$\lim_{z \to \infty} \frac{\operatorname{Re} f(z)}{z}=0 $ implies $f$ is constant.  $f$ is entire function on complex plane  I tried to bound some derivative of $e^{f(z)}$ by cauchy integral inequality hence $e^{f(z)}$ is polynomial.  But since it has exponential term depending on radius, and denominator has $R^n$ for radius $R$ so i can not bound it.  Should I try some other way?
121165
(To clarify, I would just like a hint. Please do not give me the answer to this problem. ) The solution to the following problem has really evaded me here:  Problem: Assume that $f$ is entire and that $\lim_{z \to \infty} f(z)/z  = 0.$ Prove that $f(z)$ is constant.  My Thoughts and Work So Far: We know that proving that $f'(z) =  0$ or that for some fixed $c \in \mathbb{C}$, $f(z) = c$ for all $z\in \mathbb{C}$. My first approach was to use Liouville's Theorem; If I could could show that $f$ is bounded then I am done. Since $\lim_{z \to \infty} f(z)/z  = 0$, for all $\varepsilon &gt; 0$ there exists a $N \in \mathbb{C}$ so large that if $z \geq N$ then $|f(z)/z| \leq \varepsilon$. Thus, if $C_R$ is the circle of radius $R$ centered at the point $z$, then, as long as z is large enough by Cauchy's Inequality $$ |f'(z)| \leq \frac{1}{2\pi i} \oint_{C_R} \frac{f(\zeta)}{(\zeta - z)^2}\ d\zeta \leq \bigg | \frac{1}{2\pi i} \bigg | \oint_{C_R} \bigg | \frac{f(\zeta)}{(\zeta - z)^2} \bigg | d \zeta \leq \frac{1}{2\pi}  \frac{|\zeta|\varepsilon 2\pi R}{R^2} = \frac{|\zeta|\varepsilon}{R}.  $$  Now taking the limit as $R \to \infty$ (which to me says, "let our circle around our point z dilate to an infinite radius so that it covers all of $\mathbb{C}$)  $$|f'(z)| \leq \lim_{R \to \infty} \frac{|\zeta|\varepsilon}{R} = 0.$$  Thus $f'(z) = 0$ and $z$ was arbitrary, so $f$ must be constant.   Why I Think Im Wrong: I say $z$ was arbitrary, but really it is "any $z \geq N$" which really isn't all that arbitrary.   This is where I am stuck. Am I right, wrong, close, or totally lost? Any hints would be great.  Edit: I am very sorry but I accidentally posted this before I was done typing the problem.
52170
How do you show that if $f$ is entire and if $\text{Im} f\ge0$, then $\text{Im} f$ is constant? Is this even true in the first place?
112409
This is a question from an old prelim in complex analysis. Show that if $f$ is an analytic function such that $|f(z)| \leq \frac{1}{1-|z|}$ for $|z|&lt;1$, then $|f'(0)| \leq 4$. I tried using both Cauchy's $ML$-Estimates along with the Cauchy Integral Formula and even the Schwarz-Pick Lemma, but I still do not get the desired result. Any suggestions as to how best to proceed?
28888
The question goes like this:     Let $B=\{z\in C: |z|&lt;R\}$ and $\partial B$ is the circle of radiu $R$. If $f$ is a non-constant analytic function on $B$ such that $|f|$ is a constant on $\partial B$, then $f$ must have a zero in $B$   My attempt:  From the assumption that $f$ is a non-constant analytic function on $B$, we have the maximum of $|f|$ is achieved on $\partial B$. Then I got stucked... I believe that since $\partial B$ is constant, and $f$ is a non-constant, the value of $|f|$ must variate in the interior of $B$ so there is a zero. But how to prove it mathematically?
120428
How is the function $e^z$ the only function that is entire and has the property $f'(z) = f(z)$ and that $f(0) = 1$ but how would you exactly prove it?  where $e^z = e^x(\cos y + i\sin y)$
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.
122647
The problem is:  Suppose that $f(z)$ is continuous on a domain $D$ that contains the unit circle, and that $f(z)$ satisfies: $$\vert f(e^{i\theta})\vert \leq M \; \forall \theta \, \in [0, 2\pi )$$ and $$\bigg \vert \int _{\vert z \vert=1} f(z) \text{d}z \bigg \vert =2\pi M $$ show that there exists a $c\in \mathbb{C}$ such that $f(z)=c \bar{z}$ for all $\vert z\vert=1$.  I've tried lots of different approaches such as defining a function $g(z)=\frac{f(z)}{\bar{z}}$ and trying to show that $g'(z)=0$ so that $g$ is constant on the circle, or trying to use some bounding theorems for analytic functions to show that $g$ is constant but this requires $g$ to be analytic. Some other methods were using the Cauchy-Riemann equations but I couldn't find any reason why $g$ should be analytic.  Any help would be greatly appreciated, thanks!
