Let $A$ and $B$ are Riemann surfaces. $f:A→ B$ is morphism of Rieman surface, in other word, holomorhic function, and supposed that $f$ is bijective. Then, can we say that inverse of $f$ is also holomorphic? Thank you in advance.
5436
I'm confused about the following proposition  Proposition. Let $U,V$ are open sets in $\mathbf{C}$. If $f:U\to V$ is holomorphic and bijective, then  the inverse  $f^{-1}:V\to U$ is also holomorphic.  The proof of the proposition think that the  continuity of $f^{-1}$ is obvious, but I find it is really difficult to prove  using $\epsilon-\delta$ definition. Can anyone give some hints?
87908
Prove that $f(z) = e^z$ maps $\mathbb{C} \setminus B $ onto $\mathbb{C} \setminus \{0\}$ where $B$ is a bounded subset of $\mathbb{C}$.  I have no idea how to even attempt this question so I would greatly appreciate some hints.
61516
Just based on some reading, I know that every Möbius transformation is a bijection from the Riemann sphere to itself.   I'm curious about the converse. For any holomorphic bijection on the sphere, why is it necessarily a Möbius transformation? Is there a proof or reference of why this converse is true? Thanks.  (I would appreciate an explanation at the level of someone whose just self-studying complex analysis for the first time.)
66609
Another question just now reminded me of something I realized a while ago I didn't know how to do: Say $V\subset\Bbb C$ is open, $f:V\to\Bbb C$, $f=u+iv$, and at every point of $V$ the partials of $u$ and $v$ exist and satisfy the Cauchy-Riemann equations. How do we show that $f$ is holomorphic? I mean it seems it &quot;must&quot; follow. But note if you think this is totally trivial it's possible you're wrong; for instance it's not clear how the hypothesis implies that $f'(z)$ exists.
49561
Is it true that  $ f(\bar{z})=\overline {f(z)}$, Where z is complex?  I think it holds when $f(z)$ is holomorphic since we have $f(z)=p(x,y)+iq(x,y)=p(z,0)+iq(z,0)$ Any help...
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.
9645
Let $f: B(0,1) \rightarrow \mathbb{C}$ be holomorphic and suppose $\exists\ r \in (0,1)$ such that $f$ is injective in $A = \{z \in \mathbb{C} : r &lt; |z| &lt; 1\}$. Prove that $f$ is injective.  I tried using Rouché theorem, or the identity theorem, but I don't know what to do. Any hints? :)
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.
3534
Let $\{f_n\}$ be a sequence of holomorphic functions defined in a generic domain $D \subset \mathbb{C}$. Suppose that there exist $f$ such that $f_n \to f$ uniformly.  My question is: is it true that $f$ is holomorphic too?
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.
6260
I'm trying to work through the exercises in Otto Forster's book on Riemann Surfaces. While most of them seemed not that hard, this one gives me a headache:  Let $X=\mathbb{C}\setminus\{\pm1\}$  and $Y = \mathbb{C}\setminus\{\frac{\pi}{2}+k\pi~|~k\in\mathbb{Z}\}$. Then $\sin: Y\rightarrow X$ is a covering map. This is clear.  Now consider the following curves:  $$ u: [0,1] \rightarrow X, u(t) = 1-e^{2\pi i t}$$  and  $$ v: [0,1] \rightarrow X, v(t) = -1+e^{2\pi i t} = -u(t)$$  Let $w_1: [0,1]\rightarrow Y$ be the lifting of $u\cdot v$ and $w_2$ be the lifting of $v\cdot u$ with $w_1(0)=w_2(0)=0$.  Show that $w_1(1) = 2\pi$ while $w_2(1) = -2\pi$.  As of now, I don't have many ideas how to solve this. Of course, the inverse sine is a multi-valued function so the two liftings should use different branches. But I don't see how exactly they come into play.  I'm also not sure whether the solution can be done by calculations alone or if there is some more general principle underlying. Any suggestions would be welcome.
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 $|g'(0)| \leq 1$.
