Taylor Series of Holomorphic Functions

Suppose a function is complex-differentiable at every point near some center. Suppose you only know its values on a tiny disk. Can you reconstruct the function everywhere it stays holomorphic? Can you do this using nothing but derivatives at that one center? For real functions the answer is a resounding no. Smoothness alone never guarantees convergence to the correct values. The classic example e−1/x2e^{-1/x^2} shows this clearly. For holomorphic functions of a complex variable the answer is a triumphant yes. This fact is one of the deepest structural surprises in all of analysis. This article proves that every function holomorphic on an open disk equals its Taylor series throughout that disk. This holds with an explicit, computable radius of convergence. The mechanism is Cauchy's integral formula. A single contour integral encodes the entire local behavior of a holomorphic function. Expanding its kernel in a geometric series hands us the power series coefficients for free. These coefficients are expressed as contour integrals of ff itself. This is fundamentally different from the real-variable story. There, Taylor's theorem with remainder gives only an approximation, not an identity. The big picture is that complex differentiability is an enormously stronger hypothesis than real differentiability. One derivative existing in a neighborhood forces infinitely many derivatives to exist. It forces analyticity. It forces the function's entire local structure to be captured by a single sequence of numbers. This rigidity is the engine behind nearly every subsequent theorem in complex analysis. Examples include the Identity Theorem and Zeros, the maximum modulus principle, Liouville's theorem. Another example is the classification of singularities via Laurent Series. This article delivers a complete, hypothesis-explicit statement of the Taylor expansion theorem. It includes a full proof from Cauchy's integral formula. It includes worked coefficient computations on asymmetric functions. It also offers a careful discussion of exactly where the radius of convergence comes from. Finally, it explains why each hypothesis is indispensable.

Key takeaways
  • Holomorphic on a disk ⇒\Rightarrow equal to its own Taylor series on that whole disk — no remainder term needed.
  • The proof expands the Cauchy kernel 1w−z\frac{1}{w-z} as a geometric series and integrates term by term.
  • Coefficients are an=f(n)(z0)n!=12πi∮f(w)(w−z0)n+1 dwa_n = \frac{f^{(n)}(z_0)}{n!} = \frac{1}{2\pi i}\oint \frac{f(w)}{(w-z_0)^{n+1}}\,dw.
  • The radius of convergence is at least the distance from the center to the nearest boundary point of holomorphy.
  • Holomorphic   ⟺  \iff analytic in one complex variable — a rigidity with no real-variable analogue.
  • Key idea: Cauchy's formula turns "one complex derivative exists nearby" into "the function is literally a convergent power series."

Where this lives today

Taylor expansions of holomorphic functions underlie the classification of zeros and singularities used throughout complex analysis, feeding directly into the Identity Theorem and Zeros and into Laurent Series for functions with isolated singularities. In physics, perturbative expansions of analytic Green's functions and propagators rely on exactly this convergence guarantee to justify truncating a series and trusting the remainder is controlled. In engineering, the stability analysis of linear systems via the ZZ-transform and Laplace transform exploits that transfer functions are holomorphic away from their poles, so local power series (and the radius up to the nearest pole) directly determine transient response and stability margins. In numerical analysis and scientific computing, algorithms for evaluating special functions (Bessel functions, the Gamma function, elliptic integrals) use truncated Taylor series with rigorously known convergence radii, guaranteeing a priori error bounds unavailable in the real-smooth category. In cryptography and coding theory, generating functions over finite or formal settings borrow the same coefficient-extraction philosophy — reading off coefficients via contour integrals (residues) is the complex-analytic ancestor of combinatorial coefficient extraction techniques. Finally, in statistics, characteristic functions of random variables with analytic extensions (e.g., subgaussian or exponential-family distributions) are expanded in Taylor series to generate cumulants, exploiting exactly the convergence-on-a-disk phenomenon proved here.

Ask the tutor: Why does the radius of convergence sometimes strictly exceed the distance to the nearest singularity visible in a given formula for ff?

