Find all entire functions $f(z)$ such that $|f(z)|=1$ for $|z|=1$  Hint: First show that $f(z)$ is a polynomial.  Clearly one can not use Cauchy Estimates to prove that $f(z)$ is a polynomial, the other way is to prove that $f(z)$ has a pole at infinity which I am not sure how to prove that, are there any other ways to prove that a given entire function is a polynomial ?  Thank you !
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.
67599
I can't solve this problem:  Suppose $f$ is polynomial function of even degree $n$ with always $f\geq0$.  Prove that $f+f'+f''+\cdots+f^{(n)}\geq 0$.
37155
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.
10817
The problem is as follows:     Suppose $f$ entire satisfying   $$ |f(z)| \leq A + B |z|^{3/2} $$   for some fixed $A,B &gt; 0$. Prove that $f$ is a linear polynomial.   I know I want to reduce it to a function where I can use a Cauchy bound, but I'm not sure how.
61251
Let $n\in\mathbf{N}$ be fixed, and $f$ entire and $|f^{-1}(\left\lbrace w\right\rbrace)|\leq n$ for every $w\in\mathbf{C}$. Then $f$ is a polynomial of degree at most $n$.    I try to prove this statement, and I think one can prove it as follows: consider $f(1/z)$. $0$ cannot be an essential singularity of $f(1/z)$, for the big Picard theorem would imply that on any neighborhood of $0$ $f(1/z)$ takes on all possible complex values (with at most one exception) infinitely often, but this is contrary to $|f^{-1}(\left\lbrace w\right\rbrace)|\leq n$. Then $f(1/z)$ has a pole of order $k$ say, and since $f$ is holomorphic it is a polynomial of degree $k$ (for its principal part vanishes). By the fundamental theorem of algebra, $k\leq n$.      Is it possible here to avoid using Picard's theorem?
32452
Does there exists an entire function such that for each $|z|&gt;C$ for some constant $C$, $|f(z)|=|z|+1$?  Can I assume that since $f$ has a pole in infinity, $Imf$ or $Ref$ must have a pole at infinity too? If so, I can use this fact to define an exponent function of $Imf$ or $Ref$ (the one that goes to infinity) or their minus, and show that function is a bounded entire function, and therefore constant. Which gives a contradiction to the conditions about $f$.
66307
Suppose $f:\mathbb{C}\rightarrow\mathbb{C}$ is an entire function. Let $g:\mathbb{C}\rightarrow\mathbb{C}$ be an entire function, which has no zeros.    I have shown that $\vert f(z) \vert \leq \vert g(z) \vert$ for all $z\in\mathbb{C}$ implies $f(z)=Cg(z)$ for some constant $C\in\mathbb{C}.$  I have to decide if this also holds if $g$ is allowed to have zeros. Help?
24447
I know I'm going to use Liouville's Theorem, but my main question is why is $1/|f(z)|$ entire as well if $f$ is entire? Is this just a basic property: if $f$ is entire, then $1/f$ is entire? Thanks for the help.
8689
There is an exercise in Function Theory of One Complex Variable by Greene &amp; Krantz that is very similar to a Proposition in the book, but I am having trouble getting to the conclusion.      Let $f : \mathbb{C} \to \mathbb{C}$ be a polynomial. Suppose further that $$\frac{\partial f}{\partial z} = 0$$   and $$\frac{\partial f}{\partial \bar z} = 0$$    for all $z \in \mathbb{C}$    Prove that $ f \equiv$ constant.    Now, part of the proposition (1.3.2) proves that if $p$ is a polynomial, $$p(z, \bar z) = \sum a_{lm}z^l \bar z^m$$ with $\frac{\partial f}{\partial \bar z} = 0$, then $\frac{\partial ^{l + m}}{\partial z^l \partial \bar z^m}p$ evaluated at zero is $l!m!a_{lm}$.  Is this the constant that $f$ equals? It doesn't seem like I've used the hypothesis, or reached the conclusion!  Maybe I should use the definition of the partial... $$\frac{\partial f}{\partial z} := 1/2 (\frac{\partial }{\partial z} - i \cdot \frac{\partial }{\partial z})f$$  Any hints would be greatly appreciated.
