I don't need this statement to be true globally - even if it's locally true I'm fine with that. I'm on a noncompact Riemann surface but that doesn't matter because I only want this to hold locally anyway, so we can just look at little subsets of $\mathbb{C}$. Say $h$ is an arbitrary holomorphic function.  Say $f,g$ are holomorphic functions with no common zero. Then I want to show that there exist $\phi_1, \phi_2$ holomorphic such that $h = f\phi_1 + g\phi_2$ locally (so on some neighbourhood of some fixed point). I'm happy to just work with Taylor series of all these functions (which is what I was doing) because as I said, I'm fine with this holding just locally. But I can't quite figure out how it works. I'd appreciate some help, please!
38819
Let $\Omega$ be a region in the complex plane and let $f_1$ and $f_2$ be holomorphic functions on $\Omega$ having no common zero. Show that there exist holomorphic functions $g_1$ and $g_2$ on $\Omega$ such that, $f_1g_1+f_2g_2$ is identically equal to one on $\Omega$.
95079
Suppose that $f(z)$ is analytic in the unit disk $\Delta:\,|z|&lt;1$. Then $f(z)$ has a Taylor series $\sum=\sum a_nz^n$ in the unit disk. One may assume that $\sum$ has $R=1$ as its radius of convergence for convenience.  Question 1 (local version):  If $f(z)$   is continous at some point, say $z=1$, on the unit circle $S^1$, then its Taylor series $\sum$  is  covergent at $z=1$?   Question 2 (global version):  Suppose that  $f(z)$   is continous on the closed unit disk $\overline\Delta$.  Is the  Taylor series $\sum$ covergent at every poin of $S^1$?   Of course, if the answer to the local version is "yes", then so is the answer to the global version.  EDIT. It seems that  Question 1 obviously has a negative answer. So I modify it into a new question.  Question 3 (New local version):  Suppose that  $f(z)$   is continous at some point, say $z=1$, on the unit circle $S^1$ and the coefficients of $\sum$ satisfies $a_n\to 0$.  Is the Taylor series $\sum$  covergent at $z=1$?   Note. If $f$ is analytic at $z=1$, then the answer to Question 3 is yes. This is Fatou's Theorem.
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 $|z| = 1$.
75614
Here is the question: Let $f:\mathbb{D}\rightarrow\mathbb{C}$ be an analytic function.  Suppose that there exists $\epsilon&gt;0$ such that for every $\theta$, $|\theta|&lt;\epsilon$, and $\lim_{z\rightarrow e^{i\theta}}f(z)=0$.  Prove that $f$ is identically $0$ on $\mathbb{D}$. My thoughts:  Well, honestly, I am quite stumped here.  I was thinking I could let $f$ be analytic in some other region (can I even do this?), say the lower half plane and suppose $\lim_{z\rightarrow x}f(z)=0$ whenever $-1\leq x\leq1$, by assumption.  Maybe then I could show that $f$ is identically $0$ on the bigger region, and then consider $f$ on the intersection of those two regions, and then ultimately making my way towards using the uniqueness theorem to show what I want... but I am quite stuck on how to do that... any thoughts, suggestions, etc. are greatly appreciated!  Thank you.
43808
There doesnt exist a holomorphic $f: \mathbb C\setminus0 \rightarrow \mathbb C\setminus 0 $, so that $e^{f(z)} = z $ for all $z \in \mathbb C \setminus 0 $  I tried to work with the derivative of f(z) but i didnt manage to get anything out of it.
92203
Function $f(z)$ is holomorphic at least within the unit disk.   If for all arguments $|z|&lt;1$ we have $|f(z)| \le 1$, and there exists a zero of order $n$ at $z=z_0$, then show that $|f(0)| \le |z_0|^n$.  How should I tackle this problem? Can it be generalized if for $|z|&lt;R$ we have $|f(z)| \le M$ ?    So far I only understand that there exists a holomorphic $g(z)$ with $g(z_0) \ne 0$ such that $f(z) = (z-z_0)^n g(z)$,then $|f(z)| = |z-z_0|^n |g(z)| \le 1$ and moreover $|f(0)| = |z_0|^n |g(0)|$, but that doesn't seem useful since apparently I know too little about $g(z)$. In particular why should $|g(0)| \le 1$?
44851
Suppose $f$ is a continuous complex valued function on a domain $\Omega$. Suppose $f^2$ and $f^3$ are holomorphic in $\Omega$. Show that $f$ is also holomorphic in $\Omega$.   Assume $f=u+iv$. I see that if $u,v$ are in $C^1$ then $f^2$ is holomorphic can imply $f$ is holomorphic considering Cauchy-Riemann equations. But I don't know how to get $u,v$ are in $C^1$ by the adding condition $f^3$ is holomorphic. (I can't get $u,v$ from some combinations of real and imaginary part of $f^2$ and $f^3$). Do someone know how to do this?
94143
Let the functions $f$ and $g$ be holomorphic in $U$ and continuous in $\overline{U}$. Show that $|f(z)| + |g(z)|$ attains its maximum on $\{|z| = 1\}$. Hint: consider the function   $h = e^{iα}f + e ^{iβ}g$ with suitably chosen constants $α$ and $β$.   Even with the hint I am lost. Of course $|h(z)|$ must attain its maximum when $|z|=1$, and $|h| \le |f|+|g|$, but that doesn't mean that $|f|+|g|$ couldn't attain its maximum inside the unit circle.  ny help?
