the whole circle |z| = 1 — singularities are dense all the way around, so no re-expansion can ever escape, and analytic continuation is impossible in every direction. When trouble surrounds the disk instead of sitting at a few points, the radius marks not a temporary wall but a permanent edge. Either way, the principle is the same: the radius reaches exactly as far as the function lets it.