Show that there cannot be a analytic function on $B(0,1)$ s.t if $|z|\to 1$, $|f(z)|\to \infty$. I saw a proof on here but I dont quite understand it.  The trick is to consider $1/f(z)$. Since then if $|z|\to 1$ then $1/f(z)\to 0$ and so we can continuously extend $1/f(z)$ to the closed unit ball to be zero on the boundary. Now we are done if we can extend $1/f(z)$ to be holomorphic in  the closed unit ball by maximum modulus principle. However, to do so, i need $f(z)$ to have finite zeroes in the unit ball which I cannot seem to prove. I can prove it has finite zeros if $f(z)$ was holomorphic on a compact set. But an open ball is not compact and so $f$ could have infinite zeroes, at least i think. Could someone fill in the details for me here?
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.
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!!
120910
Let $f$ be a holomorphic function in $H={\{z\in\mathbb{C}}:\Re(z)&gt;0\}$ such that it is continuous in $cl(H)$, $|f(z)|\leq1$ $\forall z\in \delta H$ and $\lim_{|z|\to\infty}f(z)=0 $. Prove that $|f(z)|\leq1$ $\forall z\in H$. I thought about taking the bounded set $H\cap D(0,R), R&gt;0$. This way we could use the maximum modulus principle to prove it but I do not really know how to relate it with the limit since I would need $|f(z)|\leq1$ $\forall z\in H\cap D(0,R)$, and I do not have it on $D(0,R)$.
58118
Let $f$ be analytic in $B(0;1)$, and suppose that $|f(z)|\leq M$ for all $z\in B(0,1)$. If $f(z_{k})=0$ for $1\leq k \leq n$. Show that  $$|f(z)| \leq M\prod_{k=1}^{n}\frac{|z-z_{k}|}{|1-\bar{z_{k}}z|}$$ for $|z|&lt;1$.  What I have done is proved the result when  $k=1$. That is if $f(z_{1})=0$, with $|f(z)|\leq M$ on $B(0;1)$, I have proved that  $$|f(z)|\leq M \frac{|z-z_{1}|}{|1-\bar{z_{1}}z|}$$ for $|z|&lt;1$.  Now I tried to consider this function:$g:B(0;1)\to \mathbb{C}$  $$g(z)=\frac{f(z)}{\prod_{k=2}^{n}\frac{|z-z_{k}|}{|1-\bar{z_{k}}z|}}$$  Now $g$ has removable singularity at the points $z_{2}, z_{3},.... z_{n}$. So after removing these singularities we can consider $g$ to be analytic. Now wanted to invoke the result I have proved above since $g(z_{1})=0$. But couldn't do so because $g(z)$ might not be bounded by $M$ on $B(0;1)$. How can I work this problem out.   Thanks in advance!!
120274
Consider the question: does there exist a holomorphic function $f$ on the unit disk in the complex plane such that $f\left({1 \over n}\right) = e^{-n}$ ? I came up with an answer but I'd like to know if there is another argument.  Here's my answer: let $f$ be such a function. $f$ must vanish at $0$ by continuity. Any holomorphic function vanishes at $0$ satisfies the following estimate : $\left| f(z) \right|\sim_{z \to 0} \alpha \,|z|^m$, where $\alpha&gt;0$ and $m$ is a positive integer (this is easily derived from the expression of $f$ as a power series). This contradicts the assumption $f\left({1 \over n}\right) = e^{-n}$.  By curiosity, I would like to know if there is another argument (maybe something about $f(z) - e^{1 \over z}$)?
81538
Suppose $f$ and $g$ are holomorphic in a region containing the disc $|z| \le 1$. Suppose that $f$ has a simple zero at $z=0$ and vanishes nowhere else in $|z| \le 1$. Let $f_\epsilon (z) = f(z)+\epsilon g(z)$. Show that if $\epsilon$ is sufficiently small, then (a) $f_\epsilon (z)$ has a unique zero in $|z| \le 1$, and (b) if $z_\epsilon$ is this zero, the mapping $\epsilon \mapsto z_\epsilon$ is continuous.    I already solved (a) by applying Rouche's theorem, but (b) is such a nuisance to me. I first tried the classical $\epsilon - \delta$ method, but it didn't work. However, I couldn't find any other ways to prove the continuity. Since $f$ has a simple zero at $z=0$, I found that $f(z)=zh(z)$ for some $h(z)$ that is holomorphic and non-zero in the unit disc and $z_\epsilon h(z_\epsilon) = -\epsilon g(z_\epsilon)$. Am I on the right track? I don't know how to proceed from this.