54122
Still quite unsure about the chapter on orientation. For instance the Mobius map indeed preserves orientation, and I can show this rigorously, however I am still having difficulty understanding what it means for a holomorphic function $f$ to preserve orientation over some region in the complex plane. Which is why I asked a question regarding this $2$ days ago, but did not receive an answer.   The problem I am still thinking about is, I cannot show or provide a counter argument for why there does or does not exist some holomorphic function $f$ on a $\{z:|z|\le1\}$ such that it sends the unit circle with the counter- clockwise orientation into the unit circle with the clockwise orientation?  From the comment from this question asked earlier, one did suggest the use of the Argument principle which states that:     Argument Principle: If $f$ is meromorphic on a simply connected region $D$ and $\Gamma$ is a simply closed curve in $D$ not passing through the roots $z_j$ nor the poles $p_k$ of $f,$ then $$\frac{1}{2\pi i}\oint_{\Gamma}\frac{f'(z)}{f(z)}dz=n-m$$   Where $n$ is the number of roots of $f$(including multiplicities) in $\Gamma$ and $m$ is the number of poles of $f$ (including multiplicities) inside $\Gamma.$     I am really curious to find out why we can apply this to try and provide an explanation for my problem. I would appreciate some help because I am quite confused with orientation at the moment, and I think this problem is good in the sense it will help me understand some important notions on holomorphic functions which send or does not send certain orientation on a region to a different orientation of the same region.   I would really appreciate some rigorous argument for this problem. Much help will be appreciated.
45078
Suppose $m,n&gt;0$, $U$ an open subset of $\mathbb{R}^{m+n}$ and let $f: U \to \mathbb{R}^n$ be continuously differentiable.  Is it possible for $f$ to be injective?    My thinking is that continuous differentiability+injectivity suggests that this is sort of getting at a converse to the Inverse Function Theorem.  We know the Jacobian is not invertible at any point in $U$, and the question is asking if this implies that $f$ cannot be invertible in any neighborhood contained in $U$.
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!!
112038
Let $f:\mathbb{C}\to\mathbb{C}$ be a complex valued function of the form $f(x,y)=u(x,y)+iv(x,y)$. Suppose that $u(x,y)=3x^2y$. Then  $f$ cannot be holomorphic on $\mathbb{C}$ for any choice of $v$.  $f$ is holomorphic on $\mathbb{C}$ for a suitable choice of $v$.  $f$ is holomorphic on $\mathbb{C}$ for all choices of $v$.  $u$ is not differentiable.    () well, I have calculated by applying CR equation, getting the option $1$ is correct? could anyone tell me  just am I right?thank you.
82525
Prove that the functions $f(z)$ and $\overline{f(\overline{z})}$ are simultaneously holomorphic.   I take this to mean that $f(z)$ is holomorphic if and only if $\overline{f(\overline{z})}$ is holomorphic.  Let $g(z)=\overline{f(\overline{z})}$. Note that $\overline{g(\overline{z})}=f(z)$. So it suffices to prove that if $f(z)$ is holomorphic, then $g(z)$ is holomorphic.  Write $f(z)=u(z)+iv(z)$. Since $f(z)$ is holomorphic, the real and imaginary parts satisfy the Cauchy-Riemann equations:  $$\frac{\partial{u(z)}}{\partial{x}} = \frac{\partial{v(z)}}{\partial{y}}, \frac{\partial{u(z)}}{\partial{y}} = -\frac{\partial{v(z)}}{\partial{x}}.$$  We have $g(z) = u(\overline{z})+i(-v(\overline{z}))$. To prove that $g(z)$ is holomorphic, we must prove that its real and imaginary parts satisfy the Cauchy-Riemann equations:  $$\frac{\partial{u(\overline{z})}}{\partial{x}} = \frac{\partial{(-v(\overline{z}))}}{\partial{y}}, \frac{\partial{u(\overline{z})}}{\partial{y}} = -\frac{\partial{(-v(\overline{z}))}}{\partial{x}}.$$  How can we obtain this from the above relations?