DefinitionTaylor Series of a Holomorphic Function

Let U⊆CU \subseteq \mathbb{C} be open, let f:U→Cf : U \to \mathbb{C} be holomorphic on UU, and let z0∈Uz_0 \in U. Suppose the open disk

D(z0,R)={ z∈C:∣z−z0∣<R }D(z_0, R) = \{\, z \in \mathbb{C} : |z - z_0| < R \,\}

is contained in UU, for some R>0R > 0. The Taylor series of ff at z0z_0 is the formal power series

∑n=0∞an(z−z0)n,an=f(n)(z0)n!.\sum_{n=0}^{\infty} a_n (z - z_0)^n, \qquad a_n = \frac{f^{(n)}(z_0)}{n!}.

We say ff equals its Taylor series on D(z0,R)D(z_0,R) if this series converges for every z∈D(z0,R)z \in D(z_0, R) and its sum equals f(z)f(z) there. Note that a priori this definition requires ff to be infinitely complex-differentiable at z0z_0 merely to write down the coefficients ana_n; the theorem below shows this hypothesis is automatic and the series in fact converges to ff throughout D(z0,R)D(z_0,R).

Intuition

The single derivative f′(z0)f'(z_0) of a holomorphic function is not an isolated local fact; it is a shadow of a global integral identity. Cauchy's integral formula says that the value of ff at any interior point of a disk is a weighted average of ff's values on the boundary circle, with weight 12πi(w−z)\frac{1}{2\pi i (w - z)}. Because that weight is a smooth (indeed holomorphic) function of zz for ww fixed on the circle, differentiating under the integral sign is legitimate to all orders, and this is precisely what forces infinite differentiability. The deeper reason the Taylor series converges to ff, not just approximates it, is a geometric-series trick. Fix zz inside the disk and ww on the boundary circle. The ratio z−z0w−z0\frac{z - z_0}{w - z_0} has modulus strictly less than 11, so

1w−z=1(w−z0)−(z−z0)=1w−z0⋅11−z−z0w−z0=∑n=0∞(z−z0)n(w−z0)n+1,\frac{1}{w - z} = \frac{1}{(w-z_0) - (z - z_0)} = \frac{1}{w - z_0} \cdot \frac{1}{1 - \frac{z-z_0}{w-z_0}} = \sum_{n=0}^{\infty} \frac{(z-z_0)^n}{(w-z_0)^{n+1}},

a convergent geometric series, uniformly in ww on the circle. Plugging this into Cauchy's formula and integrating term by term (justified by uniform convergence) turns the integral representation of ff directly into a power series representation. There is no approximation error to control — the equality is exact for every zz strictly inside the largest disk on which ff stays holomorphic.

Figure
Geometric decay of the Cauchy kernel expansion: zz inside the circle of radius RR centered at z0z_0
z₀ z w (on circle |w−z₀|=R) |z − z₀| / |w − z₀| = r < 1 1/(w−z) = Σ (z−z₀)ⁿ / (w−z₀)ⁿ⁺¹, converges since r < 1 R = radius of largest disk of holomorphy centered at z₀
TheoremTaylor's Theorem for Holomorphic Functions

Let U⊆CU \subseteq \mathbb{C} be open and let f:U→Cf : U \to \mathbb{C} be holomorphic on UU. Let z0∈Uz_0 \in U, and let

R=sup⁡{ r>0:D(z0,r)⊆U }R = \sup\{\, r > 0 : D(z_0, r) \subseteq U \,\}

(possibly R=∞R = \infty if U=CU = \mathbb{C}). Then: 1. (Smoothness) ff is infinitely complex-differentiable on UU; in particular f(n)(z0)f^{(n)}(z_0) exists for every n≥0n \ge 0. 2. (Coefficient formula) For every ρ\rho with 0<ρ<R0 < \rho < R, and every n≥0n \ge 0,

f(n)(z0)n!=12πi∮∣w−z0∣=ρf(w)(w−z0)n+1 dw,\frac{f^{(n)}(z_0)}{n!} = \frac{1}{2\pi i} \oint_{|w - z_0| = \rho} \frac{f(w)}{(w-z_0)^{n+1}}\, dw,

and this value is independent of the choice of ρ∈(0,R)\rho \in (0, R). 3. (Convergence to ff) The power series

∑n=0∞f(n)(z0)n!(z−z0)n\sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n

converges for every z∈D(z0,R)z \in D(z_0, R), and its sum equals f(z)f(z). 4. (Radius bound) The radius of convergence of this power series is at least RR.

