This question was asked in an assignment which I am solving and I am struck on this.  Let f be a non-constant entire function. Prove that the closure of $\{z:|f(z)&lt;c\}$ coincides with $\{z:|f(z)|\leq c\}$ for all $c&gt;0$.  Let $1 =\{z:|f(z)|&lt;c\}$ and $2 = \{z:|f(z)| \leq c\}$ for all $c&gt;0$. Let {$z_n$} and {$z_m$} be respective sequences for set 1 and 2 which converge to z and z' respectively . Then $f(z)&lt;c$ and f(z')$\leq$ c. But I have no reason to prove that required assertion . Can someone please tell which result should I use. Thanks!!
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Exercise: Let $f$ be entire and non-constant. For any positive real number $c&gt;0$ show that the closure of $\{ z:|f(z)|&lt;c\}$ is the set $\{z:|f(z)|\leq c\}$.   Solution: Let $\varepsilon&gt;0$ and define $A:=\{ z:|f(z)|&lt;c\}$. Then, because $f$ is continuous (via entire) and non-constant, there exists $z\in A$ such that   $$B(c;\varepsilon)\cap A \neq \varnothing .$$  That is to say that $c$ is a limit point of $A$. Since $A^-$ must contain all of its limit points, we have that $\{z:|f(z)|\leq c\}$.    Is my solution correct? I have another way of doing it using a sequence $z_n\to z$, but I would prefer to use the above if it correct. The reason I am feeling unsure about my solution is because I am not sure how we know (geometrically) that $|f(z)|$ goes all the way up to the boundary (i.e. $c$). Is it safe to say that $|f(z)$ gets infinitely close to $c$ but does not touch because $f$ is entire &amp; non-constant? FYI I am new at these concepts so please be a detailed as possible. Thank you in advance!  Note: I am using the notation $A^-$ to indicate the closure of $A$. This is the notation used in my text (Complex, Conway) and I suspect it is used as to not be confused with the complex conjugate.
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For which $a \in \mathbb{C} -\ \{0,1\}$ does the equation $f(z) = f(az) $ has a non constant solution $f$ with $f$ being analytical in a neighborhood of $0$.  My attempt:  First, we can see that any such solution must satisfy:  $f(z)=f(a^kz)$ for all $k \in \mathbb{N} $.  If $|a|&lt;1$:  The series $z_{k} = a^k$ converges to 0 which is an accumulation point, and $f(z_i)=f(z_j)$ for all $i, j\in \mathbb{N} $. Thus $f$ must be constant.  If $|a|=1$:  For all $a \neq 1$ , $f$ must be constant on any circle around $0$, so again $f$ must be constant.  My quesions are:   Am I correct with my consclusions?   Also, I'm stuck in the case where $|a|&gt;1$. Any ideas?  Thanks
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I'm going over old exam problems and I got stuck on this one. Suppose that $\mathbb{f}\colon \mathbb{D} \to \mathbb{C}$ is analytic and bounded. Let $\{a_n\}_{n=1}^\infty$ be the non-zero zeros of $\mathbb{f}$ in $\mathbb{D}$ counted according to multiplicity. Show that  $$ \sum_{n=1}^\infty \left( 1 - \left|a_n \right|\right)\lt\infty $$  I can understand that $\left|a_n \right|$ goes to $1$ since zeros are isolated, but it doesn't help showing the series is convergent. Any help will be appreciated!
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$(z_n)_{n\in \mathbb{N}}$ is a complex valued sequence such that subsequences $(z_{2n})_{n\in \mathbb{N}}$ and $(z_{2n+1})_{n\in \mathbb{N}}$ converges to same limit.  How could I show that the sequence $(z_n)_{n\in \mathbb{N}}$ converges to the same limit using $\epsilon$, $N$ definition?    For every $\epsilon &gt;0$,  $|z_{2n}-l|&lt;\epsilon$ whenever $n&gt;N_1$, and   $|z_{2n+1}-l|&lt;\epsilon$ whenever $n&gt;N_2$.  We want to show that for every $\epsilon &gt;0$, there exists $N$ such that $|z_{n}-l|&lt;\epsilon$ whenever $n&gt;N$. I tried something like  $|z_{n}-l|\le|z_n-z_{2n}|+|z_{2n}-l|$ but wasn't sure what to do next
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Suppose function $f$ is holomorphic on $\mathbb C-\{0\}$ and satisfies $$|f(z)| \le \sqrt {|z|} + \frac{1}{\sqrt {|z|}}.$$ Prove that $f$ is a constant function. I think it's related to Laurent series representation and ML inequality but I have no idea how to prove it. Can anyone give me some hints? Thanks in advance!
