Let $f$ be a holomorphic, non-constant function on $D=\{0&lt;|z|&lt;10\}$. It is given that for every $n \in \mathbb{N}$, $|f(\frac{1}{n})| \le \frac{1}{n!}$.  Show that $f$ has an essential singularity at 0.  I have no idea how to even approach this. Why is it not possible for $f$ to have a pole at 0?  Note. In the original question it was written $n \in \mathbb{C}$ but I considered this a typo.  Thanks in advance!
122566
Let $f$ be holomorphic non-constant on $D=\left\{ 0&lt;\left|z\right|&lt;10\right\}$ . Given that for all $n\in\mathbb{N}$: $\left|f\left(\frac{1}{n}\right)\right|\leq\frac{1}{n!}$ prove that $f$ has an essential singularity at $0$. Find an example of $f$ satisfying the condition.   My idea was to assume for contradiction the singularity is not essential, which means that either the limit $z\to0$ of $f$ exists or of $1/f$, and from $\left|f\left(\frac{1}{n}\right)\right|\leq\frac{1}{n!}$ we get that the limit of $f$ exists and is $0$.  I then define $h(z)=f(z)$ for $z\neq0$ and $h(0)=0$, which is holomorphic on the entire disk, and derive a contradiction from there and the uniqueness theorem. But then the suitable holomorphic function is the Gamma function, which we have not really discussed in class and I'm not sure about it's properties.  Any ways of solving this which avoids using the gamma function?
36023
I have to solve following problem     If $\Re (f)$ is bounded above or below for a function $f$ holomorphic on $\mathbb{C}$ then $f$ is constant.   My attempt: If there is $M$ such that $\Re(f) \le M$, then $\|e^{f}\|=e^{\Re(f)}\le e^{M}$. From Liouville's theorem the entire function $e^f$ is constant, that is $0=(e^f)'=f' e^f$. This means $f' =0$, so $f$ is constant. If $\Re(f)$ is bounded below, we consider $e^{-f}$ and proceed the same way.  Am I correct? Is there a solution using maximum modulus principle?
108941
Let $z_0$ be a non-removable isolated singularity of $f$. Show that $z_0$ is then an essential singularity of $\exp(f)$.      Hello, unfortunately I do not know how to proof that. To my opinion one has to consider two cases:   $z_0$ is a pole of order $k$ of $f$. $z_0$ is an essential singularity of $f$.
41450
Suppose $f$ is continuous on the closed unit disc $\overline{\mathbb{D}}$ and is holomorphic on the open unit disc $\mathbb{D}$. Must $f$ have finitely many zeros in $\mathbb{D}$? I know that this is true if $f$ is holomorphic in $\overline{\mathbb{D}}$ (by compactness of the closed unit disc), but I'm not sure of what happens when I just consider $\mathbb{D}$.
96552
Does there exist a holomorphic function $f$ defined over $D = \{ z : |z| &lt; 1 \}$  such that $|f| \rightarrow \infty$ when $|z| \rightarrow 1$?  My approach:  If such an $f$ exists, then for a given $\varepsilon &gt; 0$, there exists an $M &gt; 0$ such that $|f(z)|&gt;M$ if $1-\varepsilon &lt; |z| &lt; 1$  Since $f$ has no zeros in that set, I can consider $g = 1/f$, I would like to apply Maximum Modulus Principle and get a contradiction, but actually I am not sure I can.
46389
Let $f$ be a holomorphic function defined on $\Bbb C$ such that $f^{-1}(\Bbb R)=\Bbb R$. Prove that there exists $a,b \in \Bbb R, \; a \neq 0 \;$ such that $f(z)=az+b \;$ for every $z \in \Bbb C$.  I was given a hint: define the function $g(z)=\frac{f(z)-f(0)}{z}$. such $g$ is also holomorphic in $\Bbb C$ since the only singularity is in $0$, and it is a removable singularity since we can define $g(0)=f'(0)$. Next I thought about proving $g$ is bounded, and thus constant (using Liouville's theorem), but pretty much out of ideas. How can I finish the proof/solve in another way?
