Articles of complex analysis

Let $f:U \to V$ be a bijective holomorphic function. Show that inverse of $f$ is also holomorphic.

Suppose $U$ and $V$ be domains(i.e., open and connected) in $ \mathbb C$.Let $f\colon U \to V$ be a bijective holomorphic function. Show that the inverse of $f$ is also holomorphic. By Open Mapping Theorem it is clear that $f^{-1}$ is also continuous. Please give some ideas to complete the proof. Edit:I’m interested in a […]

Where does the iteration of the exponential map switch from one fixpoint to the 3-periodic fixpoint cycle?

In the answering of another question in MSE I’ve dealt with the iteration of $x=b^x$ where the base $b=i$. If I reversed that iteration $x=log(x)/log(i)$ then I run into a cycle of 3 periodic fixpoints. Another complex base on the unit circle however, $b = \sqrt{0.5}(1+i)$ gives only one fixpoint. I asked myself, whether there […]

Laurent series expansion of a given function

Let $\sum_{-\infty}^{\infty} a_n z^n$ be the Laurent series expansion of $f(z)=\frac{1}{2z^2-13z+15}$ in an annulus $\frac{3}{2}< \vert z \vert < 5$. What is the value of $\frac{a_1}{a_2}$? My attempt: First note that $f$ is analytic in the given annulus. $f(z)=\frac{1}{2z^2-13z+15}=\frac{1}{(2z-3)(z-5)}$ , $\quad$ $\quad$$\quad$$\quad$$\quad$$\quad$$\quad$ $=\frac{\frac{-2}{7}}{2z-3}+\frac{\frac{1}{7}}{z-5}$ , by Partial Fraction. How to goes further to find out the […]

Can this “real set” disconnect complex n-space?

Hi everyone: Suppose $E\subset \mathbb{R}^{n}$ is a closed set with empty interior, and $ D $ a domain in $\mathbb{C}^{n}$ with $n\geq2$. Can we conclude that the open set $ D\setminus E $ is connected?

let $f$ be analytic and bounded above, can I prove f is constant?

I’ve read up on Lioville’s theorem and I was wondering if this could also be proved using the theorem: let $f$ be analytic on $\mathbb{C}$ and let $K>0$ be s.t. $|f| \geq K$, could I prove using Liouville’s theorem that $f$ is constant? I tried by defining $q(z) = 1/f(z)$ and then $|q(z)| \leq 1/K$ […]

Upper bound for complex polynomial

I have a polynomial $p$ of degree $n$ satisfying $\lvert p(z) \lvert \leq c\ \ \forall z\in\partial B_1(0)$. (Isn’t this true for any polynomial?) Show $\lvert p(z)\lvert \leq c \lvert z\lvert^n \ \ \forall z\in \mathbb{C}\backslash B_1(0)$. The obvious attempt would be $|p(z)|=|p(\lvert z\lvert\frac{z}{\lvert z\lvert})|=|\sum_{i=0}^n a_i |z|^i (\frac{z}{\lvert z\lvert})^i|\leq |z|^n \sum_{i=0}^n |a_i (\frac{z}{\lvert z\lvert})^i|$ which […]

The ratio $\frac{u(z_2)}{u(z_1)}$ for positive harmonic functions is uniformly bounded on compact sets

I want to prove the following: If $E$ is a compact set in a region $\Omega \subset \mathbb C$, prove that there exists a constant $M$, depending only on $E$ and $\Omega$, such that every positive harmonic function $u(z)$ in $\Omega$ satisfies $u(z_2) \leq M u(z_1)$ for any two points $z_1, z_2 \in E$. This […]

Weierstrass product form

How to show the Weierstrass product form of the entire function $f(z)= \sinh z$ This question seem so interesting. I would like to write my some ideas, but I dont want to direct incorrectly. Please help me to learn correctly and explicitly. I asked to question in fact in order to learn the topic precisely.

Sketch complex curve $z(t) = e^{-1t+it}$, $0 \le t \le b$ for some $b>0$

Sketch complex curve $z(t) = e^{-1t+it}$, $0 \le t \le b$ for some $b>0$ I tried plotting this using mathematica, but I get two curves. Also, how do I find its length, is it just the integral? This equation doesn’t converge right? Edit: I forgot the $t$ in front of the $1$ so it’s not […]

Does convergence of power series on radius of convergence imply absolute convergence?

Let $R$ be radius of convergence of power seires $\displaystyle\sum_{k}a_kz^k$. If the power series converges for all $|z|=R$, can we say that it converges absolutely on the radius of convergence? I tried to show that ${a_k}=\frac{e^{ik^2q}}{k}$ where $q$ is an irrational, serves as a counter example. In this case, $R=1$, and the series diverges absolutely […]