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These two questions I didn't even find the way to solve So please if you can help   Suppose $f (z)$ is entire with $|f(z)| \le |\exp(z)|$ for every $z$ I want to prove that $f(z) = k\exp(z)$  for some $|k| \le 1$ Can a non constant entire function be bounded in half a plane? Prove if yes , example if not.
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I'm having trouble proving the following:  Let $f$ be analytic on $D(0,R)$ and suppose there is a real constant A for which $\operatorname{Re}f(z) &lt; A$ for all $z \in D(0,R)$. If $f(0) = 0$, show that $|f(z)| \leq \frac{2A|z|}{R-|z|}$.   I was told to first start by showing that $|\frac{f(z)}{2A-f(z)}| \leq 1$ and also use the following result, which I was able to prove:  Let $f$ be analytic on $D(0,R)$ and let $f(0)=0$. If $|f(z)| \leq M$ for all $z \in D(0,R)$, then $|f(z)| \leq M|z|/R$ for all $z\in D(0,R)$.  Thanks in advance!
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Function $f(z)$ is holomorphic at least within the unit disk.   If for all arguments $|z|&lt;1$ we have $|f(z)| \le 1$, and there exists a zero of order $n$ at $z=z_0$, then show that $|f(0)| \le |z_0|^n$.  How should I tackle this problem? Can it be generalized if for $|z|&lt;R$ we have $|f(z)| \le M$ ?    So far I only understand that there exists a holomorphic $g(z)$ with $g(z_0) \ne 0$ such that $f(z) = (z-z_0)^n g(z)$,then $|f(z)| = |z-z_0|^n |g(z)| \le 1$ and moreover $|f(0)| = |z_0|^n |g(0)|$, but that doesn't seem useful since apparently I know too little about $g(z)$. In particular why should $|g(0)| \le 1$?
110728
Let $f(z)$ be an entire function such that for some constant $K$ , $|f(z)| ≤ K$ $|z|^3&lt;$ for $|z|\ge1$ and $f(z) = f(iz)$ $∀z∈C$ , then which of the following are correct ?  (A)$ f(z) = Kz^3$ $∀ z∈\Bbb{C}$ (B) $f(z)$ is a constant function (C) $f(z)$ is quadratic function (D) No such $f$ exists   My attempt : I can deduce that since $f(z) = f(iz)$ and since $i^4=1 \implies f$ must involve only fourth powers of $z$ . Hence $f(z)=a_1z^4+b_1z^8+c_1z^{16}+ \cdots $-------(1)  Also for $|z| \ge 1 $: let C.   Then $|g(z)| = |f(z)|/|z^3| \le k$ => By the Cauchy estimate theorem we can prove that :  $f(z)/z^3$ is a constant function $= c $   $\implies f(z)=cz^3$ for $|z| &gt; 1$     ...........(2)  From (1) and (2) ; we get  there does not exist a function like this ?  My textbook answer says it's a constant function ?
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Let $c&gt;0$ be a real number. I would like to study the convergence of $a_n := c^n n!/n^n$, when $n \to \infty$, in function of $c$.   I know (from  question) that $n!&gt;(n/e)^n$, so that $c^n n!/n^n&gt;1$ for $c ≥e$. But this doesn't imply that the sequence goes to infinity. And I'm not sure what to do for $c&lt;e$. I tried usual tests (D'Alembert...), without any success. I would like to avoid using Stirling approximation.  Thank you for your help!
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Could someone advise me on how to determine all entire functions $f$ such that $|f'(z)| \leq |f(z)|, \forall z\ ?$  Hints will suffice, thank you.