123396
Let $p(z)=z^n+a_{n-1}z^{n-1}+a_{n-2}z^{n-2}+...+a_0$ be a complex polynomial satisfying $|p(z)| \leq 1$ for $|z|\leq 1$, then prove that $p(z)=z^n$  I tried with Cauchy integral formula for $n$th derivative to get $$n!=\frac{n!}{2\pi}\int_0^{2\pi}p(e^{i\theta})e^{-in\theta}d\theta$$ implies $$1\leq \frac{1}{2\pi}\int_0^{2\pi}|p(e^{i\theta})e^{-in\theta}|d\theta$$ and thereby $|p(e^{i\theta})e^{-in\theta}|=1$. (since $|p(e^{i\theta})e^{-in\theta}|$ is a non negative real continuous function $\leq 1$)  ie., $|p(z)|=|z^n|$ for $z=e^{i\theta}$, i stucked here and how can we procced?
97116
I came across this old exam problem.     Suppose $f(z)$ is entire and $|f(z)|&lt; e^{|z|}$, and also $f(z)=f(z+1)$. Show $f(z)$ is a constant.   I am able to show the singularity at infinity is not a pole.   But I can't rule out it being an essential singularity.
30425
I have problems proving the following result:     Each $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $\forall a,b \in \mathbb{R} \ : \ f_a(y) := f(a,y), \ f_b(x) := f(x,b) $ are polynomials is a polynomial with two variables.   If I consider $f$ as a function of $x$, then its derivative  $f'(x) = \frac{\partial f}{\partial b}(x)$. Similarly if we treat $f$ as a function of $y$.   I assume that $f_a(y) = \frac{\partial f}{\partial a} (y) $ and $f_b(x)=\frac{\partial f}{\partial b}(x)$.  But I am not sure if we can assume that, because  the degree of the derivative should be smaller than the degree of the original function (and it isn't).   Actually, I'm not even sure if what I'm trying to prove is true, because in the original formulation of the problems there is written $f_a(y) := (a,y), \ f_b(x):=(x,b)$. But that didn't make sense.  Could you help me here?  Thank you.
50094
Let $p(z)$ be a monic polynomial of degree $n$.   Prove that $\max\limits_{|z|=1}|p(z)|\geq 1$ and that equlity holds if and only if $p(z)=z^n$.  I first observed that $p^{(n)}(z)=n!$. Therefore by Cauchy's formula  $$n!=p^{(n)}(0)=\frac{n!}{2\pi i}\int\limits_{|z|=1}\frac{p(z)}{(z-0)^{n+1}}dz\implies 1\leq\frac{1}{2\pi}\int\limits_{|z|=1}|p(z)|dz\leq\max\limits_{|z|=1}|p(z)|$$  Now, clearly if $p(z)=z^n$ then $\max\limits_{|z|=1}|p(z)|= 1$, but what about the converse?
94686
Suppose that $P(x)$ is a polynomial of degree $n$ such that $$P(k)=\dfrac{k}{k+1}$$ for $k=0,1,\ldots,n$. Find the value of $P(n+1)$.    I could not relate this question with this one , so maybe it is not a duplicate  I have absolutely no idea, how the function became a polynomial and how to approach it. Please help!
51351
So the problem states that if $f(z)$ is entire, and satisfies the relation $f(z+i) = f(z)$ and $f(z+1) = f(z)$, show that $f(z)$ is constant. So I was thinking that since any point in $\mathbb{C}$ can be written as $\alpha * 1 + \beta * i $ we can say that $f(z + z_0) = f(z) $ in which case it is constant, but I'm having trouble breaking down the steps and using the fact that $f$ is entire, which makes me feel like I'm missing something. What should I review to figure this out?
88074
Find all entire functions such that $|f(z)|\leq |\sin(z)|$ for all $z\in\mathbb C$.  Half an hour ago I asked the same question for $|f(z)|\leq |z^2-1|$  but this time I cannot say something like $$|f(z)|\leq M|z|^m$$ so I don't know how to start. The problem is that $$ \sin(z) = \frac{e^{iz}-e^{-iz}}{2i} $$ gets really large if $z$ gets large.
68706
Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be a entire function. Suppose that there are $M$, $r&gt;0$ and $n\in \mathbb{N}$ such that $\left|f(z)\right|&lt;M\left|z\right|^n$ for all $z \in \mathbb{C}$ with $\left|z\right|\geq r$.  Show that $f$  polynomial of degree at most $n$.  Remark: I tried to follow the proof of Liuville's Theorem but I have complications with the condition  $\left|z\right|\geq r$..