69053
Show there is no analytic function in $\{|z|&lt;1\}$ such that $f(0)=1$ and $|f(z)|\ge 1+|z|^2$. I guess something went wrong along the way and I would like your assistance.   $f$ does'nt get a maximum in the domain, because if it did, it would be constant in the domain, meaning $|f(z)|=1\ge 1+|z|^2$ which doesn't hold for all z. Since $f$ isn't zero in the domain, ${1\over f}$ is analytic in the domain. but $|{1\over f(z)}|\le {1\over 1+|z|^2}\le 1={1\over f(0)}$, which indicates a local maximum making ${1\over f}$ constant, contradiction the fact the $f$ isn't. What am I really lacking here?
59973
I have a question in my assignment :     If $f$ is a nonconstant entire function such that $|f(z)|\geq M|z|^n$   for $|z|\geq R$ for some $n\in\mathbb N$ and some $M$ and $R$ in   $(0,\infty)$ show that $f$ is a polynomial of degree atleast $n$.   Now , I defined a function    $\ g(z) = \frac {1}{f(z)}\ $ such that $\ |g(z)| \le \frac{1}{M{|z|}^n}.$  Now , by using the cauchy inequality  $$|g^{(n)}(z)| \le  \frac{n!}{R^n |z|^nM}.$$    Considering that $ g(z) $ is an analytic function , it has a radius of convergence $ \infty $  $ \implies\  g^{(n)}(z) = 0.$    But if we go by this approach , then $ g^{(n)}(z) = 0 \ $ for any n . Also how can we be so sure that $ f(z) \neq 0 $ for any z ?  Is my reasoning correct or is there some other way to solve it ?
119652
Let $D \subset \mathbb{C}$ be a bounded domain and $f$ a function holomorphic in $D$ and continuous in its closure. Suppose that $|f(z)|$ is constant on the boundary of $D$ and that $f$ does not have zeroes in $D$. Prove that $f$ is a constant function.  I think that if I can prove that $f$ attains both its maximum and minimum values on the boundary, then the result follows from the maximum principle. But I've been unable to show this. Is this the right way to approach this problem? If so, how do I show this result? Thanks in advance!
74674
This is an exercise from Alhfors Complex Analysis book- to show that an analytic function with a nonessential singularity at infinity must be a polynomial. It seems like it should probably be pretty straight forward, but I must be missing something. If it has a removable singularity at infinity than it extends to an analytic function on the Riemann sphere, and so must be constant by Liouville's theorem. What if there is a pole at infinity though? This was homework some time ago, and  I never finished it :/ but have been thinking about it again recently. Thanks :)
96136
I've never really had to think about this problem before, and to be honest complex analysis isn't my strongest suit, so when I suddenly needed to know where $z^z$ is holomorphic, I didn't know where to begin to start proving my hunches. My guess is that it's holomorphic away from $0$.  More generally, given two entire functions $f$ and $g$, where is $f(z)^{g(z)}$ holomorphic? Again, my hunch is that it is away from the zeroes of $f$. This isn't a time sensitive question, so any help you can provide would be appreciated, from proofs to thoughts to hints!