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?
83190
Let $f$ be an entire function s.t $f(\mathbb{C}) \cap B_R(z_0) = \varnothing$ for some $z_0$ and some $R$. Then $f$ is constant.  I guess since the image of the whole plane isn't dense then $f$ doesn't have an essential singularity at infinity. Now I have to exclude it has a polar singularity at infinity..
110733
$f$ is an entire function with $\operatorname{Im}f \geq 0$. Then which of the followings are true: 1) $f$ is constant. 2) $\operatorname{Re}f$ is constant. 3) $f = 0$. 4) $f'$ is a non-zero constant.  That (3) &amp; (4) are wrong can be shown by using $f(z) = i$. But I'm clueless about the remaining two options.
48580
Let $f$ be an entire function such that $\mid f(z) \mid \leq A \mid z \mid$ for all $z$, where $A$ is a fixed number. Show that $f(z)=a_1z$, where $a_1$ is a complex constant.
59610
Let $\Bbb D$ be the open unit disc and $f:\Bbb D\to\Bbb C$ be an analytic function such that $|f(z)|\le |f(z^2)|$, for all $z\in\Bbb D$. Prove that $f$ is constant.   Here is my proof:   For any $0&lt;r&lt;1$, consider the closed disc $B=\{z\in \Bbb D:|z|\le r\}$. Then the given condition implies that $|f(z)|\le |f(z^{2^n})|$, for all $n\in\Bbb N$ and for $z\in B$. Since for $z\in B$, $|z^{2^n}|\le r^{2^n}\to 0\implies z^{2^n}\to 0$. So by continuity, $|f(z)|\le |f(0)|$, for all $z\in B$. So by the Maximum modulus theorem, $f$ is constant on $B$. By the uniqueness theorem, $f$ is constant on $\Bbb D$. Is this correct? Let me know if there is any fallacy.
108112
We begin by assuming that a non-constant complex polynomial $$p(z)=a_n z^n+a_{n-1}z^{n-1}+...+a_1 z+a_0,$$ where $a_0 \neq 0$, doesn't vanish in $\mathbb{C}$. Therefore $1/p$ is also an analytical function everywhere. Note that $(p(z)-a_0)/z$ is a polynomial and therefore, due to Cauchy's theorem, the following integral must be zero.  $$\int_{S_r^1}\frac{(p(z)-a_0)/z}{p(z)}dz=\int_{S_r^1}\frac{a_nz^{n-1}+a_{n-1}z^{n-2}+...+a_1}{a_nz^n+a_{n-1}z^{n-1}+...+a_1 z+a_0}dz=0.$$  On the other hand  $$\int_{S_r^1}\frac{a_nz^{n-1}+a_{n-1}z^{n-2}+...+a_1}{a_nz^n+a_{n-1}z^{n-1}+...+a_1 z+a_0}dz$$  $$=\int_{0}^{2\pi}\frac{a_n(re^{i\theta})^{n-1}+a_{n-1}(re^{i\theta})^{n-2}+...+a_1}{a_n(re^{\theta})^n+a_{n-1}(re^{i\theta})^{n-1}+...+a_1 (re^{i\theta})+a_0}(ire^{i\theta})d\theta$$  $$=\int_{0}^{2\pi}i d\theta - \int_{0}^{2\pi}\frac{a_0 i}{a_n(re^{i\theta})^n+a_{n-1}(re^{i\theta})^{n-1}+...+a_1 (re^{i\theta})+a_0}d\theta$$ $$=2\pi i -\int_{0}^{2\pi}\frac{a_0i}{p(re^{i\theta})}d\theta.$$  Let $r \rightarrow \infty$ and the integral approaches value $2\pi i$ - a contradiction. Therefore $1/p$ can't be entire and $p$ must vanish somewhere in the complex plane.  EDIT: I have changed this post and deleted the idea that the integral $\int_{\partial D_r}\frac{(p(z)-a_0)/z}{p(z)}dz=0$ could be evaluated with the fundamental theorem of calculus. Now, as it stands, this proof is essentially the same as the one that Mr. José Carlos Santos points out in his comment.
38173
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!
40890
Let $f$ be holomorphic in the unit disk $D(0,1)$, and $f(0)=0$. Suppose $|\Re f(z)|\leq 1$ for all $|z|&lt;1$. Show that  $$|\Re f(z)|\leq\frac{4}{\pi}\arctan|z|;$$ $$|\Im f(z)|\leq \frac{2}{\pi}\ln\frac{1+|z|}{1-|z|}.$$  What I could do is just to observe that $$\frac{e^{i\frac{\pi}{2}f(z)}-1}{e^{i\frac{\pi}{2}f(z)}+1}:D(0,1)\to D(0,1)$$ with $0$ being fixed. And Schwarz lemma applies. However, I could not get any information on the real and imaginary part of $f$.
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.