Power Series

Work in progress. This note is still being written and incomplete.

Power series for real numbers was covered in semester 1. This note explains the complex numbers version.

A complex power series centered at α\alpha is a series of the form

∑n=0∞cn(z−α)n\sum_{n=0}^{\infty} c_n (z-\alpha)^n

Here:

  • α\alpha: center, a fixed complex number
  • cnc_n: coefficients, fixed complex numbers
  • zz: a complex variable

Radius of Convergence

The radius of convergence ρ\rho is defined as in the real case. For complex zz, the interval of convergence becomes a disk.

There exists ρ\rho with 0≤ρ≤∞0 \le \rho \le \infty such that the series

  • converges for ∣z−α∣<ρ|z-\alpha| < \rho
  • diverges for ∣z−α∣>ρ|z-\alpha| > \rho

Ignoring boundary points, the convergence region is 1 of the following.

  • Single point
    The point α\alpha, when ρ=0\rho = 0.
  • Disk
    Dρ(α)={z∈C:∣z−α∣<ρ}D_\rho(\alpha) = \{z \in \mathbb{C} : |z-\alpha| < \rho\}, when 0<ρ<∞0 < \rho < \infty.
  • Entire complex plane
    When ρ=∞\rho = \infty.

Examples:

  • ∑n=0∞(n+1)(z−2)n3n\sum_{n=0}^{\infty} \frac{(n+1)(z-2)^n}{3^n} has ρ=3\rho = 3 and disk of convergence D3(2)D_3(2)

Algebra of Power Series

Let ff and gg have power series representations

f(z)=∑n=0∞an(z−α)n,z∈Dr1(α)f(z) = \sum_{n=0}^{\infty} a_n (z-\alpha)^n, \quad z \in D_{r_1}(\alpha) g(z)=∑n=0∞bn(z−α)n,z∈Dr2(α)g(z) = \sum_{n=0}^{\infty} b_n (z-\alpha)^n, \quad z \in D_{r_2}(\alpha)

Let r=min⁡{r1,r2}r = \min\{r_1, r_2\} and β∈C\beta \in \mathbb{C}. Then for z∈Dr(α)z \in D_r(\alpha):

  • Scalar multiple
    βf(z)=∑n=0∞βan(z−α)n\beta f(z) = \sum_{n=0}^{\infty} \beta a_n (z-\alpha)^n
  • Sum
    f(z)+g(z)=∑n=0∞(an+bn)(z−α)nf(z) + g(z) = \sum_{n=0}^{\infty} (a_n + b_n)(z-\alpha)^n
  • Product
    f(z)g(z)=∑n=0∞cn(z−α)nf(z)g(z) = \sum_{n=0}^{\infty} c_n (z-\alpha)^n

The coefficients of the product are

cn=∑k=0nakbn−kc_n = \sum_{k=0}^{n} a_k b_{n-k}

This is the Cauchy product of the series.

Examples:

  • The first 4 non-zero terms of the Maclaurin series of ezsin⁡ze^z \sin z are z+z2+z33−z530z + z^2 + \frac{z^3}{3} - \frac{z^5}{30}

Worked Examples

Radius and Boundary

Find the disk of convergence of ∑n=0∞(n+1)(z−2)n3n\sum_{n=0}^{\infty} \frac{(n+1)(z-2)^n}{3^n}.

Apply the ratio test.

lim⁡n→∞∣un+1un∣=lim⁡n→∞(n+2) ∣z−2∣3(n+1)=∣z−2∣3\lim_{n\to \infty} \left|\frac{u_{n+1}}{u_n}\right| = \lim_{n\to \infty} \frac{(n+2)\,\lvert z-2 \rvert}{3(n+1)} = \frac{|z-2|}{3}

The series converges for ∣z−2∣3<1\frac{|z-2|}{3} < 1, so ρ=3\rho = 3 and the disk is D3(2)D_3(2).

The ratio test is inconclusive on the boundary ∣z−2∣=3|z-2| = 3, since the limit equals 11. Check the terms directly.