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Let $f$ be a continuous function on the closed unit disk and holomorphic inside it, such that $f(z) = 0$ for all $ z = \exp (it) $ with $0 ≤ t ≤ \pi/2 $. Prove that $ f(z) = 0 $ for all $ z \in \overline{D}(0,1) $.  I have worked this exercise for absurdity, but it has not reached any conclusion. I would like to know a more sophisticated test, perhaps by a direct method.
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Here's my proof in my own words, does it stack up?   Showing $A$ is complete implies $A$ is closed. Let $(x_n)$ be a convergent sequence in $A$. $A$ is complete $\implies (x_n) \to p \in A$. Hence $A$ is closed. Showing $A$ is closed $\implies$ $A$ is complete. Let $(x_n)$ be a convergent sequence in $A$. $A$ is closed $\implies (x_n) \to p \in A$. As every convergent sequence is a Cauchy sequence, $(x_n)$ is a Cauchy sequence in $A$ that converges to $p \in A$ and hence $A$ is complete.   That sound ok?  As an aside, it seems to me that the definition of completeness and closed are basically identical, why are there two definitions for the same thing? Am I missing something here?
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Let $(x_n)$ be a sequence of real numbers.  Prove that if there exists $x$ such that every subsequence $(x_{n_k})$ of $(x_n)$ has a convergent (sub-)subsequence $(x_{n_{k_l}})$ to $x$, then the original sequence $(x_n)$ itself converges to $x$ .              Thanks for any help.
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Question: let $f$ be an entire function on $\mathbb{C}$ and let $g(z)=\overline{f(\bar{z})}$ then which of the following is/are correct?  (a) if $f(z) ∈\mathbb{R}$ for all $z∈\mathbb{R}$ then $f=g$  (b) if $f(z)∈\mathbb{R}$  for all $z∈\{z : Im\text{ $z$ }=0\}∪\{z : Im\text{ $z$ }=a\}$ for some $a&gt;0$ then $f(z+ ia)=f(z-ia)$ for all $z∈\mathbb{C}$  (c) if $f(z)∈\mathbb{R}$  for all $z∈\{z : Im\text{ $z$ }=0\}∪\{z : Im\text{ $z$ }=a\}$ for some $a&gt;0$ then $f(z+ 2ia)=f(z)$ for all $z∈\mathbb{C}$  (d) if $f(z)∈\mathbb{R}$  for all $z∈\{z : Im\text{ $z$ }=0\}∪\{z : Im\text{ $z$ }=a\}$ for some $a&gt;0$ then $f(z+ ia)=f(z)$ for all $z∈\mathbb{C}$  My attempt: clearly, if $z∈\mathbb{R}$ then $\bar{z}=z∈\mathbb{R}$. Hence if $f(z)∈\mathbb{R}$ then $f(\bar{z})=f(z)∈\mathbb{R}$ so that $g(z)=\overline{f(\bar{z})}=f(\bar{z})=f(z)$ for $z$ in $\mathbb{R}$ hence (a) is correct.  But as beginner i am unable to prove or discard (b), (c), (d). I am stuck on this from hours, Please help me..
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I am solving previous year question paper some competitive exam. Give me some hint to solve the following problem.  Let $f$ be an entire function. Suppose for each $a \in \mathbb{R} $ there exists at least one coefficient $c_n$ in $f(z) = \sum_{n=0}^{\infty} c_n (z-a)^n$ which is zero. Then,  a) $f^{(n)}(0) = 0$ for infinitely many $n \ge 0$  b) $f^{(2n)}(0)=0$ $ \forall n \ge 0$  c) $f^{(2n+1)}(0)=0$  $ \forall n \ge 0$  d) $f^{(n)}(0) = 0$ for all sufficiently large n.  Thanks in advance.
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Let $f$ be a non-constant analytic function on $\mathbb{D}$ satisfying $|f(z^2)| ≤ |f(z)|$ for $|z| &lt; 1.$  $(a)$ Show that $f(0) = 0.$  $(b)$ Show that $f(z)\neq 0$ for $0&lt;|z|&lt;1.$  $(c)$ Show that $f(z) = az^n,$ where $a ∈ \mathbb{C},n ∈ N.$  I know how to do $(b)$ but I'm not sure how to do the other two parts.  Some help would be great thanks.