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Derivation

Taylor Series of Analytic Functions

Statement

If \(f\) is holomorphic (complex-analytic) on an open set \(\Omega \subseteq \mathbb{C}\), and the closed disc \(\overline{D}(z_0,R) = \{\,z : |z-z_0| \le R\,\} \subset \Omega\), then for every \(z\) with \(|z-z_0| < R\) the function equals its Taylor series about \(z_0\), \[ f(z) = \sum_{n=0}^{\infty} a_n\,(z-z_0)^n, \qquad a_n = \frac{f^{(n)}(z_0)}{n!} = \frac{1}{2\pi i}\oint_{|w-z_0|=R} \frac{f(w)}{(w-z_0)^{n+1}}\,dw, \] and the series converges absolutely and uniformly on every closed sub-disc \(|z-z_0|\le r<R\). The radius of convergence \(R_c\) equals the distance from \(z_0\) to the nearest singularity of \(f\); the disc may be enlarged up to that point.

Why it matters

This is the theorem that makes complex analysis rigid: a single point-wise property, complex differentiability on an open set, forces the function to be an entire power series locally. Real smoothness (\(C^\infty\)) does not do this — \(e^{-1/x^2}\) is real-smooth yet not equal to its Taylor series — so analyticity is a genuinely stronger hypothesis with global consequences.

In physics it underlies the analytic continuation of Green's functions, dispersion relations, the convergence of perturbation series, and the interpretation of singularities of response functions (poles, branch points) as physical resonances and thresholds. The rule "radius of convergence = distance to nearest singularity" is why a perturbative expansion in a coupling can fail even when every term is finite.

Assumptions
\(f\) is holomorphic on an open neighbourhood of the closed disc.If \(f\) is merely differentiable at \(z_0\) but not on a neighbourhood, no series representation follows; holomorphy on an open set is exactly what supplies the Cauchy integral formula on the bounding circle.
The disc is simply connected and lies entirely inside the domain of holomorphy.If a singularity sits inside or on \(|w-z_0|=R\), Cauchy's formula for \(f(z)\) fails on that contour and the geometric-series step below diverges; the expansion is only valid strictly inside the largest singularity-free disc.
The contour is traversed once, positively oriented, with \(f\) continuous on it.If \(f\) is only assumed holomorphic on the open disc (not up to the boundary), one must expand on an interior circle \(|w-z_0|=\rho<R\) and let \(\rho\to R\); dropping boundary regularity changes the argument but not the conclusion on the open disc.
Term-by-term interchange of \(\sum\) and \(\oint\) is licensed by uniform convergence.Without uniform convergence of the geometric series on the contour, swapping the infinite sum and the integral is unjustified and the coefficient identification breaks.
Derivation
1
\[ f(z) = \frac{1}{2\pi i}\oint_{C}\frac{f(w)}{w-z}\,dw, \qquad C:\ |w-z_0| = R, \quad |z-z_0| < R \]
Cauchy integral formula: for \(z\) interior to \(C\) and \(f\) holomorphic on and inside \(C\), the value at \(z\) is reproduced by the boundary integral. A
2
\[ \frac{1}{w-z} = \frac{1}{(w-z_0) - (z-z_0)} = \frac{1}{w-z_0}\cdot\frac{1}{\,1 - \dfrac{z-z_0}{w-z_0}\,} \]
Purely algebraic: subtract and add \(z_0\) in the denominator, then factor out \(w-z_0\) to expose a ratio of the two displacements. A
3
\[ \left|\frac{z-z_0}{w-z_0}\right| = \frac{|z-z_0|}{R} \equiv q < 1 \quad\text{for all } w\in C \]
Because \(w\) runs on the circle \(|w-z_0|=R\) while \(z\) is fixed strictly inside, the ratio has modulus \(q<1\) uniformly in \(w\); this is the convergence condition for the next step. B
4
\[ \frac{1}{\,1 - \dfrac{z-z_0}{w-z_0}\,} = \sum_{n=0}^{\infty}\left(\frac{z-z_0}{w-z_0}\right)^{\!n} \]
Geometric series \(\frac{1}{1-\zeta}=\sum \zeta^n\), valid since \(|\zeta|=q<1\). By the Weierstrass \(M\)-test with \(M_n=q^n\), the series converges uniformly in \(w\) on \(C\). B
5
\[ f(z) = \frac{1}{2\pi i}\oint_{C} f(w)\sum_{n=0}^{\infty}\frac{(z-z_0)^n}{(w-z_0)^{n+1}}\,dw = \sum_{n=0}^{\infty}\left[\frac{1}{2\pi i}\oint_{C}\frac{f(w)}{(w-z_0)^{n+1}}\,dw\right](z-z_0)^n \]
Substitute steps 2 and 4 into step 1, then interchange \(\sum\) and \(\oint\). Legal because \(f\) is bounded on the compact contour \(C\) and the series converges uniformly there, so the swap is a uniform-limit interchange over a finite-length curve. C
6
\[ a_n \equiv \frac{1}{2\pi i}\oint_{C}\frac{f(w)}{(w-z_0)^{n+1}}\,dw = \frac{f^{(n)}(z_0)}{n!} \]
Cauchy's integral formula for derivatives (obtained by differentiating step 1 \(n\) times under the integral sign) identifies each contour integral with a Taylor coefficient. The identity of the two forms of \(a_n\) is thereby established. C
7
\[ \frac{1}{R_c} = \limsup_{n\to\infty} |a_n|^{1/n}, \qquad R_c = \operatorname{dist}\!\big(z_0,\ \{\text{singularities of } f\}\big) \]
Cauchy–Hadamard fixes the radius of convergence from the coefficients. Since the construction works for any \(R\) with the closed disc singularity-free, the series converges out to the nearest singularity, where it must diverge (else \(f\) would be holomorphic there). B
Result
\[ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}\,(z-z_0)^n, \qquad |z-z_0| < R_c = \operatorname{dist}(z_0,\text{nearest singularity}) \]