LemmaCauchy's Integral Formula

Let U⊆CU \subseteq \mathbb{C} be open, f:U→Cf : U \to \mathbb{C} holomorphic, z0∈Uz_0 \in U, and suppose D(z0,ρ)‾⊂U\overline{D(z_0,\rho)} \subset U for some ρ>0\rho > 0. Then for every z∈D(z0,ρ)z \in D(z_0, \rho),

f(z)=12πi∮∣w−z0∣=ρf(w)w−z dw,f(z) = \frac{1}{2\pi i} \oint_{|w-z_0| = \rho} \frac{f(w)}{w - z}\, dw,

the contour traversed once counterclockwise. This lemma is taken as known (it is proved from Cauchy's theorem for a keyhole contour avoiding the singularity at w=zw = z); it is the sole external input used below.

Proof

We prove parts (2)–(4) directly using Cauchy's Integral Formula and the geometric series, and part (1) falls out as a byproduct since the resulting power series is termwise differentiable to all orders. The technique is expansion of the Cauchy kernel into a uniformly convergent geometric series, followed by term-by-term integration.

Step 1: Setup. Fix any ρ\rho with 0<ρ<R0 < \rho < R. By definition of RR, the closed disk D(z0,ρ)‾\overline{D(z_0, \rho)} is contained in UU (indeed contained in D(z0,r)D(z_0, r) for ρ<r<R\rho < r < R with D(z0,r)⊆UD(z_0,r) \subseteq U). Let γ\gamma denote the circle ∣w−z0∣=ρ|w - z_0| = \rho, oriented counterclockwise. By Cauchy's Integral Formula, for every z∈D(z0,ρ)z \in D(z_0, \rho),

f(z)=12πi∮γf(w)w−z dw.(∗)f(z) = \frac{1}{2\pi i} \oint_\gamma \frac{f(w)}{w-z}\, dw. \tag{$\ast$}

Step 2: Expand the kernel. Fix z∈D(z0,ρ)z \in D(z_0,\rho) and write r=∣z−z0∣/ρ<1r = |z - z_0|/\rho < 1. For w∈γw \in \gamma we have ∣w−z0∣=ρ|w - z_0| = \rho, so

1w−z=1(w−z0)−(z−z0)=1w−z0⋅11−z−z0w−z0.\frac{1}{w-z} = \frac{1}{(w - z_0) - (z-z_0)} = \frac{1}{w-z_0}\cdot \frac{1}{1 - \frac{z-z_0}{w-z_0}}.

Since ∣z−z0w−z0∣=r<1\left|\frac{z-z_0}{w-z_0}\right| = r < 1, the geometric series

11−z−z0w−z0=∑n=0∞(z−z0w−z0)n\frac{1}{1 - \frac{z-z_0}{w-z_0}} = \sum_{n=0}^{\infty} \left(\frac{z-z_0}{w-z_0}\right)^n

converges absolutely, uniformly in w∈γw \in \gamma: the nn-th term is bounded in modulus by rnr^n, independent of ww, and ∑rn<∞\sum r^n < \infty since r<1r<1. This is precisely the Weierstrass MM-test with Mn=rnM_n = r^n. Hence

1w−z=∑n=0∞(z−z0)n(w−z0)n+1,\frac{1}{w-z} = \sum_{n=0}^\infty \frac{(z-z_0)^n}{(w-z_0)^{n+1}},

uniformly for w∈γw \in \gamma.

