Assume $|f(z)|\leq 1/|y|$ for all $z\in\mathbb{C}$. Here $f$ is entire and we express $z=x+iy$. Then is $f$ constant ?
121557
Let $f$ be an entire function, such that $$ \lvert\, f(x+yi)\rvert\le \frac{1}{\lvert y\rvert}, $$ for all $x$ and all $y\ne 0$.  Show that $f$ is bounded.
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 ?
56335
Let $f : \mathbb{C} \to \{z\in\mathbb{C}:\Im(z)\neq0\} $ entire . Show that $f$ is constant.  I took $g(z)=\frac{1}{f(z)}$ and I think that g is bounded, therefore it is constant (due to Louville's theorem) but how do I prove it? Is $|g(z)|=\frac{1}{|f(z)|}=\frac{1}{u^2+v^2}\le M$ valid?
3814
The problem as stated in R. Remmert's Theory of Complex Functions p. 62., is this:  Let $f = u + iv$ be holomorphic in the region $G \subseteq \mathbb{C}$ and satisfy $u = h \circ v$ for some differentiable function $h : \mathbb{R} \to \mathbb{R}$. Show that $f$ is constant.  I've attacked this computationally in each way I can think of. I've tried to show that $\frac{\partial f}{\partial z}$ is zero; played with the Laplace equation, knowing that $h \circ v$ and $v$ must be hamonically conjugate, etc. I even considered spinning an argument based on the maximum principle.  It seems obvious, but I have not succeeded in a proof yet. Perhaps I am missing a concept?
32672
Suppose that $f$ is analytic on some region $R\in\mathbb{C}$. If Im$(f)$ = $k\cdot$Re$(f)$ for some nonzero constant $k\in\mathbb{C}$, then show that $f$ is constant on $R$.  I know that if $f'(z)=0$ for all $z$ in some region $R$, then $f(z)$ is constant. However, I'm not sure how to apply this to the question at hand. I also think the Cauchy-Riemann equations could be helpful, but again, I'm not sure how to apply them to the question.  Any guidance would be greatly appreciated. Thank you!
121140
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.
86893
Let $f: \mathbb C \to \mathbb C$ be an analytic function, such that for all $z\in\mathbb C$, $$| \operatorname{Re}(f(z))\operatorname{Im}(f(z)) | \le 1.$$   I have to show that $f$ is constant.   I don't know how to apply Liouville's theorem to the question, but I can tell it is bounded, thus the theorem should apply. Any help would be appreciated.
77075
Let $f$ be an entire function and $L$ a line in $\mathbb{C}$ such that $f(\mathbb{C})\cap L=\emptyset$. Show that $f$ is constant function.  If $f$ is not constant  then $f(\mathbb{C})$ is dense set in $\mathbb{C}$, but how can I use that line does not intersect the image set?
86744
Let $f$ be an entire function. Suppose $f(z)$ is real if and only if $z$ is real. Prove that $f$ has at most $1$ zero. How to prove? Totally I have no idea... Please give the solution in detail since I do not understand.
6711
Let $f(z)$ be an entire function s.t $|f(z)|\leq M|z|$ ; $M&gt;0$ is a constant then $f(z)$ Is polynomial of degree 1   My try $f(z)$ is entire so it can be written as a $\sum a_n z^n$  $$|\sum a_n z^n|\leq \sum |a_n||z|^n\leq M|z|$$  So $n=1$  Seems that the move $\sum |a_n||z|^n\leq M|z|$ is problematic
60820
Is there a characterization of the nonconstant entire functions $f$ that satisfy $|f(z)|=1$ for all $|z|=1$?   Clearly, $f(z)=z^n$ works for all $n$. Also, it's not difficult to show that if $f$ is such an entire function, then $f$ must vanish somewhere inside the unit disk. What else can be said about those functions?  Thank you.
86184
Let $G\subseteq\mathbb{C}$ be a domain and $f\colon G\to\mathbb{C}$ holomorphic. Show:     $$ f'=0\mbox{ in }G\Rightarrow f\mbox{ is constant} $$      Cauchy-Riemann: $$ f'(z)=f_x(z)=u_x(x,y)+i v_x(x,y), z=x+iy=0\\ \Leftrightarrow u_x(x,y)=0~\wedge v_x(x,y)=0 $$  and $$ f'(z)=-if_y(z)=-iu_y(x,y)+v_y(x,y)=0\\\Leftrightarrow u_y(x,y)=0~\wedge~v_y(x,y)=0 $$  That means: If one derivates u and v to x, one gets 0, and therfore the x-part of these functions must be constant. The ssame for the y-part of u and v. So u and v are constant and so is then f.  Is that okay?
94122
I have to solve this problem but i have no idea how to do ...Thanks!  Let be $f:\mathbb{C}\rightarrow \mathbb{C}$ a holomorphic function. Show that if $f\circ f=f$ then f is either constant or $f(z)=z \ \forall\ z\in\mathbb{C}$
13160
How can I prove the following assertion?    Let $f$ be a holomorphic function such that $|f|$ is a constant.  Then $f$ is constant.    Edit: The more elementary the proof, the better. I'm working my way through a complex analysis workbook, and by this exercise, the only substantial thing covered has been the Cauchy-Riemann equations.
48119
I want to prove that if $f$ is an analytic function for which $\Im f$ is constant, this implies that $f$ itself is constant.  So to start off, it's not given that the function is entire or anything, ruling out Liouville. Rather, I'm guessing the maximum modulus theorem will prove useful here (?).  If I were to show using that the real part of $f$ was constant, proving the statement would be easy (just take the absolute value and investigate $e^f$), but in this case I'm none the wiser from that. Should I use the same parametrization $e^f$ and apply the Cauchy-Riemann equations somewhere? Any suggestions welcome.
11977
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!
76141
Let $D := \{z \in \mathbb{C}: |z| &lt; 1\}$ and $f\colon D \rightarrow \mathbb{C}$ be holomorphic. Suppose $\lvert f(z)\rvert \leq 1$ on $D$, show that $$\frac{|f(0)| - |z|}{1 + |f(0)||z|} \leq |f(z)| \leq \frac{|f(0)| + |z|}{1 - |f(0)||z|} \ \forall z \in D.$$  I have tried using Cauchy Integral Formula to $f$ and expanding $f$ at $0$, but I have no idea why $f(0)$ appears in the inequality in the former case. Both answers and hints are appreciated and thanks in advance!