Reading. A holomorphic function is locally identical to its Taylor series, not merely approximated by it. The single hypothesis of complex differentiability on an open set upgrades to full analyticity, and the reach of the expansion is dictated globally by where the function first misbehaves — the nearest pole, branch point, or essential singularity — regardless of how well-behaved \(f\) looks near \(z_0\).

Units check. The statement is dimensionless in the analytic-structure sense: \(z\) and \(z_0\) carry the same units (say length \(L\)), so \((z-z_0)^n\) has units \(L^n\); \(a_n = f^{(n)}(z_0)/n!\) has units \([f]\,L^{-n}\) because each derivative divides by one power of \(z\). The product \(a_n(z-z_0)^n\) then has units \([f]\), matching \(f(z)\) term by term. The convergence radius \(R_c\) carries units \(L\), consistent with a distance in the \(z\)-plane.

Limiting cases
  • Entire functions (\(e^z,\ \sin z,\ \cos z\)): no singularities anywhere, so \(R_c=\infty\) and the series converges on the whole plane.
  • Single pole, e.g. \(f(z)=1/(1-z)\) about \(z_0=0\): nearest singularity at \(z=1\), so \(R_c=1\) and \(\sum z^n\) is recovered exactly.
  • Polynomial \(f\): all \(f^{(n)}\) vanish for \(n>\deg f\); the "series" truncates to \(f\) itself, trivially convergent everywhere.
  • Expansion at the centre of symmetry: choosing \(z_0\) far from singularities maximises \(R_c\); shifting \(z_0\) toward a pole shrinks \(R_c\) linearly with the distance.
  • Real slice: restricting \(z\) to the real axis reproduces the real Taylor series, but \(R_c\) may be set by a complex singularity off the axis (e.g. \(1/(1+x^2)\) has \(R_c=1\) because of poles at \(\pm i\)).
Breaks when
  • A singularity lies inside or on the disc. If \(f\) has a pole, branch point, or essential singularity within \(|z-z_0|\le R\), Cauchy's formula on \(|w-z_0|=R\) no longer reproduces \(f(z)\) for all interior \(z\), and the geometric expansion diverges — the correct object is a Laurent series, not a Taylor series.
  • The function is only real-differentiable, not complex-analytic. A \(C^\infty\) real function such as \(f(x)=e^{-1/x^2}\) (with \(f(0)=0\)) has every derivative zero at the origin, so its formal Taylor series is identically \(0\) yet \(f\not\equiv 0\); complex differentiability is essential.
  • At the boundary \(|z-z_0|=R_c\) itself. On the circle of convergence the series may converge, diverge, or oscillate depending on direction; the theorem guarantees equality only strictly inside, so evaluating on the boundary is not licensed.
  • Multivalued functions across a branch cut. For \(\log z\) or \(z^{1/2}\), any disc must avoid the branch point; a series about \(z_0\) represents only one branch and fails to continue across the cut.
Failure modes
  • Assuming real \(C^\infty\) implies analytic. Students conflate infinite differentiability with equality to the Taylor series; the counterexample \(e^{-1/x^2}\) is the standard trap.
  • Setting \(R_c\) by the nearest real singularity only. For \(1/(1+x^2)\) the "innocent" real function has \(R_c=1\) due to complex poles at \(\pm i\); ignoring the complex plane gives the wrong radius.
  • Interchanging \(\sum\) and \(\oint\) without justification. Skipping the uniform-convergence check (step 5) is a logical gap even though the answer happens to be right.
  • Using \(|(z-z_0)/(w-z_0)|<1\) with \(z\) on the contour. The ratio is \(<1\) only when \(z\) is strictly interior; forgetting the strictness invalidates the geometric series at the boundary.
  • Confusing the Taylor and Laurent expansions. Trying to Taylor-expand about a pole, or omitting negative powers when a singularity is enclosed, yields nonsense; the correct tool is the Laurent series.
  • Reading \(a_n=f^{(n)}(z_0)/n!\) as requiring \(f\) to be defined by that formula. The contour form of \(a_n\) is often easier and avoids computing high derivatives directly.
Discussion

