Let $f$ be a meromorphic function on $\mathbb C$ such that $|f(z)|\ge|z|$ at each $z$ where $f$ is holomorphic. Then which of the following is/are true?   The hypothesis are contradictory. Such an $f$ is entire. There is a unique $f$ satisfying the given conditions. There is an $A\in\mathbb C$ with $|A|\ge1$ such that $f(z)=Az$ for each $z\in\mathbb C.$   I think 2 and 4 are correct. But I can't proceed rigorously. Neither do the discussion  is complete.
121878
Let $f$ be a meromorphic function on $\mathbb{C}$ such that $|f(z)|\geq |z|$ at each $z$, where $f$ is holomorphic. Then, which of the following is/are true.   The hypothesis are contradictory, so no such $f$ exists. Such an $f$ is entire. There is a unique $f$ satisfying the given conditions. There is an $A\in \mathbb{C}$ with $|A|\geq 1$ such that $f(z)=Az$ for each $z\in \mathbb{C}$   I think $1$ and $3$ are false but not sure. And what about $2$ and $4$? I am unable to find suitable examples. Please help me here.
49608
Let $f,g$ be entire functions such that $g \not\equiv 0.$ If $fg\equiv0$ in $\mathbb{C},$  could anyone advise me how to show $f \equiv0$ in $\mathbb{C} \ ?$  Thank you.
90731
I am preparing for qualifying exams, and this is a question from the Penn State Qualifying Exam for Fall 2015. It is stated as follows  Let $\epsilon &gt; 0$ and let $f$ be holomorphic (analytic) on the disk $S = \{z \in \mathbb{C} \ | \ |z| &lt; 1 + \epsilon \}$. Suppose that $f(z)$ is real valued whenever $|z| = 1$. Prove that $f$ is constant.  I have tried a few things in showing this to be true, but I keep finding holes in my logic. My largest issue is that the portion of the domain in which $f$ is real valued is not open, and so I'm not able to use many of the theorems I otherwise feel would be helpful (Open Mapping Theorem, Cauchy-Riemann Equations, Maximum Modulus, etc.).   Most of my attacks towards this problem have centered around showing things for the unit disk $\mathbb{D}$ instead of $S$, because I figure if I can show that $f$ is constant on $\mathbb{D}$, then I can use the Identity Theorem to extend it to $S$. However, since I don't know anything about the specific values that $f$ obtains, I can't use Schwarz' Lemma either.  I played with the idea, also, of suggesting that, if we consider the closure of $\mathbb{D}$, then $f(\partial{\mathbb{D}}) \in \mathbb{R}$. Since $\partial\mathbb{D}$ is closed and bounded and $f$ is analytic, it's image should also be bounded. That would put, for some $M \in \mathbb{R}$, $f(\partial{\mathbb{D}}) \in [-M,M]$, which is bounded and obtains a maximum (since $f$ is continuous and real valued here). I know that this set isn't open, but there is a Corollary in my text (Complex Analysis - Freitag) that says the following:  "If $K$ is a compact subset of the domain $D$ and $f:D \rightarrow \mathbb{C}$ is analytic, then the restriction of $f|K$ being a continuous function has a maximal modulus on $K$. By the Maximum Modulus Principle, we can moreover affirm that the maximal modulus value is necessarily taken on the boundary of $\mathbb{D}$."  The proof of that statement follows from the Open Mapping Theorem. Would it be appropriate to use that in this case, then? If $K = \partial \mathbb{D}$, even though it's technically not an open set, can I say that $f$ obtains its maximum on $\mathbb{D}$, state that it's constant, and then extend this to $S$ using the Identity Theorem? My issue here is that the constant itself isn't actually in $\mathbb{D}$,  I appreciate any help for this problem (or any tips for showing that complex functions will be constant, as these types of problems show up often).  Edit: My function $f$ is not necessarily entire, so Liouville's Theorem does not apply
56325
I came across the following problem that says:  Let $f$ be an analytic function defined on $\mathbb{D}=\{z\in\mathbb{C}:|z|&lt;1\}$ such that the range of $f$ is contained in the set $\mathbb{C}\setminus (-\infty,0]$. Then  $f$ is necessarily a constant function. there exists an analytic function $g$ on $\mathbb{D}$ such that $g(x)$ is a square root of $f(z)$ for each $z\in\mathbb{D}$. there exists an analytic function $g$ on $\mathbb{D}$ such that $\operatorname{Re}g(z)\geq 0$ and $g(z)$ is a square root of $f(z)$ for each $z\in\mathbb{D}$. there exists an analytic function $g$ on $\mathbb{D}$ such that $\operatorname{Re}g(z)\leq 0$ and $g(z)$ is a square root of $f(z)$ for each $z\in\mathbb{D}$.   I have to determine which options are correct. Can someone help in the right direction? Thanks in advance for your time.
