Counterexamples in complex analysis

Question

In contrast to other topics in analysis such as functional analysis with its vast amount of counterexamples to intuitively correct looking statements (see here for an example), everything in complex analysis seems to be very well-behaved (for example holomorphic functions are always analytic). But is this maxim always right? Do you know any holomorphic functions which behave in a way one wouldn't expect at first sight?

EDIT: As you can see in the answers, I came up with something myself. But I would be glad if you knew more examples where strange stuff happens.

Answer 1

Voronin's Universality Theorem about the Riemann Zeta Function really surprised me the first time I heard about it.

This theorem essentially says that the Zeta Function in some sense encodes all the possible behaviors of holomorphic functions and also all the possible behaviors of curves.

Answer 2

I used to think that the functions in complex analysis were in general too "nice" in the sense of rigidly controlled to leave much room for counter-intuitive situations, until I started reading Pommerenke's book on Boundary Behaviour of Conformal Maps. As he states in the preface:

"Then the conformal map has many unexpected properties, for instance almost all the boundary is mapped onto almost nothing and vice versa."

Admittedly, he is talking about situations where the domain is not necessarily bounded by a piecewise smooth curve, and considers meromorphic and not only holomorphic functions, but the extend to which the boundary behaviour of such "nice" functions can be badly behaved came as a bit of a shock to me at the time.

Answer 3

Now I recall this function which struck me as particularly weird.

Consider $$f(z)=\sum_{k=0}^\infty z^{n!}.$$

This power series converges for $|z|<1$, but for every $z=e^{i2\pi q}$ with $q \in \Bbb Q$ we see that there is a $k$ with $z^n=1$ for each $k \ge n$, so that $$\lim\limits_{r \rightarrow 1}f(re^{i2\pi q})=\lim\limits_{r \rightarrow 1}(const+\sum_{k=0}^\infty r^{n!})=+\infty$$

Thus, the set of singularities is dense on the unit circle and $f$ cannot be extend beyond the unit circle.

This came to my mind when I saw the prime zeta function, which cannot be extended beyond $Re(z)=0$ because it has pole at each $\rho/n$, where $\rho$ is a nontrivial Zeta function zero and $n$ is a positive integer.

Answer 4

From another math.SE question, a function $f(z) = \sum a_n z^n$ which is convergent on the interior of the unit disc and extends continuously to the entire closed disc, but doesn't converge at all the points on the boundary.