Complex Power Series
On the real line a
power series converges
on an interval centred at its base point. Move the very same series into the complex
plane and the picture upgrades: it converges on a disk. A
complex power series about z_0 is
\sum_{n=0}^{\infty} a_n (z - z_0)^n = a_0 + a_1 (z - z_0) + a_2 (z - z_0)^2 + \cdots,
with complex coefficients a_n and a complex variable
z. The new feature is purely geometric: convergence is governed by
the modulus |z - z_0|, a single non-negative real number,
so the set where the series converges is symmetric in every direction — a disk, not a line
segment.
The disk of convergence
Step 1 — convergence depends only on |z - z_0|.
The natural test is for absolute convergence, and
|a_n (z - z_0)^n| = |a_n|\,|z - z_0|^n involves only the distance
from z_0. So if the series converges at one point at distance
\rho, it converges at every point closer than
\rho.
Step 2 — that forces a circular threshold. There is a single radius
R \in [0, \infty], the radius of convergence, with
|z - z_0| < R \;\Rightarrow\; \text{converges}, \qquad |z - z_0| > R \;\Rightarrow\; \text{diverges}.
The open set |z - z_0| < R is the disk of
convergence. (On the boundary circle |z - z_0| = R anything
can happen, point by point.)
Step 3 — find R from the
ratio or
root test.
Applying the ratio test to |a_n|\,|z - z_0|^n gives the limit
|z - z_0|\cdot\lim |a_{n+1}/a_n|, which is below
1 exactly when |z - z_0| < R, where
\frac{1}{R} = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| \qquad\text{or}\qquad \frac{1}{R} = \lim_{n\to\infty}\sqrt[n]{|a_n|}.
Three worked radii
The complex geometric series. Take a_n = 1, so the
series is \sum z^n. The ratio is
|a_{n+1}/a_n| = 1, hence 1/R = 1 and
R = 1. Inside the unit disk it sums in closed form, exactly like its
real cousin:
\sum_{n=0}^{\infty} z^n = \frac{1}{1 - z}, \qquad |z| < 1.
The exponential series. For
\sum z^n / n! we have a_n = 1/n!, so
\left|\frac{a_{n+1}}{a_n}\right| = \frac{n!}{(n+1)!} = \frac{1}{n+1} \to 0 \;\Rightarrow\; \frac{1}{R} = 0 \;\Rightarrow\; R = \infty.
The series converges on the whole plane — it is
e^z. The factorial-coefficient series. For
\sum n!\, z^n the ratio
(n+1)!/n! = n+1 \to \infty, so
1/R = \infty and R = 0: it converges
only at the single point z = 0.
For a complex power series \sum a_n (z - z_0)^n:
-
there is a radius R \in [0, \infty] with the series converging
for |z - z_0| < R and diverging for
|z - z_0| > R;
-
the convergence set is a disk (not an interval) because it depends only on
the distance |z - z_0|;
-
1/R = \lim |a_{n+1}/a_n| = \lim \sqrt[n]{|a_n|};
-
examples: \sum z^n has R = 1,
\sum z^n/n! has R = \infty,
\sum n!\,z^n has R = 0.
The real radius of convergence always felt slightly arbitrary: why should
1/(1 + x^2) = 1 - x^2 + x^4 - \cdots stop converging at
|x| = 1 when the function itself is perfectly smooth there? The
complex view answers it at a glance. Convergence fills a disk, and that disk grows until it
hits trouble. The number R is not a property of the formula on the
line — it is the radius of a circle in the plane, and we will soon see precisely what stops
it from growing any larger.
Move a test point in and out of the disk
The shaded disk has radius R about the origin
z_0 = 0. Set the radius with the R slider and move the test
point z with the x and y sliders. The readout
reports |z| and flags whether the series converges (inside),
diverges (outside), or sits on the uncertain boundary circle.