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Darij, such stuff is usually Gauss-Lucas in disguise and this case is no exception, though one needs to use once the version for polar derivative $D_1f(z)=(1-z)f'(z)+nf(z)$ of a polynomial $f$ of degr …

Let's create a proof a la Koosis. All the techniques used below can be found in his book "The Logarithmic Integral". Take $a>1$ and put $f(z)=a\prod_j(1+z/z_j)^{p_j}$. That is a nice analytic functi …

It took me some time to find a solution that satisfies both requirements: a) If should be based on the power series expansion b) It should use no tools heavier than contour integration. So, let $g( …

OK, shameless cheating, as promised. Part 1. Let's start with something. We need a function bounded in $\Re z>1$ and growing not too fast on each vertical line whose zeroes are somewhere on the left …

That is a rather tough puzzle (took me two full days) with a rather short solution. The first step is the differential equation Fred already mentioned: $$ (1-z^2)H'(z)-(1+z)H(z)+2zH(z^2)=0\,. $$ Now d …

Strange that I missed this one. On the other hand I haven't visited MO often lately. Anyway, here is the solution (It is always tempting to renew our old discussion on what problems a minimally intell …

It can get arbitrarily ugly. Indeed, approximate $1/z$ by a polynomial $p$ in the domain $K\subset\mathbb D$ whose complement is connected but goes from $0$ to the boundary along a long winding narrow …

Yes, it is dense. Indeed, if $g$ is an $L^2$ function supported on $[0,1]$ such that $g$ is orthogonal to every $f+\mathscr Hf$ with $f$ compactly supported on $(0,+\infty)$, then $g-\mathscr Hg=0$ on …

Unfortunately, no, as requested: Take any sequence $\delta_j\in(0,1)$ decaying to $0$, choose small $\mu_j>0$ such that $\prod_j \delta_j^{\mu_j}=e^{-1}$ and put $f_n(z)=e^n\prod_j B_{\delta_j}(z)^{[\ …

Also, the classical Fabry gap theorem tells you that the unit circumference is the natural boundary. Meanwhile, all "elementary" functions can be analytically continued along almost every path on the …

If all you want is a compositional square root of something like $e^z-1$ analytic in some disk around the origin, I would go for $e^z-1-\frac 34 z=\frac z4+h(z)$. Then, putting $f(z)=\frac z2+g(z)$, w …

The answer is "Quite often yes, but the error terms are seldom as good as in the meromorphic case". The reason the asymptotics for the meromorphic functions works is that we know the exact coefficient …

$Hello$, Tomasz! (for some reason the MO prohibits saying "Hi" or "Hello" in the normal text mode). Nice to see you back. Apparently you are still asking the same question whether a function $H$ close …

The set can be infinite (but only countable). For an example choose any $t_j$ linearly independent over $\mathbb Q$ and let $V$ be the set of all $v$ such that $vt_j\mod 1 \notin (\frac 12-a_j,\frac 1 …

Disclaimer: I learned the trickery below from N.K.Nikolskii, who was giving us a special topics course in complex analysis when I was a fourth year undergraduate student. I have no idea whether it can …