The theorem is the precise statement of why complex analysis is "rigid." A holomorphic function is determined on its entire domain by its values on any small disc, or even any set with a limit point, via the identity theorem — a corollary of the Taylor expansion. This is dramatically unlike the real-variable world, where smooth functions can be freely modified on one region without affecting another (bump functions exist in the real category but not the holomorphic one).

Physically, the coefficients \(a_n\) are the derivatives \(f^{(n)}(z_0)/n!\), so the local Taylor data encode the full analytic function. This is the mathematical backbone of analytic continuation: a response function measured or computed in one regime can be continued into another, and the singularities encountered along the way — poles, branch cuts — carry physical content (bound states, resonances, particle-production thresholds). Dispersion relations in optics and particle physics are exactly the statement that the analytic structure links the real and imaginary parts of a response.

The rule \(R_c = \operatorname{dist}(z_0,\text{nearest singularity})\) also explains the practical limits of perturbation theory. Expanding an observable in a coupling \(g\) is a Taylor expansion in the complex \(g\)-plane; a singularity at some complex \(g_\star\) caps the radius of convergence, so the series can diverge even if the physics at real \(g\) is perfectly finite. Dyson's argument that QED perturbation theory is at best asymptotic is a physical instance: unphysical \(g<0\) makes the vacuum unstable, placing a singularity arbitrarily close to \(g=0\).

At the deepest level, the proof shows that holomorphy on an open set, term-by-term uniform convergence, and Cauchy's formula for derivatives conspire so that the Taylor coefficients are simultaneously (i) local — built from derivatives at \(z_0\) — and (ii) global — computable as contour integrals sensitive to the whole disc. This dual character is what lets one prove Liouville's theorem, the maximum modulus principle, and Morera's theorem as near-immediate consequences. It also foreshadows the Cauchy estimates \(|a_n|\le M/R^n\) with \(M=\sup_C|f|\), which quantitatively bound how fast coefficients can grow and drive the sharp form of Cauchy–Hadamard.

Common misconceptions. (1) "If all derivatives exist, the Taylor series equals the function" — true in the complex-analytic category, false in the real \(C^\infty\) category. (2) "The radius of convergence is a property of the real formula" — it is a property of the analytic function in the complex plane; complex singularities can limit a manifestly real, smooth expression. (3) "Convergence extends to the boundary" — the theorem asserts equality only strictly inside \(R_c\).

Worked examples
1
Expand \(f(z)=\dfrac{1}{2-z}\) about \(z_0=0\) and give its radius of convergence.
Setup: the only singularity is the simple pole at \(z=2\). A
2
\[ \frac{1}{2-z} = \frac{1}{2}\cdot\frac{1}{1 - z/2} = \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{z}{2}\right)^{\!n} = \sum_{n=0}^{\infty}\frac{z^n}{2^{\,n+1}} \]
Factor to geometric form and expand; valid where \(|z/2|<1\). A
3
\[ a_n = \frac{1}{2^{\,n+1}}, \qquad \frac{1}{R_c} = \lim_{n\to\infty}\left(\frac{1}{2^{\,n+1}}\right)^{1/n} = \frac{1}{2} \;\Rightarrow\; R_c = 2 \]
Cauchy–Hadamard; the radius equals the distance \(|2-0|=2\) to the pole, confirming the general rule. B
\[ \frac{1}{2-z} = \sum_{n=0}^{\infty}\frac{z^n}{2^{\,n+1}}, \qquad R_c = 2 \]

Reading. The expansion converges on \(|z|<2\), exactly up to the nearest (and only) singularity at \(z=2\). If \(z\) carries units of length, so does \(R_c=2\) (in the same units), and each term \(z^n/2^{n+1}\) inherits the units of \(f=1/(2-z)\), namely inverse length.

