Let $D = \{z\in\mathbb C \vert\vert {z}\vert \lt 1\}$ be the open unit disc in the complex plane, and let $f(z)$ be the following power series of the complex variable $z$: $$f(z)=z+z^2+z^4+\cdots+z^{2^n}+\cdots.$$ By considering the limit $\lim_{x\to 1^{-}} f(x) $ as $0\lt x \lt 1 $ approaches $1$ from the left. How to show that there does not ecist a holomorphic function $g(z)$  defined on a neighborhood $U$ of $z=1$ such that $f(z)=g(z)$ for all $z\in U\cap D$ And how to deal with the limit $\lim_{x\to 1^{-}} f(x) $ as $0\lt x \lt 1 $
123921
Another Qual question here,  For the function $$\sum_{n=0}^\infty z^{2^n}$$,  Prove the following:  i) $f$ converges to a function analytic in the open unit disk $D$,  ii) $f(z) =z+f(z^2)$ and  iii) $f(z)$ can not be analytically continued past any point on the unit circle.  I can even see (ii) very easily, but I can not see how can I prove rigorously (i) and  (iii). Help please.
79633
I'm going over old exam problems and I got stuck on this one. Suppose that $\mathbb{f}\colon \mathbb{D} \to \mathbb{C}$ is analytic and bounded. Let $\{a_n\}_{n=1}^\infty$ be the non-zero zeros of $\mathbb{f}$ in $\mathbb{D}$ counted according to multiplicity. Show that  $$ \sum_{n=1}^\infty \left( 1 - \left|a_n \right|\right)\lt\infty $$  I can understand that $\left|a_n \right|$ goes to $1$ since zeros are isolated, but it doesn't help showing the series is convergent. Any help will be appreciated!
21547
Let $D \subset \mathbb{C}$ be the open unit disk, $f:D \to \mathbb{C}$ a holomorphic function and $d=\sup \{|f(z)-f(w)|:z,w\in D \}$. How can I show that $|f'(0)|\leq \frac{d}{2}$?  I tried using Schwarz's lemma on the function $g:D \to \mathbb{C}$ defined by $g(z)=\frac{f(z)-f(0)}{d}$, but I am only getting $|f'(0)|\leq d$.
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 $?
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!!
33619
Well, after spending hours on this problem, I'm still stuck, so I thought I'd turn to you guys. The problem statement is as follows.  Let $f$ be a complex-valued function that is $C^1$ in the disk $|z - z_0| &lt; R$ for some $R&gt;0$, and let $C_r$ be the circle $|z - z_0| = r$. Show that $$\lim_{r\to 0} \frac{1}{r^2}\int_{C_{r}}f(z)dz = 2 \pi i\frac{\partial f}{\partial \bar{z}}(z_0).$$  I tried parametrizing the circle in the usual way and differentiating to see if I get something useful, but no luck. I just can't get the LHS to give me something to do with $\frac{\partial f}{\partial \bar{z}}$, so that I could then put the RHS under the integral and show that the difference between the two can be less than arbitrary $\epsilon &gt; 0$ by the continuity of partial derivatives.  Thanks in advance for your help!
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.
113388
Let $f_n$ be continuous on the open set $U$ and let $f_n \rightarrow f$ uniformly on compact sets.  and if  $U \supseteq \{z_n\}$ and $z_n \rightarrow z_0 \in U$ and $f_j$ are holomorphic, then what i want to show is for $0 &lt; k \in \mathbb{Z}$,  \begin{align} \left(\frac{\partial}{\partial z}\right)^k f_n(z_n) \rightarrow \left(\frac{\partial}{\partial z}\right)^kf(z_0) \end{align}  My trial is similar treatment for $\lim_{n\rightarrow \infty} f_n(z_n) = f(z)$. But i am not sure about considering derivatives.
106718
Let $f$ be a continuous function on the closed unit disk and holomorphic inside it, such that $f(z) = 0$ for all $ z = \exp (it) $ with $0 ≤ t ≤ \pi/2 $. Prove that $ f(z) = 0 $ for all $ z \in \overline{D}(0,1) $.  I have worked this exercise for absurdity, but it has not reached any conclusion. I would like to know a more sophisticated test, perhaps by a direct method.
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!
67183
Consider the following complex power series :$$\sum_{n=1}^\infty\frac{z^n}{n}$$ The radius of convergence of this series is $1$ and the series is divergent for $z=1$. I want to know what are the values of $z\in C:=\lbrace z\in\mathbb{C}: |z|=1\rbrace$, the circle of convergence, for which the given series converges.
70110
$\textbf{Problem.}$ Suppose $f$ is a holomorphic function on $\{z\in\mathbb{C}:|z|&lt;1\}$, the open unit disk, with the property that Re$f(z)&gt;0$ for every point $z$ in the disk. Prove that $|f'(0)|\leq 2\text{Re}f(0)$.  This is a problem that appeared on a qualifying exam in some graduate school. What I tried is, define $\displaystyle\varphi(z)=\frac{z-1}{z+1}$, then this sends the open right half plane onto the open unit disk, so let $\hat{f}=\varphi\circ f(z)$, then its image is contained in the open unit disk. And I wanted to use the Schwarz lemma, but then the origin had to be fixed, so define $\displaystyle\psi(z)=\frac{\hat{f}(0)-z}{1-\overline{\hat{f}(0)}z}$ and let $\tilde{f}=\psi\circ\hat{f}$, then the image of this is still contained in the open unit disk and it fixes the origin so the Schwarz lemma $|(\tilde{f})'(0)|\leq 1$ could be applied. But after all the calculation, I got $$|f'(0)|\leq\frac{|f(0)+\overline{f(0)}+2|f(0)|^{2}|^{2}}{2|f(0)+1|^{2}|f(0)|}$$ but, for example, if the magnitude of the imaginary part of $f(0)$ is quite bigger than the magnitude of the real part of $f(0)$, then this doesn't result the desired inequality but actually means that the desired inequality is wrong instead. I also thought about composing some different function to the RHS of $\hat{f}$ and use the Schwarz lemma but it didn't seem to work.  Maybe I should try something else, but what could be tried instead?
