Let $\mathbb{D}$ denote the unit disk in the complex plane $\mathbb{C}$. I want to find a conformal mapping of the slit disk $\mathbb{D}-[0, 1)$ onto $\mathbb{D}$.    Here's what I know so far: The conformal mapping $$\phi(z)= e^{i \theta} \frac{z-a}{1- \bar{a}z}, |z|&lt;1$$ with $a \in \mathbb{C},|a|&lt;1, 0 \leq \theta \leq 2 \pi$ maps $\mathbb{D}$ onto $\mathbb{D}$. I'm also familiar with the standard mappings from the upper half plane onto the unit disk and vice versa. Since we are considering a slit disk to be mapped onto the unit disk, would that entail the composite of $\phi$ with an appropriate branch of the square root function? I'm still trying to come up with an explicit mapping of some sort, but haven't gotten very far yet. Any suggestions would be helpful.
121418
I am trying to find a conformal map of  slit unit disk (slit in negative real axis) i.e $${{z: |z|&lt;1, z\notin (-1,0]}}$$ on to the unit disk that takes $\sqrt2 -1 $ to $0$.  This is what I think,  I can see $\sqrt{z} $ taking the slit disk to disk ( half disk) in the right half plane, and then rotating counterclockwise gives the disk in the upper half plane and the mapping $(\frac {1-z}{1+z} )^2$ maps to the upper half plane and the map $\frac {z-i}{z+i}$ maps to open unit disk.  I tried to compose and get the image of $\sqrt2 -1 $ under this composition. I am getting some frustrating crap.   If I knew that this $\sqrt2 -1 $ gets mapped to some $\alpha$ then I would compose the above function (composition function) with the map $(\frac {z-\alpha}{1-\bar\alpha z} )$ to get the image of $\sqrt2 -1 $ as $0$.  I also think this map is not unique.  So the question is if my work  correct? If not what am I doing wrong? Can someone give me the explicit formula for this. Thanks in advance.
97399
Specifically, the question is as follows:     Let $\mathbb{D}$ be the open unit disk in $\mathbb{C}$ and suppose that $f:\bar{\mathbb{D}}\rightarrow\mathbb{C}$ is a continuous function such that both $\Re f$ and $\Im f$ are harmonic.       (a) Show that   $$f(re^{i\theta})=\frac{1}{2\pi}\int_{-\pi}^\pi f(e^{it})P_r(\theta-t)dt$$   for all $re^{i\theta}$ in $\mathbb{D}$.       (b) Show that $f$ is holomorphic in $\mathbb{D}$ if and only if   $$\int_{-\pi}^\pi f(e^{it})e^{int}dt=0$$   for all $n\geq1$.   I believe (a) is a direct consequence of Poisson's representation theorem for harmonic functions, stated as follows:  Let $\mathbb{D}$ be the open unit disk and $f:\partial\mathbb{D}\rightarrow\mathbb{R}$ be continuous. Then there exists a continuous function $u:\bar{\mathbb{D}}\rightarrow\mathbb{R}$ such that $u(z)=f(z)$ if $z\in\partial\mathbb{D}$, $u$ is harmonic on $\mathbb{D}$, and $u(re^{i\theta})=\frac{1}{2\pi}\int_{-\pi}^\pi P_r(\theta-t)f(e^{it})dt$ and $u$ is unique.  It seems to me that this can just be applied to the real and imaginary parts of the function $f$ from the question to derive the desired result, but maybe I'm making a careless mistake?  For (b), it's not obvious to me how part (a) assists in the solution. This is a homework question, so I am looking for assistance with understanding it, but not a full solution. Thanks.
63791
Consider the complex mapping defined by $$f_a(z)=\frac{z-a}{\bar az-1}$$ for some constant $a.$ I want to show that $f$ maps the unit circle to itself. I know that the unit circle is defined by $|z|=1,$ or equivalently $z=e^{i \theta}.$ This gives me $$f_a(z)=f_a(e^{i \theta})=\frac{e^{i \theta}-a}{\bar a e^{i \theta}-1}$$  I'm not sure how to proceed from here though.   Additional question: I also want to show that of $a$ lies inside the unit disk then $f$ maps the unit disk to itself. In order to answer this I have shown that $$|\bar a z-1|^2 - |z-a|^2 =(1-|a|^2)(1-|z|^2)&lt;1$$  How am I able to use this to deduce the result?
123213
Let $S^1\subseteq\mathbb{C}$ be the unit circle and let $U\subseteq S^1$ be open. How to show that there exist a continuous function $$\theta:U\to\mathbb{R}$$ such that $$e^{i\theta(z)}=z$$ for all $z\in U$ if and only if $U\neq S^1$?  I have been able to show the $\Rightarrow$ direction but not the other one.  $(\Rightarrow)$ If such a continuous $\theta:S^1\to\mathbb{R}$ exist, then it is injective and so $\theta(S^1)$ is a subset of $\mathbb{R}$ that is homeomorphic to $S^1$. Now, $S^1$ is compact and connected so $\theta(S^1)$ is a closed interval $[a,b]$. But removing a point from $[a,b]$ gives two connected components, while removing a point from $S^1$ still gives a connected space. Hence, $[a,b]\not\cong S^1$, so $\theta$ doesn't exist.  How would you show the $\Leftarrow$ direction?  Edit:  This is an exercise in topology with no assumed prior knowledge of complex analysis. Note that we only need to show continuity of $\theta$ (not analyticity). The question is phrased in terms of complex numbers for simplicity of notation, but it is to be interpreted in terms of subspace of $\mathbb{R}^2$.