105364
$\textbf{Problem.}$ Suppose $f$ is holomorphic on the unit disk $\mathbb{D}$. Show there are points $a_n\in \mathbb{D}$, $a\in \partial \mathbb{D}$, and $b\in \mathbb{C}$ such that $a_n\to a$ and $f(a_n)\to b$, as $n\to \infty.$   $\textbf{Attempt at a Solution:}$ First, if $f$ has an essential singularity at some point $a\in \partial\mathbb{D}$, I think the statement follows easily.  To contradiction suppose that the statement is false. That is, suppose that $f(a_n)\to \infty$ for all sequences $\{a_n\}$ in $\mathbb{D}$ where $a_n\to a$, $a\in \partial\mathbb{D}.$  Then $$\frac{1}{f(z)}\to 0, \hspace{1cm} \mbox{as} \hspace{1cm}|z|\to 1.$$  $\textbf{ Case 1}$: If $f$ is nowhere vanishing on $\mathbb{D}$, then $1/f(z)$ would be holomorphic in $\mathbb{D}$ and can be extended continuously on $\bar{\mathbb{D}}$. Then, by the maximum principle this would imply that $1/f(z)\equiv 0$ on all of $\mathbb{D}$. But this contradicts the fact that $f$ is holomorphic (non-constant) in $\mathbb{D}.$  $\textbf{Case 2}$: Suppose $f$ has a finite number of zeros in $\mathbb{D}$. That is, let $\{b_1,...,b_n\}$ be points of $\mathbb{D}$ such that $f(b_i)=0.$   Let $r=\min \{|z-b_i| \forall z\in \partial\mathbb{D}\}$, and consider the annulus $$A=\{z\in\mathbb{C}: r&lt;r_1&lt;|z|&lt;1\}$$  Then, $1/f(z)$ is holomorphic on $A$, and can be continuously extended to $\bar A$. (Note that, $1/f(z)$ is already holomorphic for $|z|=r_1$.  Let $$g(z)=\frac{1}{f(z)}\frac{1}{f(r_1/z)}.$$ Then $g(z)=0$ for all $z\in \partial \bar{A}$. By the maximum principle it follows that $g(z)\equiv 0$ on $A$, and by the identity theorem it follows that $1/f(z)\equiv 0$ on $A$. But this, again, contradicts the fact that $f$ is holomorphic on $\mathbb{D}$ and thus on $A$ as well.   $\textbf{Case 3:}$ This is the case where i'm not so sure what i'm doing is correct.   Suppose, now, that $f$ has an infinite number of zeros in $\mathbb{D}$. That is, let $E=\{b_i\in \mathbb{D}: g(b_i)=0\}$. Then, $E$ cannot have an accumulation point in $\mathbb{D}.$  My idea was to take neighborhoods, $\mathbb{B}_{r_i}(b_i)$, and consider $$g(z)=\frac{1}{f(z)}\prod_{w\in A_w}\frac{1}{f(z/w)},$$ where $A_w=\{w\in \mathbb{D}: |w-b_i|=r_i\},$ for all $z\in \overline{\mathbb{D}\setminus\bigcup \mathbb{B}_{r_i}(b_i)}.$  Then, I would argue similarly as in Case 2.  There is probably a much shorter method.  I'd appreciate any thoughts or suggestions!!
89502
It is true that given a domain $D$ and a function $f$ that is holomorphic and not vanishing in $D$, then a branch $\varphi$ (with $\varphi$ continuous) of $f^{1/n}$ in $D$ is holomorphic in $D$ ().   My question is, does the statement remain true if we remove the non-vanishing condition? I haven't been able to find a counterexample or a proof on that more general case.
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?
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.
50367
At $z=0$ the  function $f(z)=\exp({z\over 1-\cos z})$ has    $1$. A removable singualrity  $2$. A pole  $3$. An essential singularity  $4$. Laurent series around $z=0$ has infinitely man positive and negative power of $z$  I see $\lim_{z\to\infty}f(z)=\infty\cup -\infty$ if you approach to $0$ from left and right side of the real line, so $f$ can not be bounded near $0$ so it has $f$ essential singularity?
72288
I have some trouble solving this problem: Does there exist a holomorphic function $f$ on $\mathbb C\setminus \{0\}$ such that $$|f(z)|\geq \frac{1}{\sqrt{|z|}}$$ for all $z\in\mathbb C \setminus \{0\}$? I don't know where to start. My intuition is that you would get a problem with the singularity near $0$, but I am not sure how to prove it. Any help would be appreciated! Thanks!