Step 3: Multiply by f(w)f(w) and integrate term by term. Since ff is continuous on the compact set γ\gamma (being holomorphic on U⊇γU \supseteq \gamma), it is bounded there, say ∣f(w)∣≤M|f(w)| \le M for all w∈γw \in \gamma. Multiplying the uniformly convergent series in Step 2 by the bounded function f(w)f(w) preserves uniform convergence on γ\gamma: the tail

∣f(w)∑n=N∞(z−z0)n(w−z0)n+1∣≤Mρ∑n=N∞rn→0 as N→∞,\left| f(w) \sum_{n=N}^\infty \frac{(z-z_0)^n}{(w-z_0)^{n+1}} \right| \le \frac{M}{\rho} \sum_{n=N}^\infty r^n \to 0 \text{ as } N \to \infty,

uniformly in ww. Uniform convergence of a series of continuous functions on a curve of finite length permits term-by-term integration (this is a standard consequence of uniform convergence: the partial sums converge uniformly to the limit function, and the integral of a uniform limit over a fixed rectifiable curve is the limit of the integrals, since ∣∮γ(gN−g)∣≤length⁡(γ)⋅sup⁡γ∣gN−g∣→0\left|\oint_\gamma (g_N - g)\right| \le \operatorname{length}(\gamma)\cdot \sup_\gamma|g_N - g| \to 0). Applying this to (∗)(\ast):

f(z)=12πi∮γf(w)∑n=0∞(z−z0)n(w−z0)n+1 dw=∑n=0∞(12πi∮γf(w)(w−z0)n+1 dw)(z−z0)n.f(z) = \frac{1}{2\pi i}\oint_\gamma f(w)\sum_{n=0}^\infty \frac{(z-z_0)^n}{(w-z_0)^{n+1}}\, dw = \sum_{n=0}^\infty \left(\frac{1}{2\pi i}\oint_\gamma \frac{f(w)}{(w-z_0)^{n+1}}\, dw\right)(z-z_0)^n.

Define

an(ρ):=12πi∮γf(w)(w−z0)n+1 dw.a_n(\rho) := \frac{1}{2\pi i}\oint_\gamma \frac{f(w)}{(w-z_0)^{n+1}}\, dw.

We have shown: for every z∈D(z0,ρ)z \in D(z_0,\rho),

f(z)=∑n=0∞an(ρ) (z−z0)n.(∗∗)f(z) = \sum_{n=0}^\infty a_n(\rho)\,(z-z_0)^n. \tag{$\ast\ast$}

Step 4: The coefficients are independent of ρ\rho, and equal f(n)(z0)/n!f^{(n)}(z_0)/n!. The right side of (∗∗)(\ast\ast) is a convergent power series in zz on D(z0,ρ)D(z_0,\rho). A convergent power series is infinitely differentiable term by term inside its disk of convergence (a fact established when studying Power Series and Analyticity), and its Taylor coefficients at the center are recovered by repeated differentiation: differentiating (∗∗)(\ast\ast) nn times and evaluating at z=z0z = z_0 kills all terms except the nn-th, giving

f(n)(z0)=n! an(ρ).f^{(n)}(z_0) = n!\, a_n(\rho).

In particular ff is infinitely differentiable at z0z_0 (this proves part (1) at the point z0z_0; since z0∈Uz_0 \in U was arbitrary, ff is infinitely differentiable throughout UU), and

an(ρ)=f(n)(z0)n!a_n(\rho) = \frac{f^{(n)}(z_0)}{n!}

for every valid ρ∈(0,R)\rho \in (0,R). In particular the left side does not depend on ρ\rho, which proves part (2), including the independence-of-ρ\rho claim.

Step 5: Convergence to ff on the full disk D(z0,R)D(z_0,R). Fix any z∈D(z0,R)z \in D(z_0, R). Since ∣z−z0∣<R|z - z_0| < R, choose ρ\rho with ∣z−z0∣<ρ<R|z-z_0| < \rho < R; such ρ\rho exists by definition of RR as a supremum. Then z∈D(z0,ρ)z \in D(z_0,\rho), and Steps 1–4 (applied with this ρ\rho) give exactly

f(z)=∑n=0∞an(ρ)(z−z0)n=∑n=0∞f(n)(z0)n!(z−z0)n.f(z) = \sum_{n=0}^\infty a_n(\rho)(z-z_0)^n = \sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n.

Since z∈D(z0,R)z \in D(z_0,R) was arbitrary, the series converges to f(z)f(z) at every point of D(z0,R)D(z_0,R). This proves part (3).