The nn-th term is un=(n+1)(z−2)n3nu_n = \frac{(n+1)(z-2)^n}{3^n}. Taking the modulus gives

∣un∣=(n+1)∣z−2∣n3n=(n+1)(∣z−2∣3)n|u_n| = (n+1)\frac{|z-2|^n}{3^n} = (n+1)\left(\frac{|z-2|}{3}\right)^n

On the boundary ∣z−2∣=3|z-2| = 3, so ∣z−2∣3=1\frac{|z-2|}{3} = 1 and

∣un∣=n+1→∞|u_n| = n+1 \to \infty

A convergent series needs un→0u_n \to 0. Here ∣un∣|u_n| grows without bound, so the series diverges at every boundary point, regardless of which point on the circle zz is.

All 3 Regions

  • ∑n! zn\sum n!\, z^n
    The ratio is (n+1)∣z∣(n+1)|z|, which tends to ∞\infty for every z≠0z \neq 0. So ρ=0\rho = 0 and the region is the point 00.
  • ∑zn\sum z^n
    The ratio is ∣z∣|z|, so ρ=1\rho = 1 and the region is D1(0)D_1(0).
  • ∑znn!\sum \frac{z^n}{n!}
    The ratio is ∣z∣n+1→0\frac{|z|}{n+1} \to 0 for every zz. So ρ=∞\rho = \infty and the region is the entire plane.

Sum of 2 Series

Let f(z)=11−z=∑znf(z) = \frac{1}{1-z} = \sum z^n and g(z)=11+z=∑(−1)nzng(z) = \frac{1}{1+z} = \sum (-1)^n z^n, both valid in D1(0)D_1(0).

f(z)+g(z)=∑n=0∞(1+(−1)n)zn=2∑n=0∞z2n=21−z2f(z) + g(z) = \sum_{n=0}^{\infty} \left(1 + (-1)^n\right) z^n = 2\sum_{n=0}^{\infty} z^{2n} = \frac{2}{1-z^2}

The odd coefficients cancel and the even coefficients equal 22. The series is valid in D1(0)D_1(0).

Cauchy Product of Exponential and Sine

Find the first 4 non-zero terms of the Maclaurin series of ezsin⁡ze^z \sin z.

The coefficients are

an=1n!,b1=1,  b3=−16,  b5=1120,b0=b2=b4=0a_n = \frac{1}{n!}, \quad b_1 = 1, \; b_3 = -\frac{1}{6}, \; b_5 = \frac{1}{120}, \quad b_0 = b_2 = b_4 = 0

Apply cn=∑k=0nakbn−kc_n = \sum_{k=0}^{n} a_k b_{n-k}.

c0=0,c1=a0b1=1,c2=a1b1=1c_0 = 0, \quad c_1 = a_0 b_1 = 1, \quad c_2 = a_1 b_1 = 1 c3=a2b1+a0b3=12−16=13c_3 = a_2 b_1 + a_0 b_3 = \frac{1}{2} - \frac{1}{6} = \frac{1}{3} c4=a3b1+a1b3=16−16=0c_4 = a_3 b_1 + a_1 b_3 = \frac{1}{6} - \frac{1}{6} = 0 c5=a4b1+a2b3+a0b5=124−112+1120=−130c_5 = a_4 b_1 + a_2 b_3 + a_0 b_5 = \frac{1}{24} - \frac{1}{12} + \frac{1}{120} = -\frac{1}{30} ezsin⁡z=z+z2+z33−z530+⋯e^z \sin z = z + z^2 + \frac{z^3}{3} - \frac{z^5}{30} + \cdots

Cauchy Product with Geometric Series

Find the first 5 non-zero terms of the Maclaurin series of ez1−z\frac{e^z}{1-z}.

Here an=1n!a_n = \frac{1}{n!} and bn=1b_n = 1, so cn=∑k=0n1k!c_n = \sum_{k=0}^{n} \frac{1}{k!}.

c0=1,c1=2,c2=52,c3=83,c4=6524c_0 = 1, \quad c_1 = 2, \quad c_2 = \frac{5}{2}, \quad c_3 = \frac{8}{3}, \quad c_4 = \frac{65}{24} ez1−z=1+2z+52z2+83z3+6524z4+⋯\frac{e^z}{1-z} = 1 + 2z + \frac{5}{2}z^2 + \frac{8}{3}z^3 + \frac{65}{24}z^4 + \cdots

The series is valid in D1(0)D_1(0).

Written by October 8, 2026 3 min read
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