Power series in Calculus 2 converged on an interval centered at $a$. With complex $z$, the interval becomes a disk, a region you already know how to read. The complex picture also explains something that the real line keeps hidden: why the radius is what it is.
Key idea
A power series $\sum_{n \ge 0} c_n(z - a)^n$ has a radius of convergence $R$ (with $0 \le R \le \infty$): it converges absolutely for $|z - a| < R$ and diverges for $|z - a| > R$ (on the circle itself anything can happen). The ratio test finds it: if $\left|\frac{c_{n + 1}}{c_n}\right| \to \lambda$, then $R = \frac{1}{\lambda}$ (with $R = \infty$ when $\lambda = 0$). Inside its disk the sum is analytic, and it can be differentiated term by term.
Why it works
Compare with a geometric series. If the series converges at some $z_1$, its terms are bounded: $|c_n(z_1 - a)^n| \le K$. Then for $|z - a| < |z_1 - a|$,
$$|c_n(z - a)^n| \le K\left|\frac{z - a}{z_1 - a}\right|^n,$$
a convergent geometric series. So convergence at one point gives convergence on the whole disk through that point, and the set of convergence is a disk (plus perhaps some boundary points). The ratio test is this same comparison done with $\frac{|c_{n + 1}||z - a|}{|c_n|} \to \lambda|z - a| < 1$.
Worked examples
Worked example: A geometric series
Find the radius of convergence of $\displaystyle\sum_{n \ge 0}\frac{z^n}{(3 + 4i)^n}$.
- It is geometric with ratio $\frac{z}{3 + 4i}$, convergent exactly when $\left|\frac{z}{3 + 4i}\right| < 1$.
- That is $|z| < |3 + 4i| = 5$: $R = 5$.
Answer: $5$
Worked example: The ratio test
Find the radius of convergence of $\displaystyle\sum_{n \ge 1}\frac{n^2z^n}{2^n}$, and of $\displaystyle\sum_{n \ge 0}\frac{z^n}{n!}$.
- First: $\left|\frac{c_{n + 1}}{c_n}\right| = \frac{(n + 1)^2}{n^2}\cdot\frac{1}{2} \to \frac{1}{2}$, so $R = 2$. The polynomial factor $n^2$ does not change the radius.
- Second: $\frac{n!}{(n + 1)!} = \frac{1}{n + 1} \to 0$, so $R = \infty$: the exponential series converges everywhere.
Answer: $2$ and $\infty$
Worked example: The radius from the singular points
What is the radius of convergence of the Maclaurin series of $\dfrac{1}{z^2 - 2z + 5}$?
- The bottom is $0$ at $z = 1 \pm 2i$ (quadratic formula).
- Both are at distance $|1 \pm 2i| = \sqrt{5}$ from the center $0$.
- $R = \sqrt{5}$. (The next lesson proves that the radius reaches exactly to the nearest singular point.)
Answer: $\sqrt{5}$
Common mistake
Expecting the Maclaurin series of $\frac{1}{1 + x^2}$ to converge for every real $x$. The function is smooth on the whole real line, so this seems natural. But the series $1 - x^2 + x^4 - \cdots$ diverges for $|x| \ge 1$. The reason is complex: at $z = \pm i$ the function blows up, and the disk of convergence cannot reach past them. To find a radius, look for singular points in the whole complex plane, not only on the real line.