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## Hot answers tagged cv.complex-variables

14

Extending the argument by GH from MO, $\zeta(P(s))$ has a pole for any $s$ such that $P(s)=1$, while $P(\zeta(s))$ has unique pole for $s=1$. Therefore if $\zeta(P(s))=P(\zeta(s))$, then $P(s)=1$ has unique solution $s=1$ and $P(x)-1=c(x-1)^n$ for some complex $c$ and positive integer $n$, and we get $\zeta(1+c(s-1)^n)=1+c(\zeta(s)-1)^n$. For $s=1+x$ with ...

12

It doesn't seem to me that either article follows the line of reasoning as you have presented it. Indeed, we do not take the integral over the union of integrals $(x_n,(1+\varepsilon)x_n)$. We do get that the integral over those intervals is infinite, but you correctly note this does not give a contradiction. Instead, the argument goes as follows. Let me ...

12

Main areas are dynamics of automorphisms (for example, Henon maps), dynamics of endomorphisms, dynamics of foliations, and local dynamics. Eric Bedford, Tien-Cuong Dinh, John Fornaess, Misha Lyubich, Nessim Sibony, John Smiley, and their students and collaborators. One-dimensional holomorphic dynamics, functions of several complex variables (especially ...

9

It is straightforward to see that if $P\in\mathbb{C}[x]$ satisfies $P(\zeta(s))=\zeta(P(s))$, then $P(1)=1$. Note that if $P(\zeta(s))=\zeta(P(s))$ holds for all real $s>1$, then it also holds for all complex $s\neq 1$ by the uniqueness of analytic continuation. Indeed, $\zeta(s)$ has a pole at $s=1$, hence $P(\zeta(s))=\zeta(P(s))$ also has a pole at $s=... 9 No. If$P$has a positive leading coefficient, then letting$s$go to infinity and using continuity of$P$we get$P(1)=1$. If$P$has a negative leading coefficient, then$\zeta(P(s))$has zeros at arbitrarily large$s$, while$P(\zeta(s))$does not. 6 Here is a simple argument which applies to polynomials in$C[x]$, and even to most rational functions in$C(x)$.$\zeta$is a meromorphic function in the plane. So the Nevanlinna characteristic$T(r,\zeta)$is defined. It is a positive increasing function, a kind of transcendental analog of the degree of a rational function. Your equation implies that$(\deg ...

6

The identity $$\sum_{n=1}^\infty\frac{\mu(n)\log n}{n}=-1$$ was conjectured by Möbius (1832) and proved by Landau (1899). It is a consequence of the prime number theorem. Not surprisingly, the rate of convergence is determined by the (known) zero-free region of $\zeta(s)$. In particular, $$S(x)=-1+O_\epsilon(x^{\alpha-1+\epsilon})$$ holds for any $\epsilon&... 4 Let$z_0\in\Bbb{R}$be arbitrary. The matrices$A_0:=A(z_0)$and$B_0:=B(z_0)$are normal; in view of$A_0^*=A^\#(\bar{z_0})=A^\#(z_0)$and$B_0^*=B^\#(\bar{z_0})=B^\#(z_0)$they commute with their Hermitian adjoints. They also commute with each other. So$A_0$and$B_0$could be simultaneously diagonalized by a unitary matrix. That matrix also diagonalizes$...

4

A partial solution: As mentioned by @MargaretFriedland, the desired $M_a(r)$ is the absolute maximum of $g(t):=\sqrt{{\rm{e}}^{2r\cos t}+2a{\rm{e}}^{r\cos t}\cos(r\sin t)+a^2}$. Notice that this is an even function, so it suffices to maximize over $[0,\pi]$. If $t\in\left[\frac{\pi}{2},\pi\right]$, then $\cos t<0$. So ${\rm{e}}^{2r\cos t}\in [0,1]$ and, ...

4

An answer before the numerics start: First, note that for $w=u+iv \in \mathbb{C}$ and a fixed $a \in \mathbb{R}$ we have $|w+a|=\sqrt{u^2+v^2+2au+a^2}=\sqrt{|w|^2 + 2au+a^2}$ . Next, consider the image under the complex exponential of the circle $|z|=r>0$. Using polar coordinates, we get $|\exp (r\cos t+i r\sin t)|=e^{r\cos t}$. Thus to get $M_a(r)$ we ...

4

You should be able to find anwers to your question in a well witten book by Franc Forstnerič: Stein Manifolds and Holomorphic Mappings -The Homotopy Principle in Complex Analysis, https://www.springer.com/gp/book/9783319610573 In particular look at sections 9.10, 9.11 and Theorem 9.11.5. The book contains all the background material you need and also ...

4

The unique measure of maximal entropy $\mu_f$ supported on the Julia set of a rational map $f$ of degree $d \geq 2$ is indeed the unique balanced measure for $f$, i.e., the only probability measure $\mu$ not charging the exceptional set and satisfying $f^*\mu =d \cdot \mu$. As you already noticed in the comments, uniqueness of a measure with this property ...

2

The measure of maximal entropy is the unique measure that is "fully invariant" in your sense. I believe that this already follows from the original proofs - indeed, it is well-known that if you take a point mass at some non-exceptional point, and keep pulling back, you will converge to the measure of maximal entropy. This should be enough to deduce the claim....

1

I think this an issue of choice of branch of square root. Note that the paper asks you to use the branch holomorphic in $\mathbb{C} \setminus E$ and behaves as $z$ as $z \to \infty$. Let's call this choice of branch $w(z)$ and call the compact set delimited by $[-1, 1] \cup E$ by $K$. Then the map we are looking for can be defined by  z + w(z) = \left\{ ...

1

Assuming that a closed formula is not accessible, possibly maxmod, which calculates the maximum modulus of a complex polynomial on the unit disk, would speed a numerical attack. (Polynomial order of a few hundred is not a problem.)

1

The way that the paper by Freire-Lopes-Mañé makes sense of $f^*\mu=d\mu$ is the following: ''For any Borel subset $A$ of $\Bbb{C}_\infty$ with $f\restriction_A$ injective, one has $\mu(f(A))=d.\mu(A)$.'' (See p. 46 of this paper.) One observation is that such an ergodic measure $\mu$ is either supported on the Julia set or is one of those measures with ...

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