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.
104157
If $f(z)$ is analytic in the unit disk and $f(0)=0$, show that $$f(z)+f(z^2)+\cdots+ f(z^n)+\cdots$$ converges locally uniformly to an analytic function in the unit disk.  I am thinking to apply Weierstrass $M$ test for convergence or since all the $f(z^n)$ are analytic in the unit disk, so each of them can be expressed as the taylor series. Then rewriting the series again might work. But I am not sure. Any 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 $D(0,R)$.
14038
I'm reading Gong Sheng's  to get some basic understanding.  On $\S 2.4$ page 61 Theorem 2.15 (Hurwitz Theorem) it says     Theorem 2.15 (Hurwitz Theorem) Let $\{f_j\}$ be a sequence of holomorphic functions on $U\subseteq \mathbb C$ that converges uniformly to a function $f$ on    every compact subset of $U$. If $f_j$ is never equal to zero on $U$ for all $j$, then    $f$ is either identically zero or never equal to zero on $U$.      Proof: For an arbitrary point $z \in U \subseteq \mathbb C$, choose a simple closed curve $\gamma$ in $U$ such that the inside of $\gamma$ contains $z$.    Since $f_j$ is holomorphic on $U$, by Cauchy integral formula we have $$f_j(z)=\frac{1}{2\pi i}\int_\gamma \frac{f_j(\zeta)\,d \zeta}{\zeta-z}$$   Since $\{f_j\}$ converges uniformly on every compact subset of $U$, we have    $$\lim\limits_{j\to \infty} f_j(z) = \lim\limits_{j\to \infty} \frac{1}{2\pi j} \int_\gamma \frac{f_j(\zeta)\,d \zeta}{\zeta-z}  = \frac{1}{2\pi j} \int_\gamma \lim\limits_{j\to \infty} \frac{f_j(\zeta)\,d \zeta}{\zeta-z} $$   It follows that $$f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta) \, d \zeta}{\zeta - z}$$   Hence, $f(z)$ is a holomorphic function.    Similarly, we can prove that $\{f_j'(z)\}$ converges uniformly to $f'(z)$ on every compact subset of $U$.       If $f(z)$ is not identically zero, then by Theorem 2.13, the zeros of $f$ are  discrete. Let $\gamma$ be a curve that does not pass through these zeros. Then    $$\frac{1}{2\pi i} \int_\gamma \frac{f_j'(\zeta)}{f_j(\zeta)\, d \zeta} \to \frac{1}{2\pi i} \int_\gamma \frac{f'(\zeta)}{f(\zeta)\, d \zeta}$$   as $j\to \infty$.   By the assumption and Theorem 2.14, we have    $$\frac{1}{2\pi i} \int_\gamma \frac{f_j'(\zeta)}{f_j(\zeta)\, d \zeta} = 0$$   Therefore  $$\frac{1}{2\pi i} \int_\gamma \frac{f'(\zeta)}{f(\zeta)\, d \zeta} = 0$$   and $f(z)$ has no zero on $U$.    I wonder why $f(z)$ is a holomorphic function once $f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta) \, d \zeta}{\zeta - z}$?  So far from that book I only know such ways to determine a function is holomorphic   By definition, that is $f(z)$ is holomorphic in $U$, iff $\forall z\in U$, $\lim_{h\to 0} \frac{f(z+h)-f(z)}{h}$ exists, here $h \in \mathbb C$.  Cauchy-Riemann equation, that is, if $f(z) \in \mathscr L ^1 (U)$ and fulfills the Cauchy-Riemann equation $\frac{\partial f}{\partial x} = -i \frac{\partial f}{\partial y}$, then $f$ is holomorphic on $U$, here $z=x+iy$. Power series expansion (Taylor series): $f(z)$ is holomorphic in $U$ iff $f$ has a power series expansion $\forall z \in U$. Morera theorem:  If $f(z)$ is continuous on $U$ and  the integral of $f$ along any rectifiable closed curve is zero, then $f(z)$ is  holomorphic on $U$.    It seems none of such 1-4 could get to the conclusion that $f(z)$ is a holomorphic function?
114990
Let $0&lt;\delta&lt;r$ and $R&gt;0$, and $a\in\mathbb C$. I want to show that there exists $C&gt;0$ such that for all holomorphic functions $f\colon B(a,r)\to B(0,R)$, we have $$ \vert f(z)-f(w)\vert\leq C\vert z-w\vert $$ for $z,w\in B(a,\delta)$.  I'm not sure how to approach this. I was wondering if I could use that $$ \left\vert\frac{f^{(n)}(a)}{n!}\right\vert\leq\frac{\Vert f\Vert_{C(a,\rho)}}{\rho^n} $$ for some $0&lt;\rho&lt;r$. I also have an expression from biholomorphic maps from $B(0,1)$ to itself, but I'm not sure if I could use that here. Any hints?