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The proof of Sullivan's theorem uses in very substantial way that the boundary of the domain is contained in the Julia set. The main part of the proof for simply connected domains goes like this. Cons …

Analytic continuation of probability densities and PDF to the complex plane has no direct probabilistic meaning but this is an important tool in proving many results which do have probabilistic meanin …

Asymptotics for $|\alpha_k|$ is not sufficient to make conclusions about asymptotics of $f$. You need to know separately, the asymptotics on the positive and negative ray. Let us enumerate the roots $ …

G. H. Hardy, Divergent series (Oxford, Clarendon Press, 1956) discusses and proves this theorem, see p. 79 and 190-191.

It is a product of two meromorphic functions: $$f(z)=g(z)h(z),$$ where $g$ is meromorphic in $C$ and $g$ is meromorphic in $\overline{C}\backslash\{0\}$, so that $h(z)=h_1(1/z)$ where $h_1$ is meromor …

The answer is negative. For the non-injective case, the reason is that non-constant complex analytic functions are open, discrete maps, while real analytic functions can be neither open nor discrete ( …

You should explain more precisely what do you mean by "extension in terms of prime ends". In the case that $\gamma_j$ are simple curves (images of $[0,1]$ under injective continuous maps), each point …

Cosine has one attracting fixed point $a\approx0.7390851$, and both critical values $\pm1$ are attracted to it. Then it follows from general theorems of dynamics of entire functions that it has one co …

In Calculus textbooks $$\int_\gamma f(z)|dz|$$ is called the line integral of the first kind, In arbitrary dimension it is frequently written as $$\int_\gamma f(x)ds,$$ where $ds$ is the length elemen …

For infinite degree, the definition of "branched covering" can be somewhat ambiguous. But $$z\mapsto \cos z: \mathbb{C}\to \mathbb{C}$$ $$\wp: \mathbb{C}\to S$$ are a simple examples of branched cove …

One restriction on the type follows from Jensen's formula which can be written as $$\frac{1}{2\pi}\int_{-\pi}^{\pi}\log|f(re^{i\theta})|d\theta= \int_0^r\frac{n(t)}{t}dt+\log|f(0)|,$$ assuming that $f …

The answer is negative. Since every neighborhood of a point on the Julia set is mapped onto the whole Julia set by some iterate of the rational function, it follows that if a small piece of the Julia …

Essentially there is nothing else. Indeed, the group of deck transformations can be identified with a group of conformal automorphisms of the plane which acts discontinuously. There is a complete clas …

Your function has simple poles at $0,-\beta$ and $k_0$. So strictly speaking your integral diverges. The standard way to bypass this difficulty is to consider it as a principal value. Then the residue …

Edited. $f=u+iv$, where $u$ is even and $v$ is odd is equivalent to $f(-x)=\overline{f(x)}$. A function has this property if and only if its Fourier transform is real. Boundedness of support of $f$ is …