Questions tagged [holomorphic-functions]

For questions on holomorphic functions, complex-valued functions of one or more complex variables that are complex differentiable in a neighborhood of every point in its domain.

A holomorphic function is a complex-valued function of one or more complex variables that is complex differentiable in a neighborhood of every point in its domain. The existence of a complex derivative in a neighborhood is a very strong condition, for it implies that any holomorphic function is actually infinitely differentiable and equal to its own Taylor series (analytic). Holomorphic functions are the central objects of study in complex analysis.

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What is it that makes holomorphic functions so rigid?

This semester I took a complex analysis class and, as far as I've seen, holomorphic functions on the complex plane have very "powerful" properties; For example, the identifying theorem, or the Casorati-Weierstrass theorem, Morera's Theorem, or even…
JustDroppedIn
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Proving that a doubly-periodic entire function $f$ is constant.

Let $f: \Bbb C \to \Bbb C$ be an entire (analytic on the whole plane) function such that exists $\omega_1,\omega_2 \in \mathbb{S}^1$, linearly independent over $\Bbb R$ such that: $$f(z+\omega_1)=f(z)=f(z+\omega_2), \quad \forall\,z\in \Bbb…
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Is every entire function is a sum of an entire function bounded on every horizontal strip and an entire function bounded on every vertical strip?

Is it true that every entire function is a sum of an entire function bounded on every horizontal strip (horizontal strip is a set of the form $H_y:=\{x+iy : x \in \mathbb R \}$ ) and an entire function bounded on every vertical strip (vertical…
user228168
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Why is it called a holomorphic function?

Why is it called a Holomorphic function? The "Holo" means "entire" and "morphē" means "form" or "apparence", cf wiki. I understand the "entire", because a holomorphic function is differentiable on the entire complex plane, but why "form", or…
roi_saumon
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Is there a holomorphic and bijective function of the unit disc onto the complex plane?

Is there a holomorphic and bijective function between the open unit ball of $\mathbb{C}$ and $\mathbb{C}$? The usual homeomorphisms $\psi(z):=\frac{z}{1+|z|}$ and his composition with the conjugate map are not holomorphic on $\mathbb{C}$.
Federico Fallucca
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Holomorphic function satisfying $f^{-1}(\Bbb R)=\Bbb R$ is of the form $f(z)=az+b$

Let $f$ be a holomorphic function defined on $\Bbb C$ such that $f^{-1}(\Bbb R)=\Bbb R$. Prove that there exists $a,b \in \Bbb R, \; a \neq 0 \;$ such that $f(z)=az+b \;$ for every $z \in \Bbb C$. I was given a hint: define the function…
user401516
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Milne Thomson method for determining an analytic function from its real part

What is the logic behind taking $z = {\bar {z}}$ while finding analytic function in Milne-Thomson Method ? I mean we write $ { f(z)=u(x,y)+iv(x,y)} $ as $ {\displaystyle f(z)=u\left({\frac {z+{\bar {z}}}{2}}\ ,{\frac {z-{\bar…
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Definitions of analytic, regular, holomorphic, differentiable, conformal: what implies what and do any imply that a function is a bijection?

I'm looking back at some complex analysis and have gotten myself a little muddled in all of the definitions analytic/ regular/ holomorphic/ differentiable/ conformal... In particular, at the moment I'm thinking about conformal functions $f(z)$ on…
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Holomorphicity of the square of a function

I am trying to figure out whether the following statement is true. If $f$ is such that $f^2$ is holomorphic on an open set $\Omega \subset \mathbb{C}$ then $f$ itself is holomorphic on $\Omega$. My feeling would be that since the function mapping…
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Holomorphic function with constant modulus in the boundary of an annulus can be written as $f(z)=cz^n$ for $c\in \mathbb{C}$ and $n\in \mathbb{Z}$

I'm preparing myself to take a qualifying exam for my math PHD. I was trying to solve the previous exams and I found a complex analysis problem I couldn't solve: Let $0
Knull
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When is composition of meromorphic functions meromorphic

When I compose a meromorphic and a holomorphic function, I get a meromorphic function. Are there other cases when a composition of two meromorphic functions is meromorphic? For example, if I compose a holomorphic and a meromorphic function? Or does…
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$f: \mathbb{D} \to \mathbb{D}$ holomorphic, $f(\frac{1}{2}) + f(-\frac{1}{2}) = 0$. Prove $|f(0)| \leq \frac{1}{4}$

Here's the problem I'm having trouble with: Let $f: \mathbb{D} \to \mathbb{D}$ be a holomorphic function with $f(\frac{1}{2}) + f(-\frac{1}{2}) = 0$. Prove $|f(0)|\leq \frac{1}{4}$. I suspect that I somehow need to use some variant of Schwarz's…
Matija Sreckovic
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Preimage set of complex polynomial is connected.

Let $f(z) \in \mathbb{C}[z]$ with deg$(f) \geq 1$. Then $\{z \in \mathbb{C} : |f(z)| > 1\}$ is connected. I know that $f(z)$ is holomorphic, so continuous on $\mathbb{C}.$ I dont think that preimage of connected set under continuous map is always…
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Quasiregularity almost everywhere (removability)

The three equivalent definitions of quasiregular mapping that I am using are these ones: Let $U\subset\mathbb{C}$ be an open set and $K < \infty$. Then: A mapping $g:U\to\mathbb{C}$ is $K$-quasiregular if and only if $g = f\circ\phi$ for some…
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Prove or disprove that every Holomorphic function preserving unboundedness is a polynomial.

Roughly speaking as the title says: Prove or disprove that every holomorphic function preserving unboundedness is a polynomial. Let $f : \Bbb C \to \Bbb C$ be a holomorphic function such that for every unbounded set $ V \subset \Bbb C$ the image…
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