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Holomorphic Means Analytic: Taylor's Theorem

On the real line, smooth and analytic are different worlds. In the complex plane they collapse into one: a single complex derivative on an open set forces a convergent Taylor series. Here is the theorem that makes that miracle precise.

Two roads that turn out to be one

In the first two guides of this rung you built a power series from the outside in: pick coefficients a_n, form sum a_n (z - z_0)^n, and ask where it converges. The answer was a clean disk |z - z_0| < R, the radius of convergence, inside which the series behaves like a dream — you can even differentiate it term by term and the result is again a convergent power series. So a power series is automatically a beautifully smooth function on its disk. That was one road.

Now walk the other road, the one from the previous rung. There you started with a holomorphic function f — just complex-differentiable on an open set — and proved, via the Cauchy integral formula, that it is secretly infinitely differentiable. Taylor's theorem is the bridge that says these two roads meet: every holomorphic function, near each of its points, IS a convergent power series. The function you started with and the series you wrote down are literally the same thing.

What the theorem actually claims

Here is the statement, stripped to its bones. Suppose f is holomorphic on an open set, and z_0 is a point of that set. Let D be the largest open disk centred at z_0 that fits inside the set where f lives. Then for every z in D, f(z) equals a convergent power series in (z - z_0), and the coefficients are not mysterious — each a_n is just a derivative of f at the centre, divided by a factorial. These are the Taylor coefficients, and the series is f's Taylor series about z_0.

f(z) = sum_{n>=0} a_n (z - z_0)^n      for all z in D

        a_n = f^(n)(z_0) / n!

  a_0 = f(z_0)
  a_1 = f'(z_0)
  a_2 = f''(z_0) / 2
  a_3 = f'''(z_0) / 6 = f'''(z_0) / 3!
  ...

  D = largest open disk about z_0 inside f's domain
Taylor's theorem: a holomorphic f equals its own Taylor series on the largest disk that fits.

Notice how much this gives you for free. Because a power series can be differentiated term by term, having one means f has derivatives of every order automatically — this is the infinite differentiability you met before, now seen as a side effect of the series rather than a separate theorem. And the coefficient formula a_n = f^(n)(z_0)/n! is forced: if a function equals a power series, that series can only be its Taylor series. There is no freedom, no second power series competing for the job.

Why it is true: the engine under the hood

You do not have to take Taylor's theorem on faith — its proof is a short, beautiful trick built from the Cauchy integral formula. That formula already lets you recover f(z) at an interior point from the values of f on a surrounding circle: f(z) is an integral of f(w)/(w - z) around the circle, scaled by 1/(2 pi i). The whole proof is to take that kernel 1/(w - z) and expand it as a geometric series in (z - z_0).

  1. Start from f(z) = (1/2 pi i) times the integral of f(w)/(w - z) dw around a circle |w - z_0| = r, with z inside it.
  2. Rewrite the kernel: 1/(w - z) = 1/[(w - z_0) - (z - z_0)] = 1/(w - z_0) times 1/[1 - (z - z_0)/(w - z_0)].
  3. Since z is inside and w is on the circle, |(z - z_0)/(w - z_0)| < 1, so expand the last factor as a geometric series sum [(z - z_0)/(w - z_0)]^n.
  4. Swap the (uniformly convergent) sum and the integral, and collect the powers of (z - z_0). What multiplies (z - z_0)^n is exactly a_n = f^(n)(z_0)/n!, by the generalized Cauchy formula for derivatives.

Two honest remarks about that proof. The swap of sum and integral in the last step is not a formality — it is licensed precisely because the geometric series converges uniformly on the circle for a fixed interior z, the kind of uniform convergence on compact sets you saw makes term-by-term operations legal. And the radius r of the circle can be pushed outward to anywhere f stays holomorphic, which is exactly why the Taylor series is valid on the whole largest disk D, not just a timid little neighbourhood.

A tiny worked example, and where the disk stops

Take f(z) = 1/(1 - z), holomorphic everywhere except the single bad point z = 1. Expand about z_0 = 0. The derivatives are f^(n)(z) = n!/(1 - z)^(n+1), so f^(n)(0) = n!, and the Taylor coefficients are a_n = f^(n)(0)/n! = 1. The series is sum z^n — the plain geometric series — which converges exactly for |z| < 1. And 1 is precisely the distance from the centre 0 to the bad point. The disk did not stop arbitrarily; it stopped at the nearest place where f breaks.

Here is a sharper version of the same lesson. Consider f(z) = 1/(1 + z^2), expanded about 0. As a real function of x this looks perfectly tame — smooth and bounded on the whole real line, with no visible reason for trouble — yet its Taylor series converges only for |x| < 1, and generations of calculus students found that radius baffling. The complex view explains it instantly: f has singular points at z = i and z = -i, each at distance 1 from the origin, and the disk simply cannot reach past them. The behaviour that is mysterious on the real line is obvious in the plane.

Reading the theorem honestly

A few cautions keep this powerful result from being misremembered. First, "f equals its Taylor series" is a statement only inside the disk D; on the boundary circle anything can happen — the series may converge, diverge, or converge to a different-looking value, and sorting out the edge is a delicate separate story. Second, the disk is centred where YOU choose: re-expanding the same holomorphic f about a different point z_1 gives a different Taylor series with its own, possibly larger, radius. The function is one object; its Taylor series is local.

Third, and most important, do not blur which direction is the deep one. "A convergent power series is holomorphic" is the easy half — it follows straight from term-by-term differentiation. The astonishing half is the converse: "holomorphic implies a power series," the claim that analytic and holomorphic are the same class. That converse is special to the complex plane and rests on the Cauchy integral formula; it has no real-variable analogue, as our smooth-but-not-analytic example showed. When someone says holomorphic equals analytic, they are quoting a theorem, never a definition.

With Taylor's theorem in hand, the rest of this rung almost writes itself. Because a holomorphic function is locally a power series, its zeros cannot pile up — they must sit apart from one another, which is the seed of the identity theorem in the next guide. And because the disk of validity reaches exactly to the nearest singularity, the radius becomes a measuring tape for where trouble lives, the subject of the guide after that. Everything downstream flows from this one equality: holomorphic means analytic.