Step 6: Radius of convergence is at least RR. Part (3) shows the series ∑an(z−z0)n\sum a_n(z-z_0)^n (with an=f(n)(z0)/n!a_n = f^{(n)}(z_0)/n!) converges (to f(z)f(z), in particular converges as a numerical series) for every zz with ∣z−z0∣<R|z-z_0| < R. By definition, the radius of convergence RconvR_{\text{conv}} of a power series is the supremum of ∣z−z0∣|z - z_0| over points of convergence, so Rconv≥RR_{\text{conv}} \ge R. This proves part (4). ■\blacksquare

∎
CorollaryHolomorphic ⇒\Rightarrow Analytic

If ff is holomorphic on an open set U⊆CU \subseteq \mathbb{C}, then ff is (complex-)analytic on UU: around every point of UU, ff agrees with a convergent power series on some disk. Consequently f∈C∞(U)f \in C^\infty(U), and all such power series are unique (their coefficients are forced to be the Taylor coefficients, by term-by-term differentiation as in Step 4 above).

Worked examples

Example 1: A rational function with an off-axis pole. Let f(z)=12−izf(z) = \dfrac{1}{2 - iz}, holomorphic wherever 2−iz≠02 - iz \ne 0, i.e. everywhere except z=−2iz = -2i. Expand ff about the asymmetric center z0=1z_0 = 1. Write 2−iz=(2−i)−i(z−1)2 - iz = (2 - i) - i(z-1), so

f(z)=1(2−i)(1−i(z−1)2−i)=12−i∑n=0∞(i2−i)n(z−1)n,f(z) = \frac{1}{(2-i)\left(1 - \frac{i(z-1)}{2-i}\right)} = \frac{1}{2-i}\sum_{n=0}^\infty \left(\frac{i}{2-i}\right)^n (z-1)^n,

valid whenever ∣i(z−1)2−i∣<1\left|\frac{i(z-1)}{2-i}\right| < 1, i.e. ∣z−1∣<∣2−i∣=5|z - 1| < |2-i| = \sqrt{5}. Check against the theorem: the only singularity of ff is at z=−2iz = -2i, and the distance from the center z0=1z_0 = 1 to this singularity is

∣1−(−2i)∣=∣1+2i∣=12+22=5.|1 - (-2i)| = |1 + 2i| = \sqrt{1^2 + 2^2} = \sqrt{5}.

So R=5R = \sqrt{5} is exactly the radius up to which ff is holomorphic on D(1,R)D(1,R), and the theorem guarantees convergence to ff on all of D(1,5)D(1,\sqrt5) — matching the radius of convergence 5\sqrt5 computed directly from the geometric series. This illustrates part (4) sharply: here the radius of convergence equals RR exactly (not merely ≥R\ge R), because the singularity at −2i-2i obstructs any larger disk of holomorphy. The coefficients are

an=12−i(i2−i)n.a_n = \frac{1}{2-i}\left(\frac{i}{2-i}\right)^n.

As a numerical check for n=1n=1: a1=i(2−i)2a_1 = \dfrac{i}{(2-i)^2}. Compute (2−i)2=4−4i+i2=3−4i(2-i)^2 = 4 - 4i + i^2 = 3 - 4i, so

a1=i3−4i=i(3+4i)(3−4i)(3+4i)=3i−49+16=−4+3i25.a_1 = \frac{i}{3-4i} = \frac{i(3+4i)}{(3-4i)(3+4i)} = \frac{3i - 4}{9+16} = \frac{-4+3i}{25}.

Compare with the direct derivative formula: f′(z)=i(2−iz)2f'(z) = \dfrac{i}{(2-iz)^2}, so f′(1)=i(2−i)2=i3−4if'(1) = \dfrac{i}{(2-i)^2} = \dfrac{i}{3-4i}, matching a1a_1 exactly (recall a1=f′(z0)/1!a_1 = f'(z_0)/1!).