122605
Let $f$ be holomorphic on $\overline{N(0,1)}$. Suppose $|f(z)| \leq 1$ for every $|z| = 1$. Without using fixed point theorem, show that there exists $z \in \overline{N(0,1)}$ such that $f(z) = z$.  My attempt:  Using Rouché's Theorem, I managed to show that if $|f(z)| &lt; 1$ for every $|z| = 1$, then there is a unique $z$ such that $f(z) = z$.  However, I am not sure how I should proceed if $|f(z)| = 1$ for some $|z| = 1$.
38912
I am struggling with the following exercise:  Let $G \subseteq \mathbb{C}$ be a domain and let $K \subset G$ be compact with non-empty interior. Let further be $f: G \rightarrow \mathbb{C}$ a holomorphic and non-constant function with $\lvert f(z) \rvert = C$ for all $z \in \partial K$ and some constant $C \in \mathbb{R}$. Show that $f$ must have zeroes in $K$.  My idea is this: I think it should be true that the the maximum value of $f$ on $K$ has to be on the boundary of $f(K)$ and is due to our assumption equal to $C$. So the disk centered at $0$ with radius $C$ should contain $K$. All the values of $f$ in the interior of this disk have to be smaller than $C$ , however. I guess that we may now conclude that $f$ has to be equal to $0$ for some point in $K$. Could you tell me if I am on the right track?
99254
We know that if $f(z)$ and $g(z)$ are entire functions such that $g(z)\ne0$ and $|f(z)|\le|g(z)|$ for all $z\in\mathbb{C} $ then by Liouville's theorem $$ f=ag$$ for some constant $a\in \mathbb{C} $ .  Now my question is this that similar to above argument, if $f(z)$ and $g(z)$ are Meromorphic functions such $|f(z)|\le|g(z)|$ for all $z\in\mathbb{C} $ then I want to show $$ f=ag$$ for some constant $a\in \mathbb{C} $ .  I am thinking in this way that because because poles and zeros of Meromorphic functions are isolated , by the Riemann's theorem on removable singularities and By using analytic continuation to eliminate removable singularities also we can have $$ f=ag$$ for some constant $a\in \mathbb{C} $ .  Is this true way?
18801
Suppose $f(z)$ is holomorphic in $|z|&lt; 1$ and $\operatorname{Re}f(z) &gt; 0$ in this region. Furthermore, suppose $f(0) = a &gt; 0$. Then why does $|f'(0)| \leq 2a$?  The hint I have is to consider $g(z) = \frac{f(z) - a}{f(z) + a}$ and then show $|g'(0)| \leq 1$. If I do have $|g'(0)| \leq 1$, then the result follows as $|f'(0)| = 2a|g'(0)| \leq 2a$, but I have no idea on how to show $|g'(0)| \leq 1$.
83217
Let $G\subset \mathbb{C}$ be a non-empty open set and $f$ be a function holomorphic on $G$. Let $g: G\times G\to \mathbb{C}$ be a function defined as $$g(z,w)= \begin{cases}        \frac{f(z)-f(w)}{z-w}, &amp; z\ne w \\       f'(z), &amp; z=w     \end{cases}$$  I need to prove that $g$ is continuous on $G\times G$.  My approach:  Let $\varepsilon &gt; 0$. Then $\exists \delta_1, \delta_2&gt;0$ such that $\| (z,w)-(z_0,w_0) \|^2\le|z-z_0|+|w-w_0|&lt;\delta_1+\delta_2$ implies that $|f(z)-f(z_0)|&lt;\epsilon$, $|f(w)-f(w_0)|&lt;\epsilon$, since $f$ is continuous on $G$.  Now, $\| g(z,w)-g(z_0,w_0) \|=\left| \frac{f(z)-f(w)}{z-w} - \frac{f(z_0)-f(w_0)}{z_0-w_0} \right|\le \left| \frac{f(z)-f(w)}{z-w}\right| + \left|\frac{f(z_0)-f(w_0)}{z_0-w_0} \right|$  Since $f$ is holomorphic on $G$, it is continuous on $G$, thus $\exists \delta'&gt;0$ such that $| {f(z)-f(w)}|&lt;\varepsilon'|z-w|&lt;\varepsilon'\delta'$ whenever $|z-w|&lt;\delta'$.  And here's where I'm stuck, because it's not clear how to deal with $\left|\frac{f(z_0)-f(w_0)}{z_0-w_0} \right|$, because $z_0, w_0$ are fixed points, so we can't just make them approach each other. Of course, my first thought was to somehow rearrange the inequality in such a way as to obtain $\left| \frac{f(z)-f(z_0)}{z-z_0}\right| + \left|\frac{f(w)-f(w_0)}{w-w_0} \right|$, but I don't see how to do so.  Your help would be appreciated.