97979
Is there an analytic function $f$ on $B(0,1)\subset\mathbb{C}$ such that $f(1/n)=e^{-n}$ for $n=2,3,4,...$? I know the following doesn't work:  Let $g(z)=\exp(-1/z)$. Then, $f=g$ on a sequence with a limit point in $B(0,1)$ and so $f=g$ on $B(0,1)$. Since $g$ is not $\mathbb{C}$-differentiable at $0$, neither is $f$ and so such a function cannot exist.  This is not the solution because you cannot use the identity principle with a non analytic function like $\exp(-1/z)$ is not analytic at $z=0$. Any help using the identity principle another way?
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.
18635
Let $A \subset \mathbb{C} $ be an open set containing the closed unit disc. Let $f$ be an analytic function from $A$ to $\mathbb{C}$ such that $|f(z)|=1$ if $|z|=1$.  Does it follow that $f(z) = a z^{n} \frac{cz^{m}-b}{1-cz^{m}\bar{b}} $ for some $a,b,c \in \mathbb{C}$ s.t. $|c|=1$ $|a|=1$, $|b|&lt;1$ and some $n,m\ge 0 $?
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.
43299
Suppose $A=\{z\in \mathbb{C}: 0&lt;|z|&lt;1\}$ and $B=\{z\in \mathbb{C}: 2&lt;|z|&lt;3\}$.   Show that there is no one -to-one analytic function from A to B. Any hints? Thanks!
71420
Suppose $f$ is holomorphic in the unit disk $D=\{z: |z|&lt;1\}$ and $|f(z)|\leq |f(z^2)|$ on $D$. Show that $f$ is a constant function.   My Try:  I wanted to use Strong Maximum Principle, but failed to prove that $f$ is bounded on $D$. I want to try this problem in my own. Can anybody please give me just a hint?
66441
Show that if $f(z)=\sum\limits_{n\geq0}a_nz^n$ is analytic in $\{z\in\mathbb{C}:|z|&lt;1\}\cup\{1\}$ and $\forall n\geq0:a_n\geq0$ then the radius of convergence of the power series is strictly larger than $1$.  My approach:   Let $D=\{z\in\mathbb{C}:|z|\leq1\}$.  We have that $f(1)$ converges, so $\sum\limits_{n\geq 0} a_n&lt;\infty$. Moreover since the $a_n$'s are all positive we have that $|a_n|=a_n$, therefore $\sum\limits_{n\geq 0} |a_n|=\sum\limits_{n\geq 0} a_n&lt;\infty$.  Let $f_n(z)=a_nz^n$, then $\forall z\in D, n\geq 0$ we have:  $$|f_n(z)|=|a_nz^n|=|a_n|\cdot|z^n|=a_n\cdot |z|^n\leq a_n\cdot 1^n\leq a_n$$  Then, applying Weierstrass M-test we get that the power series is convergent in $D$.  Now, I need to "find" an open ball $B$ such that $D\subset B$ and the power series is convergent in it.  Edit  $f$ is analytic in $1$, so there is an open set $U$ that contains $1$ such that   $$f(z)=\sum\limits_{n\geq 0} b_n(z-1)^n$$  Does it mean that the series is convergent in $U$?   I can see that we have that the two series coincide in $D\cap U$.
93591
Working on a problem in Gamelin's book.  "Show that if $f(z)$ is a continuous function on a domain $D$ such that $f(z)^N$ is analytic on $D$ for some integer $N$, then $f(z)$ is analytic on $D$."  He starts a hint with the statement "Show that the zeros of $f(z)$ are isolated." This makes me think. I know that if $f$ is analytic and nonconstant, then the zeros are isolated. Is there some sort of converse to this theorem? I.e., for example, if continuous and zeros are isolated then analytic?  His next hint is: "At a zero of $f(z)$, write $f(z)^N=(z-z_0)^mh(z)$ where $h(z_0)\ne 0$ and show that $N$ divides $m$." However, the book has covered nothing on winding numbers thus far. Also, how will the fact that $N$ divides $m$ help to show that $f(z)$ is analytic?
16886
I want to show the following:  If $f(z) $ is a continuous  function on a connected open subset of the complex plane and $f(z)^2$ is an analytic function, then $f(z)$ is analytic.  Clearly if $f(z) \neq 0$ then $$\frac{f(z+h)-f(z)}{h}=\frac{f(z+h)^2-f(z)^2}{h}.\frac{1}{f(z+h)+f(z)}$$  which shows $f^\prime(z) $ exists if $f(z)\neq0$.  What do I do when $f(z) =0$?
22279
Suppose one has a function $f$ in the disc algebra ie: $f$ is continuous on $|z|\leq1$ and holomorphic in $|z|&lt;1$. I wondered, can $f$ always be extended to a holomorphic function on some region containing the boundary (ie: $T$ the unit circle). If not can you give me an example? Thank you,