Example 2: The complex logarithm branch, expanded off-center. Let Log(z)\mathrm{Log}(z) denote the principal branch of the logarithm, holomorphic on U=C∖(−∞,0]U = \mathbb{C}\setminus(-\infty,0], with Log(1)=0\mathrm{Log}(1) = 0. Expand about z0=2z_0 = 2 (note: 2≠12 \ne 1, an asymmetric, non-obvious center). First compute derivatives: f(z)=Log(z)f(z) = \mathrm{Log}(z), f′(z)=1/zf'(z) = 1/z, and for n≥1n \ge 1,

f(n)(z)=(−1)n−1(n−1)! z−n.f^{(n)}(z) = (-1)^{n-1}(n-1)!\, z^{-n}.

(Check: f(1)(z)=z−1f^{(1)}(z) = z^{-1} matches with n=1n=1: (−1)0⋅0!⋅z−1=z−1(-1)^0 \cdot 0! \cdot z^{-1} = z^{-1}. ✓. Differentiating again, f(2)(z)=−z−2f^{(2)}(z) = -z^{-2}, matching n=2n=2: (−1)1⋅1!⋅z−2=−z−2(-1)^1\cdot 1!\cdot z^{-2} = -z^{-2}. ✓.) At z0=2z_0 = 2: f(2)=ln⁡2f(2) = \ln 2, and for n≥1n \ge 1,

f(n)(2)=(−1)n−1(n−1)! 2−n.f^{(n)}(2) = (-1)^{n-1}(n-1)!\,2^{-n}.

So the Taylor coefficients are a0=ln⁡2a_0 = \ln 2 and for n≥1n \ge 1,

an=f(n)(2)n!=(−1)n−1(n−1)!n! 2n=(−1)n−1n 2n.a_n = \frac{f^{(n)}(2)}{n!} = \frac{(-1)^{n-1}(n-1)!}{n!\, 2^n} = \frac{(-1)^{n-1}}{n\, 2^n}.

The Taylor series is

Log(z)=ln⁡2+∑n=1∞(−1)n−1n 2n(z−2)n.\mathrm{Log}(z) = \ln 2 + \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n\,2^n}(z-2)^n.

Now determine RR: the domain of holomorphy U=C∖(−∞,0]U = \mathbb{C}\setminus(-\infty,0] has, relative to center z0=2z_0=2, nearest excluded boundary point at z=0z=0 (distance 22), since the entire negative real axis is farther or equal for points near 00, and indeed the closest point of the slit (−∞,0](-\infty,0] to 22 is the origin itself (the slit lies along the negative axis, and among points x≤0x \le 0, the value minimizing ∣2−x∣|2-x| is x=0x=0, giving distance 22). So R=2R = 2. Independently, the ratio test on the coefficients confirms this: ∣an+1an∣=n2(n+1)→12\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{n}{2(n+1)} \to \dfrac{1}{2}, so the radius of convergence is 1/(1/2)=21/(1/2) = 2, exactly matching RR. This example also shows a case (unlike Example 1) where the obstruction to holomorphy is a whole ray, not an isolated pole, yet the theorem's distance-to-boundary computation still governs RR correctly — since the nearest obstructing point still determines the radius.

CounterexampleWhy "holomorphic on the disk" cannot be weakened to "real-differentiable" or "C∞C^\infty as a function of x,yx,y"

Consider f:R→Rf: \mathbb{R} \to \mathbb{R} (the real-variable analogue, to isolate exactly which hypothesis is failing) defined by

f(x)={e−1/x2x≠00x=0.f(x) = \begin{cases} e^{-1/x^2} & x \ne 0 \\ 0 & x = 0.\end{cases}