1
Expand \(f(z)=\dfrac{1}{1+z^2}\) about \(z_0=0\); predict \(R_c\) from the singularities, then confirm.
Setup: \(1+z^2=0\) at \(z=\pm i\), both at distance \(1\) from the origin. A
2
\[ \frac{1}{1+z^2} = \frac{1}{1-(-z^2)} = \sum_{n=0}^{\infty}(-1)^n z^{2n} = 1 - z^2 + z^4 - z^6 + \cdots \]
Geometric series with ratio \(-z^2\), valid for \(|{-z^2}|=|z|^2<1\). A
3
\[ |z|^2 < 1 \iff |z| < 1 \;\Rightarrow\; R_c = 1 = \operatorname{dist}(0,\pm i) \]
The convergence condition reproduces the distance to the complex poles \(\pm i\), even though the function is real and smooth on the whole real axis. B
\[ \frac{1}{1+z^2} = \sum_{n=0}^{\infty}(-1)^n z^{2n}, \qquad R_c = 1 \]

Reading. The radius is set by the off-axis singularities at \(\pm i\); a purely real analysis of \(1/(1+x^2)\) would find nothing on the real line to explain why the series stops converging at \(|x|=1\). This is the canonical demonstration that convergence is governed by the complex plane. Units: \(z\) dimensionless here (or measured relative to a scale), \(R_c=1\) in the same normalisation.

Problems
  1. Find the Taylor series of \(f(z)=e^{z}\) about \(z_0=0\) and state its radius of convergence.
    Solution Since \(f^{(n)}(z)=e^z\), all derivatives at \(0\) equal \(1\), so \(a_n=1/n!\). Thus \(e^z=\sum_{n=0}^{\infty} z^n/n!\). By Cauchy–Hadamard, \(1/R_c=\lim (1/n!)^{1/n}=0\), so \(R_c=\infty\): the function is entire (no singularities), and the series converges on all of \(\mathbb{C}\).
  2. Expand \(f(z)=\dfrac{1}{3-z}\) about \(z_0=1\) and give \(R_c\).
    Solution Write \(3-z=2-(z-1)=2\left(1-\tfrac{z-1}{2}\right)\). Then \(\dfrac{1}{3-z}=\dfrac{1}{2}\sum_{n=0}^{\infty}\left(\dfrac{z-1}{2}\right)^n=\sum_{n=0}^{\infty}\dfrac{(z-1)^n}{2^{\,n+1}}\), valid for \(|z-1|<2\). The only singularity is at \(z=3\), distance \(|3-1|=2\) from \(z_0=1\), so \(R_c=2\), matching the general rule.
  3. The function \(f(x)=e^{-1/x^2}\) (with \(f(0)=0\)) is real-\(C^\infty\). Compute its formal Taylor series about \(x=0\) and explain what goes wrong.
    Solution Every derivative at the origin vanishes: \(f^{(n)}(0)=0\) for all \(n\) (each derivative is a rational function times \(e^{-1/x^2}\), and \(e^{-1/x^2}\to 0\) faster than any power as \(x\to 0\)). Hence the formal Taylor series is \(\sum 0\cdot x^n \equiv 0\), which converges (to \(0\)) everywhere but does not equal \(f\) for any \(x\ne 0\). The resolution: as a complex function \(e^{-1/z^2}\) has an essential singularity at \(z=0\), so it is not holomorphic there and the Taylor theorem does not apply. This is the standard demonstration that real smoothness is weaker than complex analyticity.
  4. Determine the radius of convergence of the Taylor series of \(f(z)=\dfrac{z}{(z-2)(z+3i)}\) about \(z_0=0\), without computing the coefficients.
    Solution The singularities are the simple poles at \(z=2\) (distance \(2\)) and \(z=-3i\) (distance \(3\)). The nearest to \(z_0=0\) is \(z=2\), so \(R_c=\min(2,3)=2\). No coefficient computation is needed: the radius is the distance to the closest singularity. (The point \(z=0\) is a removable zero of the numerator cancelling nothing in the denominator, so it is not a singularity.)
  5. Show that if \(f\) is entire and bounded, \(|f(z)|\le M\) for all \(z\), then \(f\) is constant (Liouville's theorem), using the Cauchy coefficient estimates that follow from this derivation.
    Solution Expand \(f\) about any \(z_0\); since \(f\) is entire, the expansion holds on \(|z-z_0|<R\) for every \(R\). From \(a_n=\dfrac{1}{2\pi i}\oint_{|w-z_0|=R}\dfrac{f(w)}{(w-z_0)^{n+1}}\,dw\), bound the integrand by \(M/R^{n+1}\) and the contour length by \(2\pi R\): \[ |a_n| \le \frac{1}{2\pi}\cdot\frac{M}{R^{n+1}}\cdot 2\pi R = \frac{M}{R^{n}}. \] For \(n\ge 1\), let \(R\to\infty\): the right side \(\to 0\), so \(a_n=0\). Only \(a_0\) survives, hence \(f(z)=a_0\) is constant. This shows how the contour form of the Taylor coefficients delivers the Cauchy estimates and, immediately, Liouville's theorem.