81136
Let $f:\mathbb{C}\to\mathbb{C}$ be a holomorphic function such there exists a $c\in\mathbb C$ with $|f(z)|\leq c\cdot |z|^n$ for all $z\in\mathbb C$. Show that there exists an $a\in\mathbb C$ with $f(z)=a\cdot z^n$.   Proof.  My general idea is to show that the taylor series only has one non-zero coefficient:  $\displaystyle |a_k|\leq\frac{k!}{2\pi}\oint_{|w|=r}\frac{|f(w)|}{|w|^{k+1}}dw\leq \frac{k!}{2\pi}\oint_{|w|=r}\frac{c\cdot|w|^n}{|w|^{k+1}}dw\leq \\\displaystyle\leq \frac{c\cdot k!}{2\pi}\oint_{|w|=r}|w|^{n-k-1}dw$  Now using the standard estimate since $\oint=2\pi r$:  $\displaystyle \leq c\cdot k!\cdot r^{n-k}\longrightarrow\begin{cases}0,\ k&gt;n\\c\cdot n!,\ k=n\\ \infty,\ k&lt;n\end{cases}$ as $r\to\infty$  In conlusion the taylor series for now looks like $f(z)=\sum_{k=0}^{n-1}a_k z^k+ a_n z^n$ with $|a_n|\leq n!\cdot c$. What do I do with the remaining summands?  Edit: I would say that for $k&lt;n$ we have $|z|^k&gt;|z|^n$ for $z\in D_1(0)$. So since the estimate has to hold for all $z\in \mathbb C$ we can't have that either.
21134
One of my assignment questions which I could not solve was following.  Question : If f is a holomorphic function on a region X which is not identically zero show that the zeroes of the function form an atmost countable set .  I was confused while solving the question . So I search MSE and found this solution :  Answer of User : Daniel Roberts says:&quot;If the set {z∈U|f(z)=0} were uncountable, then it would have an accumulation point &quot;  My question: Why any uncountable set must have a limit point ? Can anyone please give a rigorious proof.
13160
How can I prove the following assertion?    Let $f$ be a holomorphic function such that $|f|$ is a constant.  Then $f$ is constant.    Edit: The more elementary the proof, the better. I'm working my way through a complex analysis workbook, and by this exercise, the only substantial thing covered has been the Cauchy-Riemann equations.
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 $|g'(0)| \leq 1$.
78557
If $g(w)$ and $f(z)$ are holomorphic functions, show that $g(f(z))$ is also holomorphic.   From the assumptions we have that for some $A,B$,  $$\lim_{h\rightarrow 0}\frac{g(f(z)+h)-g(f(z))}{h}=A,$$  $$\lim_{h\rightarrow 0}\frac{f(z+h)-f(z)}{h}=B.$$  We want to prove that there exists $C$ with  $$\lim_{h\rightarrow 0}\frac{g(f(z+h))-g(f(z))}{h}=C.$$  The second equation means, by definition of the limit, that for any $\varepsilon &gt; 0$, there exists $\delta$ such that $0&lt;|h|&lt;\delta$ implies $$\left|\frac{f(z+h)-f(z)}{h}-B\right|&lt;\varepsilon.$$  I don't know how to get to $\dfrac{g(f(z+h))-g(f(z))}{h}$ from here.
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.
123447
Let $f$ be a function holomorphic in an open set containing the closed unit disc $D(0,1)$, except at the point $z_0$ with $|z_0|=1$, where $f$ has an isolated singularity. If $a_n$ are the coefficients of the Taylor expansion of $f$ centered at the origin, show that   $$\displaystyle\lim_{n \to\infty}\frac{a_n}{a_{n+1}}=z_0$$  I considered first the case: $z_0$=simple pole. Then i can write the Laurent series for $f$ at $z_0$  $$f(z)=\frac{c}{z-z_0}+c_0+c_1(z-z_0)+\ldots$$ I put  $$g(z)=f(z)-\frac{c}{z-z_0}$$  Then $g$ is holomorphic in an open set containing the closed unit disc $\overline{D(0,1)}$ and we have $$g(z)=\sum a_n z^n+\frac{c}{z_0}\sum(\frac{z}{z_0})^n=\sum(a_n+\frac{c}{z_0^{n+1}})z^n$$ The power series of $g$ has radius of convergence $&gt;1$, hence i can substitute 1 in RHS of last equation, and i get a numeric convergent series. The necessary condition for convergence says that its general term has to be infinitesimal $$\displaystyle\lim_{n\to\infty}(a_n+\frac{c}{z_0^{n+1}})=0$$ from which follows $\frac{a_n}{a_{n+1}}\rightarrow z_0$  the question is:how can i adapt this argument to deal with cases where z_0 is not a simple pole?
788
Let $f$ be holomorphic in a domain $U$, then $f$ has at most countably zeros.  How can I prove this statement? I have seen it in Sheng Gong  explicitly, but without any indications.
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 :)