Power Series and Analytic Functions

Part 9, Chapter 9: Complex Analysis, Holomorphic Functions

Learning objectives

  • Compute the radius of convergence of a complex power series via root or ratio test
  • Expand holomorphic functions as Taylor series within their disk of analyticity
  • Use Laurent series to expand functions on annuli around isolated singularities
  • Classify singularities (removable, pole, essential) by their Laurent series

Holomorphic functions ARE power series. That is the rigidity statement: every holomorphic function on a domain equals its Taylor series in a disk around each point, with radius determined by the nearest singularity. Around an isolated singularity, the more general Laurent series adds negative powers of (z−z_0)(z - z_0), and the tail of those negative powers classifies the singularity: zero negative terms means removable, finitely many negative terms means a pole, infinitely many means essential. This is the algebraic engine behind the residue calculus.

Power series and radius of convergence

A power series centred at z_0z_0 has the form

sumn=0inftyan(z−z0)n.\sum_{n=0}^{\infty} a_n (z - z_0)^n.n=0inftya_n(z−z_0)n.

The radius of convergence RR is the supremum of ∣z−z_0∣|z - z_0| for which the series converges. The Cauchy-Hadamard formula gives it:

frac1R=limsupntoinfty∣an∣1/n.\frac{1}{R} = \limsup_{n \to \infty} |a_n|^{1/n}.

Inside the disk ∣z−z_0∣<R|z - z_0| < R the series converges absolutely (and uniformly on compact subsets) to a holomorphic function. On the boundary ∣z−z0∣=R|z - z_0| = R, convergence is delicate, some series converge everywhere on the boundary, others nowhere. Outside the disk the series diverges. The radius RR equals the distance from z0z_0 to the nearest singularity of the limit function.

Taylor series of holomorphic functions

If ff is holomorphic on a disk DD centred at z_0z_0, then ff equals its Taylor series there:

f(z)=sumn=0inftyfracf(n)(z0)n!(z−z0)n.f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!} (z - z_0)^n.n=0inftyfracf(n)(z_0)n!(z−z_0)n.

This is genuinely an equality, not merely an asymptotic expansion. The radius of convergence equals the distance from z_0z_0 to the boundary of the maximal disk where ff remains holomorphic. For example, frac11−z\frac{1}{1 - z} has Taylor series 1+z+z2+cdots1 + z + z^2 + \cdots around z=0z = 0 with R=1R = 1, the nearest singularity is at z=1z = 1.

The widget visualizes the disk of analyticity. Try f(z)=1/(1−z)f(z) = 1/(1 - z): the Taylor series at 00 converges inside ∣z∣<1|z| < 1 and diverges outside. The singularity at z=1z = 1 on the boundary blocks any extension.

Laurent series and isolated singularities

If ff is holomorphic on an annulus r<∣z−z_0∣<Rr < |z - z_0| < R (typical case: ff has an isolated singularity at z0z_0), then ff has a Laurent series there:

f(z)=sumn=−inftyinftyan(z−z0)n,f(z) = \sum_{n = -\infty}^{\infty} a_n (z - z_0)^n,n=−inftyinftya_n(z−z_0)n,

with both positive and negative powers. The negative-power tail sumn=1inftyfraca−n(z−z0)n\sum_{n=1}^{\infty} \frac{a_{-n}}{(z - z_0)^n}n=1inftyfraca−n(z−z0)n is the principal part. The coefficient a−1a_{-1}−1 is the residue (used in the residue theorem).

The principal part classifies the singularity at z_0z_0:

  • Removable singularity: all a−n=0a_{-n} = 0−n=0. Defining f(z0)=a0f(z_0) = a_0 makes ff holomorphic at z0z_0. Example: sinz/z\sin z / z at z=0z = 0.
  • Pole of order mm: a−mneq0a_{-m} \neq 0−mneq0 but a−n=0a_{-n} = 0−n=0 for n>mn > m. The principal part is a polynomial in 1/(z−z_0)1/(z - z_0) of degree mm. Example: frac1(z−1)2\frac{1}{(z - 1)^2} has a pole of order 2 at z=1z = 1.
  • Essential singularity: infinitely many a−nneq0a_{-n} \neq 0−nneq0. By Picard's great theorem, near such a singularity ff takes every complex value (with at most one exception) infinitely many times. Example: e1/ze^{1/z} at z=0z = 0.
    Where this shows up
    • Numerical analysis: Pade approximations and continued-fraction extrapolations rely on the Laurent and Taylor structure of meromorphic functions to compute extreme-precision approximations of special functions (gamma, zeta, polylog) far outside their naive Taylor disks.
    • Control theory: Transfer functions are rational and live entirely in the complex plane. The pole-zero diagram, the locations of a−1neq0a_{-1} \neq 0−1neq0 residues, encodes system stability, transient response, and resonance frequencies.
    • Optics: The Gouy phase shift of a focused Gaussian beam is computed from the Laurent expansion of the beam's complex amplitude around the focus, an essential-singularity-like behaviour gives the pi/2\pi/2 phase anomaly observed experimentally.
    • Statistical physics: The radius of convergence of a partition-function series gives the location of the nearest critical point in the complex temperature plane (the Lee-Yang programme). Phase transitions are diagnosed by where the Taylor series breaks.