This function is a standard example: it is C∞(R)C^\infty(\mathbb{R}), and one can show by induction (using the chain rule and the fact that e−1/x2e^{-1/x^2} decays faster than any polynomial blows up) that f(n)(0)=0f^{(n)}(0) = 0 for every n≥0n \ge 0. Consequently every Taylor coefficient at z0=0z_0 = 0 vanishes, and the Taylor series of ff at 00 is the identically zero series, which trivially converges everywhere — to 00, not to f(x)f(x) for any x≠0x \ne 0, where f(x)=e−1/x2>0f(x) = e^{-1/x^2} > 0. This is not a complex-analytic function of a complex variable; it is only being differentiated in the real sense, and this is exactly the point of the counterexample. If one asks "what if we drop holomorphy and merely assume f:U⊆C→Cf: U \subseteq \mathbb{C} \to \mathbb{C} has all partial derivatives in the real sense (i.e., is C∞C^\infty as a map R2→R2\mathbb{R}^2 \to \mathbb{R}^2), without the Cauchy–Riemann equations," the conclusion of the theorem fails completely: infinite differentiability no longer forces the function to agree with its Taylor series at any point besides the center. The engine of the proof — expanding the Cauchy kernel via Cauchy's Integral Formula — has no analogue in the real category, because there is no contour-integral representation of f(x)f(x) in terms of its boundary values the way there is for holomorphic f(z)f(z). This shows precisely why the hypothesis "ff holomorphic" (not merely infinitely real-differentiable) is indispensable: it is what supplies Cauchy's Integral Formula, and without it the entire mechanism of Steps 1–3 of the proof collapses.

Scope, edge cases & common traps

SituationWhat the theorem saysCommon error
UU is a disk itself, U=D(z0,R)U = D(z_0,R)Series converges to ff on all of UUNone — this is the clean case
UU properly contains a disk, with boundary singularities elsewhereRR = distance to nearest obstruction to holomorphy, not to UU's boundary in some other directionConfusing RR with dist(z0,∂U)\text{dist}(z_0,\partial U) measured toward an irrelevant direction — always take the infimum distance
ff has a singularity strictly outside D(z0,R)D(z_0,R) but the series still "numerically converges" farther via analytic continuation of a different branchRadius of convergence can sometimes exceed the na\"ive RR of the given formula for ff, if ff extends holomorphicallyAssuming the coefficient formula's RR is always the true radius of convergence; it is only a lower bound (part 4 says "at least RR")
Multivalued expressions like z\sqrt{z}, Log(z)\mathrm{Log}(z)Must fix a single-valued branch on a slit domain before applying the theoremApplying the theorem to a "function" that is not well-defined and single-valued on UU
Real-differentiable but not holomorphic functions (e.g. f(z)=zˉf(z) = \bar z)Theorem does not apply at allAttempting to write a "Taylor series" using Wirtinger-style derivatives without holomorphy — the resulting series need not converge to ff

The mathematician behind it

Origins · 1 minAugustin-Louis Cauchy and how this was discovered
AC
Augustin-Louis Cauchy
1789–1857 · Founder of rigorous complex analysis, École Polytechnique and the Collège de France, Paris

Cauchy built the theory of complex integration essentially single-handedly in the 1820s, introducing the integral theorem and integral formula that now bear his name. Working during a period when analysis was being rebuilt on rigorous epsilon-delta foundations, Cauchy recognized that contour integration was the right tool to unlock the rigidity of complex-differentiable functions — a rigidity utterly absent from the real theory he had also helped to found. His integral formula, from which the Taylor expansion theorem follows almost immediately, remains the central computational and theoretical engine of complex analysis two centuries later.

History
- 1814 — Cauchy submits his first memoir on definite integrals between complex limits, planting the seeds of contour integration.
- 1825 — Cauchy proves what is now called Cauchy's Integral Theorem for holomorphic functions on simply connected domains.
- 1831 — Cauchy derives the Integral Formula and, from it, the convergence of Taylor series for holomorphic functions, during his exile in Turin.
- 1841 — Cauchy further develops the calculus of residues, connecting Taylor expansions to the broader theory that becomes Laurent Series.
- 1851 — Riemann's dissertation reformulates complex differentiability geometrically via the Cauchy–Riemann equations, complementing Cauchy's integral approach.
Why does complex differentiability force infinite differentiability, when real differentiability does not?

