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I assume that you are assuming $\sigma > 3\delta$, otherwise, the problem doesn't make sense. Can't you just apply the Argument Principle? Fix a $z_0\in S_{\sigma-3\delta}$ and an $R>>0$ so big th …

You can prove by induction that, if $f$ satisfies your equation, then there exist holomorphic functions $f_0,\ldots,f_{n-1}$ on the domain of $f$ such that $$ f = f_0(z) + f_1(z)\ \bar z + \cdots + f_ …

Actually, the correct notion of maps preserving harmonicity is that of 'harmonic morphisms': A map $f:(M,g)\to (N,h)$ between Riemannian manifolds is a harmonic morphism if it pulls back $h$-harmonic …

The answer is basically 'no', there is no 'elementary method' involving elementary operations and quadrature (i.e., finding antiderivatives of known holomorphic functions) that will give you a solutio …

I think that, even if you include the boundary $S$, they are all equivalent. In saying so, I'm assuming that the final 'it' in your first sentence refers to the plane, not to $U$ and that you are con …

Since it doesn't really involve the complex structure, can't we use the standard real coordinates on $\mathbb{C}^2=\mathbb{R}^4$ as $x_0,x_1,x_2,x_3$ and consider the harmonic polynomial $$ f = x_0 + …

Here is a revised and somewhat expanded version of my answer, with a preparatory 'toy version' to help orient the reader. A simple warmup problem: Before discussing a quantitative variant of the Sch …

Let me point out a somewhat different answer. Rbega explicitly mentions $$ Q = \left(\frac1{z^3} + \frac1{z^2}\right)\ (dz)^2 $$ as an example of the sort of meromorphic quadratic differential that i …

For your specific question, note that the domain you describe $\mathbb{D}$, consisting of those $z = x+iy$ for which $y > 1/(1+x^2)$, when regarded as a domain in the extended complex plane, $\mathbb{ …

It's easy to derive a third-order (nonlinear) differential equation for $u(x,y)$ that satisfies your conditions (1) and (2): Namely, set $\theta(x,y) = \arctan\bigl(u_y(x,y)/u_x(x,y)\bigr)$ and then …

The reason the complex Hessian (actually, it ought to be called the 'Hermitian Hessian', since it defines an Hermitian form at every point, but 'the complex Hessian' is entrenched in the literature) i …

I don't know a reference, but this desired normal form is, indeed, attainable. Here is the argument: Assume given a Kähler form $\omega$ defined on a neighborhood of $0\in\mathbb{C}^n$ and that ther …

Edit: (21 May 2017) I have modified my answer to cover the case that the OP meant to ask, i.e., the assumption is that the closure of an orbit has nonempty interior. Now that you have added the assu …

An even stronger statement is true: If $[X,D]\subset D$ and $X$ belongs to $\Gamma(D)$, then $X = 0 $. The reason is that, because $D$ is a contact $2n$-plane field (since your CR structure is stric …

It is not a finite dimensional Lie group. For example, all of the maps $$ \Phi\bigl(z,[a,b]\bigr) = \bigl(z, [a+p(z)b,\ b]\bigr), $$ where $p:\mathbb{C}^\ast\to \mathbb{C}$ is holomorphic, belong to …