In real calculus, a function with derivatives of every order can still fail to equal its Taylor series. Complex analysis has no such gap. Cauchy's formula turns into a power series directly, and the disk where it converges is as large as it could possibly be.
Key idea
If $f$ is analytic on the disk $|z - a| < R$, then on that whole disk
$$f(z) = \sum_{n \ge 0}\frac{f^{(n)}(a)}{n!}(z - a)^n.$$
So the radius of convergence of the Taylor series at $a$ is at least the distance from $a$ to the nearest point where $f$ fails to be analytic (and for functions like rational ones, exactly that distance).
Why it works
Fix $z$ in the disk and take a circle $C$ around $a$ that encloses $z$. In Cauchy's formula $f(z) = \frac{1}{2\pi i}\oint_C\frac{f(w)}{w - z}dw$, expand the kernel as a geometric series:
$$\frac{1}{w - z} = \frac{1}{(w - a) - (z - a)} = \sum_{n \ge 0}\frac{(z - a)^n}{(w - a)^{n + 1}}, \qquad \text{valid since } |z - a| < |w - a|.$$
Integrate term by term: the coefficient of $(z - a)^n$ is $\frac{1}{2\pi i}\oint_C\frac{f(w)}{(w - a)^{n + 1}}dw = \frac{f^{(n)}(a)}{n!}$. The only requirement was a circle inside the domain, so the series works out to the nearest bad point.
Worked examples
Worked example: A new center
Find the Taylor series of $\dfrac{1}{1 - z}$ about $a = i$ and its radius of convergence.
- Write $1 - z = (1 - i) - (z - i)$, so $\frac{1}{1 - z} = \frac{1}{1 - i}\cdot\frac{1}{1 - \frac{z - i}{1 - i}}$.
- Geometric series: $\frac{1}{1 - z} = \sum_{n \ge 0}\frac{(z - i)^n}{(1 - i)^{n + 1}}$, valid for $|z - i| < |1 - i|$.
- Radius $\sqrt{2}$: the distance from $i$ to the singular point $1$.
Answer: $\sum \frac{(z - i)^n}{(1 - i)^{n + 1}}$, radius $\sqrt{2}$
Worked example: A coefficient
Find the coefficient of $z^3$ in the Maclaurin series of $e^{(1 + i)z}$.
- $e^w = \sum\frac{w^n}{n!}$ with $w = (1 + i)z$; the $z^3$ term is $\frac{(1 + i)^3}{3!}z^3$.
- $(1 + i)^3 = -2 + 2i$, so the coefficient is $\frac{-2 + 2i}{6} = -\frac{1}{3} + \frac{1}{3}i$.
Answer: $-\frac{1}{3} + \frac{1}{3}i$
Worked example: Radius from the nearest singular point
Find the radius of convergence of the Taylor series of $\dfrac{e^z}{z^2 + 4}$ about $a = 1$.
- $e^z$ is entire; the bad points are $\pm 2i$.
- Distances from $1$: $|2i - 1| = |-2i - 1| = \sqrt{5}$.
- Radius $\sqrt{5}$.
Answer: $\sqrt{5}$
- $e^z = \sum \frac{z^n}{n!}$, $\sin z = \sum \frac{(-1)^nz^{2n + 1}}{(2n + 1)!}$, $\cos z = \sum \frac{(-1)^nz^{2n}}{(2n)!}$: valid for every $z$.
- $\frac{1}{1 - z} = \sum z^n$: valid for $|z| < 1$, the distance from $0$ to the pole at $1$.
- Substituting into these (for example $z^2$ or $-z$ for $z$) gives most series you need, valid on the corresponding region.
Common mistake
Reusing a known series without changing the center. $\frac{1}{1 - z} = \sum z^n$ is the series about $0$, valid for $|z| < 1$. About $a = i$ you need powers of $z - i$, and the radius changes to $|1 - i| = \sqrt{2}$. Check: the center of a Taylor series is wherever the powers $(z - a)^n$ vanish, and its radius is the distance from that center to the nearest singular point.