The key is Cauchy's Integral Formula, which expresses f(z)f(z) as a contour integral 12πi∮f(w)w−zdw\frac{1}{2\pi i}\oint \frac{f(w)}{w-z}dw whenever ff is holomorphic near the contour. The integrand, as a function of zz, is holomorphic in zz to arbitrarily high order because the kernel 1/(w−z)1/(w-z) is smooth in zz away from ww, and differentiation under the integral sign is justified by uniform estimates on the compact contour. Differentiating nn times produces the formula f(n)(z)=n!2πi∮f(w)(w−z)n+1dwf^{(n)}(z) = \frac{n!}{2\pi i}\oint \frac{f(w)}{(w-z)^{n+1}}dw, valid for every nn. No real-variable analogue of this integral representation exists, since the real Cauchy–Riemann-free calculus has no contour-integral formula connecting a function's boundary values to its interior values.

Does the theorem say the Taylor series converges everywhere ff is defined, or only on the disk D(z0,R)D(z_0,R)?

Strictly on the disk D(z0,R)D(z_0,R), where RR is the distance to the nearest point at which ff fails to extend holomorphically. Outside this disk the series may diverge, even if ff itself (via a different formula or analytic continuation) is defined farther away in some directions. The series "sees" only the nearest singularity in any direction from z0z_0, because the geometric-series argument in the proof requires the ratio ∣z−z0∣/ρ|z-z_0|/\rho to stay below 11 for a circle ρ\rho inside the full domain of holomorphy.

How do you compute the radius of convergence in practice without doing the full contour integral?

Two equivalent routes are used in the worked examples. First, find every singularity of ff (poles, branch points, essential singularities) and compute the distance from z0z_0 to the closest one; this distance is RR, guaranteed by part (4) to be a valid convergence radius. Second, directly analyze the coefficients an=f(n)(z0)/n!a_n = f^{(n)}(z_0)/n! using the ratio test or root test: Rconv=(lim sup⁡n∣an∣1/n)−1R_{\text{conv}} = \left(\limsup_n |a_n|^{1/n}\right)^{-1}. In well-behaved examples (rational functions, log⁡\log, exponentials) these two computations agree exactly, since the nearest singularity fully determines the coefficient asymptotics.

Is the Taylor series of a holomorphic function unique?

Yes. If f(z)=∑an(z−z0)n=∑bn(z−z0)nf(z) = \sum a_n (z-z_0)^n = \sum b_n(z-z_0)^n for all zz in some disk around z0z_0, then differentiating both convergent power series nn times and evaluating at z0z_0 forces an=f(n)(z0)/n!=bna_n = f^{(n)}(z_0)/n! = b_n for every nn. This uniqueness is what makes "the Taylor series of ff at z0z_0" a well-defined object rather than one choice among many possible expansions.

What goes wrong if ff has a pole or branch point exactly at the center z0z_0?

Then ff is not holomorphic at z0z_0, so the hypotheses of the theorem fail outright — f(n)(z0)f^{(n)}(z_0) may not even exist, and there is no Taylor series at that center. In this situation, if ff is still holomorphic in a punctured neighborhood of z0z_0, the right tool is not a Taylor series but a Laurent Series, which allows negative-power terms to capture the singular behavior at z0z_0 itself.

How does this theorem relate to the Identity Theorem?

The Taylor expansion theorem shows a holomorphic function is completely determined near any point by its derivatives (equivalently, its Taylor coefficients) at that single point. The Identity Theorem and Zeros pushes this further: if two holomorphic functions on a connected domain agree on a set with an accumulation point (not just at one point with all derivatives matching), they must agree everywhere on that domain. Both results exploit the same underlying rigidity — that local data at a point, correctly packaged, propagates to determine global behavior — but the Identity Theorem requires additionally that the domain be connected and uses analytic continuation along paths.

Can this theorem fail if UU is not open, e.g. if ff is only defined and differentiable at isolated points?

Yes — the entire argument requires ff to be holomorphic on an open neighborhood of z0z_0, specifically on a disk D(z0,ρ)D(z_0,\rho) with ρ>0\rho>0, so that Cauchy's Integral Formula applies on a genuine closed contour around z0z_0 lying entirely within the domain. If UU is not open (say ff is defined only on a single ray or a discrete set), there is no contour to integrate over, Cauchy's formula does not apply, and the notion of "holomorphic at z0z_0" itself is not even meaningful without an open neighborhood.

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