5224
I recently came across the result that if $g$ is holomorphic in a neighbourhood of a point $z_0$ and  $$\sum_{n=0}^{\infty }{g^{(n)}(z_0)} $$ converges then $g $ is an entire function. Now, my doubt is if the above series is absolutely convergent then can we say something more about $g$, like say $g$ is a polynomial? I could not find any non-polynomial  entire function( I tried with functions like $e^z$, $\sin z $, et al), whose successive derivatives converge absolutely as series. Any help please?
48559
Let $\mathbb{D}=\{ z\in\mathbb{C}:|z|&lt;1\}$. Which of the following are correct?   There exist a holomorphic function $f:\mathbb{D}\rightarrow\mathbb{D}$ with $f(0)=0$ and $f'(0)=2$ There exist a holomorphic function $f:\mathbb{D}\rightarrow\mathbb{D}$ with $f(3/4)=3/4$ and $f'(2/3)=3/4$ There exist a holomorphic function $f:\mathbb{D}\rightarrow\mathbb{D}$ with $f(3/4)=-3/4$ and $f'(3/4)=-3/4$ There exist a holomorphic function $f:\mathbb{D}\rightarrow\mathbb{D}$ with $f(1/2)=-1/2$ and $f'(1/4)=1.$   Option one is not true by Schwartz-lemma. But i don't know how to think about other options. I  don't know which theorem or result gives condition of existence of this type of holomorphic function. Please help me to solve this problem. If possible please solve this problem. Thanks in advance.
31277
Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for all $z \in \mathbb{C}$, Prove that $f(z)=cg(z)$.   My try :  I consider $h(z)=\frac{f(z)}{g(z)}$. If I prove that $h(z)$ is entire, using the fact that $|h(z)|\leq 1$, the result follows immediately from Liouville's theorem.  To prove that $h(z)$ is entire, I have to prove that $h(z)$ has removable singularities at possible zeros of $g(z)$.  Suppose $g(z_0)=0 \Rightarrow |g(z_0)|=0 \Rightarrow |f(z_0)|=0$, BUT what if $z_0$ be a root of of order $k$ of $g(z)$ (i.e. $f^{(n)}(z_0) = 0 $ for all $ 0 \leq n \leq k-1$) and be a zero of order $m$ of $f(z)$ and $k&gt;m$ ?  And if that's not the case, how to rigorously show that $z_0$ is a removable singularity ?  Thank you all in advance.
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.
72847
Let $G$ be a convex area and $f:G \rightarrow \mathbb C$ be a holomorphic function (with continious derivative). Prove: If $|f'(z)-1|\lt 1,$ then f is injective on $G$  Could anyone give me a hint ? I dont know how to tackle this problem
50342
Let $\mathbb{D}=\{z \in \mathbb{C} \ | \ |z|&lt;1 \} $ be the open unit disk in the complex plane. I would like to see explicit examples of the following phenomena:   a holomorphic function $f$ on $\mathbb{D}$ which extends continuously to the boundary but has no holomorphic extension beyond any boundary point (i.e. on sets of the form $\mathbb{D} \cup B(z_0,r)$ for some $r&gt;0$ and $z_0 \in \partial \mathbb{D}$) a holomorphic function $f$ on $\mathbb{D}$ which is bounded on $\mathbb{D}$ but has no holomorphic extension beyond any boundary point (i.e. on sets of the form $\mathbb{D} \cup B(z_0,r)$ for some $r&gt;0$ and $z_0 \in \partial \mathbb{D}$)   Such functions should exist according to this  to a previous . Thanks for any reference/advice.
119136
Let $U \subset \mathbb{C}$ be an open set such that {$ z\in\mathbb{C};|z|\leq2$}$ \subset U$ and $f:U\rightarrow\mathbb{C}$ a holomorphic function.   Show that there exists infinitely many natural numbers $n \in \mathbb{N}$ such that $f(\frac{1}{n})\neq(\frac{1}{n+1})$  So i have been told to start by comparing with the function $g(z)=(\frac{z}{z+1})$  however i am still not getting very far with this. Would i have to prove by a contradiction?   Completely stuck and unsure, some help would be fantastic, many thanks!
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 $f$.