      • Pause and think: What is the radius of convergence of sumn!,zn\sum n! \, z^n? Apply the ratio test: ∣an+1/an∣cdot∣z∣=(n+1)∣z∣toinfty|a_{n+1}/a_n| \cdot |z| = (n+1)|z| \to \inftycdot∣z∣=(n+1)∣z∣toinfty for any zneq0z \neq 0. So R=0R = 0: the series only converges at z=0z = 0. Compare to sumzn/n!\sum z^n/n!, which has R=inftyR = \infty, convergent everywhere, and equals eze^z.

      Try it

      • Compute the radius of convergence of sumngeq1zn/n2\sum_{n \geq 1} z^n / n^2ngeq1zn/n2. (Hint: ∣an∣1/n=(1/n2)1/nto1|a_n|^{1/n} = (1/n^2)^{1/n} \to 1n∣1/n=(1/n2)1/nto1, so R=1R = 1.)
      • Classify the singularity at z=0z = 0 of f(z)=(1−cosz)/z2f(z) = (1 - \cos z) / z^2. (Hint: expand cosz=1−z2/2+z4/24−cdots\cos z = 1 - z^2/2 + z^4/24 - \cdots.)
      • Find the Laurent expansion of f(z)=1/(z(z−1))f(z) = 1/(z(z - 1)) on 0<∣z∣<10 < |z| < 1. (Hint: partial fractions, then geometric series in zz.)
      • True or false: if ff has an essential singularity at z0z_0, there exists a sequence zntoz0z_n \to z_0 with f(zn)toinftyf(z_n) \to \inftyn)toinfty and a different sequence wntoz0w_n \to z_0 with f(wn)to0f(w_n) \to 0n)to0. (Picard predicts even more: every value is hit infinitely often.)

        A trap to watch for

        The Laurent expansion of a function depends on which annulus you are in. The function f(z)=1/(z(z−1))f(z) = 1/(z(z-1)) has TWO different Laurent expansions: one on 0<∣z∣<10 < |z| < 1 (negative powers from the 1/z1/z pole, positive powers from the 1/(1−z)1/(1-z) geometric series) and a completely different one on ∣z∣>1|z| > 1 (negative powers from both). Beginners conflate these. Always specify the annulus before computing the Laurent series, and remember that the geometric series sumzn\sum z^n converges only for ∣z∣<1|z| < 1; for ∣z∣>1|z| > 1 rewrite 1/(1−z)=−1/zcdot1/(1−1/z)=−sum1/zn+11/(1 - z) = -1/z \cdot 1/(1 - 1/z) = -\sum 1/z^{n+1}.

        What you now know

        You can apply the Cauchy-Hadamard formula, expand holomorphic functions in Taylor series, work in annuli with Laurent series, and classify singularities by their negative-power tail. The next section is the geometric side of complex analysis: conformal maps, which preserve angles and let us transport Laplace's equation between geometrically very different domains.

        Mark section complete →

        References

        • Garrity, T. (2002). All the Mathematics You Missed: But Need to Know for Graduate School. Cambridge University Press, §9.6-9.7.
        • Ahlfors, L. V. (1979). Complex Analysis (3rd ed.). McGraw-Hill, ch. 5.
        • Stein, E. M., Shakarchi, R. (2003). Complex Analysis. Princeton University Press, ch. 2 and 3.
        • Conway, J. B. (1978). Functions of One Complex Variable I (2nd ed.). Springer, ch. 5.
        • Brown, J. W., Churchill, R. V. (2014). Complex Variables and Applications (9th ed.). McGraw-Hill, ch. 5.

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