Power series and analyticity

In one sentence

Holomorphic and analytic are the same thing in one complex variable, the radius of convergence is the distance to the nearest singularity, and the identity theorem makes holomorphic functions so rigid that agreeing on a small set forces agreeing everywhere.

Why the exam asks it

Because the equivalence of "differentiable once" and "equal to its power series" is the single largest difference between real and complex analysis, and because the identity theorem's hypothesis — an accumulation point inside the domain — is precisely the sort of qualifier that gets dropped in an option.

The idea

A power series $\sum a_n (z - z_0)^n$ has a radius of convergence $R$ given by

$$\frac1R = \limsup_n |a_n|^{1/n} \qquad \text{(Cauchy–Hadamard)},$$

converging absolutely for $|z - z_0| < R$, diverging for $|z-z_0| > R$, and saying nothing about $|z - z_0| = R$. Inside the disc it is holomorphic, and may be differentiated and integrated term by term, with the same radius each time.

The equivalence

$f$ holomorphic on an open set $\iff$ $f$ is analytic there, that is, locally the sum of a convergent power series.

One direction is easy; the other is Cauchy's theory, and it is the reason complex analysis behaves so differently. In real analysis these are genuinely different classes: $e^{-1/x^2}$ is smooth with every derivative zero at the origin, so its Taylor series is $0$ and does not represent it. There is no complex counterpart. Any option asserting a complex smooth-but-not-analytic function is false.

The radius is a distance

For $f$ holomorphic on a domain, the Taylor series at $z_0$ has radius exactly the distance from $z_0$ to the nearest point where $f$ fails to be holomorphic.

This resolves the classic puzzle. On $\mathbb{R}$, $\frac{1}{1+x^2}$ is smooth, bounded, positive everywhere, and its Maclaurin series $\sum (-1)^n x^{2n}$ diverges for $|x| > 1$. Nothing on the real line explains the number $1$. In $\mathbb{C}$ the function is $\frac{1}{1+z^2}$, singular at $\pm i$, and $|{\pm i}| = 1$. The obstruction was never on the line.

The same idea reads off radii without computing coefficients: the radius of the series for $\frac{1}{z-3}$ about $0$ is $3$; about $1$ it is $2$. And $e^z$, $\sin z$, $\cos z$ are entire, so their series have infinite radius.

The identity theorem

Let $f, g$ be holomorphic on a connected open set $D$. If $\{z : f(z) = g(z)\}$ has an accumulation point in $D$, then $f \equiv g$ on $D$.

Equivalently: the zeros of a nonzero holomorphic function on a connected domain are isolated.

Both qualifiers are load-bearing.

Connected. On a disconnected domain the theorem applies componentwise, so $f$ can be $0$ on one component and $1$ on another.

Accumulation point in $D$. Take $f(z) = \sin(1/z)$ on $\mathbb{C}\setminus\{0\}$. Its zeros are $1/(n\pi)$, infinitely many, accumulating at $0$ — which is not in the domain. So $f$ is not identically zero, and nothing is violated. Contrast $\sin(1/z)$ with any function holomorphic at $0$: there the accumulation point would be inside and the function would have to vanish identically.

A frequent application: two holomorphic functions agreeing on a segment of the real line, or on a sequence converging to an interior point, agree everywhere on a connected domain. That is how identities like $\sin^2 + \cos^2 = 1$ transfer from $\mathbb{R}$ to $\mathbb{C}$ for free.

Zeros and order

If $f$ is holomorphic and not identically zero near $z_0$ with $f(z_0) = 0$, then $f(z) = (z - z_0)^m g(z)$ with $g(z_0) \ne 0$ and $m \ge 1$ finite — the order of the zero. The finiteness of $m$ is exactly the statement that zeros are isolated.

Where intuition breaks

"Smooth but not analytic." No such complex function. In the real case $e^{-1/x^2}$ is the standard example, and the instinct carries over wrongly.

"$\frac{1}{1+x^2}$ has no singularity on $\mathbb{R}$, so its series should converge everywhere." Radius $1$, forced by $\pm i$. The real line is a slice of the real situation, not the whole of it.

"$f$ vanishes on an infinite set, so $f \equiv 0$." $\sin(1/z)$ on the punctured plane. The set must have an accumulation point in the domain. Infinitely many zeros is not enough.

"$f = g$ at $1/n$ for every $n$, so $f = g$." True if $f, g$ are holomorphic at $0$ — the accumulation point is then in the domain. On $\mathbb{C}\setminus\{0\}$, false.

"The series converges on its circle of convergence." Cauchy–Hadamard says nothing about the boundary. $\sum z^n$ diverges everywhere on $|z| = 1$; $\sum z^n/n^2$ converges everywhere on it; $\sum z^n/n$ converges at every boundary point except $z = 1$. All three have radius $1$.

"Differentiating a power series shrinks the radius." The radius is unchanged under term-by-term differentiation and integration.

"A holomorphic function with infinitely many zeros in a disc is zero." Only if the zeros accumulate inside the domain — in a compact subset they must, so on a closed disc inside the domain this is true. The distinction is whether the accumulation point is reachable.

The exam's angle

1. To find a radius, locate the singularities and measure. Almost never compute $\limsup|a_n|^{1/n}$ — find the nearest pole to the centre.
2. For identity-theorem options, ask where the accumulation point is. Inside the domain, the conclusion holds; on the boundary or at a puncture, it does not.
3. Watch for "connected". A disconnected domain breaks every rigidity statement in this topic.
4. Boundary behaviour is a separate question from radius, and needs a real-series test at that point.
5. If an option describes a complex function that is smooth and not analytic, it is false — no further work needed.

The night before

• Holomorphic $\iff$ analytic. No complex $e^{-1/x^2}$.
• Cauchy–Hadamard: $1/R = \limsup |a_n|^{1/n}$. Nothing about the boundary.
• Radius $=$ distance from the centre to the nearest singularity. $\frac1{1+z^2}$ about $0$ has radius $1$ because of $\pm i$.
• Term-by-term differentiation and integration preserve the radius.
• Identity theorem: agreement on a set with an accumulation point in a connected domain forces agreement everywhere.
• Zeros of a nonzero holomorphic function are isolated and of finite order.
• $\sin(1/z)$ on $\mathbb{C}\setminus\{0\}$: infinitely many zeros, accumulating at the puncture, not identically zero.
• Boundary cases at $|z|=1$: $\sum z^n$ diverges everywhere, $\sum z^n/n^2$ converges everywhere, $\sum z^n/n$ converges except at $z=1$.