124015
Let $[z_1,z_2,z_3,z_4]$ denote the  of the complex numbers $z_1,z_2,z_3,z_4\in \mathbb{C}$. Show that the distinct points $z_1,z_2,z_3,z_4\in\widehat{\mathbb{C}}$ lie on a generalized circle if and only if $[z_1,z_2,z_3,z_4]\in\mathbb{R}$.    I saw this statement claimed in , and wondered how one would go about showing it. I know of some facts we could work with:   First, the definition: A  in $\mathbb{C}$ is either a circle or a line in $\mathbb{C}$. I know that a Mobius transformation maps generalized circles to generalized circles. I know that if $f$ is a Mobius transformation and $z_j'=f(z_j)$, $j=1,2,3,4$, then $[z_1',z_2',z_3',z_4']=[z_1,z_2,z_3,z_4]$.   I have posted a potential proof as an answer below. Please let me know if you think it's complete. Thank you!
18552
$f$ is an analytical function on the unit disk $\mathbb{D}$ for which there is a $C &gt; 0$ such that  $\forall z \in \mathbb{D} \colon |f(z)| \leq \dfrac{C}{1-|z|}$.  I know have to prove that    $\forall z \in \mathbb{D} \colon |f'(z)| \leq \dfrac{4C}{(1-|z|)^2}$.  I tried using the Cauchy integral formula but I can't find a fitting circle to get this result.
41780
I haven't found a similar question on here, though I suspect the question may be rather well-covered.   I want to find a conformal map from the vertical strip $\{z:-1&lt;Re(z)&lt;1\}$ onto the unit disc.  Under the exponential map the region is taken to the annulus with radii $e$ and $e^{-1}$, but I'm not sure how useful this will be.  Can anyone advise on what the conformal map may be?  Thanks.
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?
61181
How can one show there is no holomorphic function $f$ on the open unit disc $\mathbb{D}$ such that it extends continuously to $\frac{1}{z}$ on $\partial\mathbb{D}$? I mean $f$ takes value $\frac{1}{z}$ when $|z|=1$ and continuous here.  Thank you.
11455
Let $D$ be a bounded domain and let $f$ be analytic function from $D$ into $D$. Show that if $z_{0}$ $\in D$ is a fixed point for $f$ , then $|f' (z_{0} )| \leq 1$.  WHAT I WAS THINKING:   To have a conformal mapping from $D$ to the unit disc &amp; after composing it with $f$ ; I was thinking to use Pick's Lemma. So, can anyone tell me any conformal map from any bounded domain to the unit disc?? OR Is there any other way out to solve this problem??
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.
36930
Let the domain $O=\mathbb{C}-(-\infty,0)$, the point $z_0 \in O$ and the circle $\gamma=C(0,r&lt;|z_0|)$ in the positive direction. Compute $\int_C \frac{\log z}{z-z_0} dz$.   How do I solve the problem with the Cauchy theorem knowing that the point $z_0$ is outside the interior of $\gamma$? Do I have to use the primitive of $\log z$, i.e. $\log' z = \frac{1}{z}$?  Thank you to bear in mind that I have not yet seen the Residue theorem.
31338
It is known as a fact that conformal maps on $\mathbb{R}^n \rightarrow \mathbb{R}^n$ for $n&gt;2$ are rotations, dilations, translations, and special transformations while conformal maps for $n=2$ are from a much wider class of maps, holomorphic/antiholomorphic maps. I was wondering to know if there is any topological or geometrical description for this.   To show what I mean, consider this example: in $\mathbb{R}^n$ for $n&gt;2$ interchanging particles can only change the wave function to itself or its minus. It is related to the fundamental group of $\mathbb{R}^n-x_0$ ($x_0$ is a point in $\mathbb{R}^n$ and $\pi_1(\mathbb{R}^n-\{x_0\})=e$ for $n&gt;2$) but this is not true for $n=2$.   I want to know whether exists any topological invariant or just any geometrical explanation that is related to the fact that I mentioned about conformal maps on $\mathbb{R}^n$.
31554
Let $n&gt;1$ , consider $\mathbb R^n$ with standard volume form $\omega _0 =dx_1 \wedge ...\wedge dx_n$ ; let $r,R$ be two positive real numbers with $R&gt;r$ . Then is it true that for every $a,b \in B(0;r)$ , $\exists$ a diffeomorphism $f:\mathbb R^n \to \mathbb R^n $ such that $f(a)=b ; f(x)=x , \forall x \in \mathbb R^n $ with $||x||&gt;R$ and $f^*\omega _0=\omega _0$ ?   As $f^*(dx_1 \wedge ...\wedge dx_n)=(det f' )(dx_1 \wedge ...\wedge dx_n)$ , so basically given $R&gt;r&gt;0$ and $||a||,||b||&lt;r$ , we have to find a diffeomorphism $f : \mathbb R^n \to \mathbb R^n  $ with $f(a)=b$ ; $f|_{\mathbb R^n \setminus B[0;R]}=Id$ and $det f'(p)=1 , \forall p \in \mathbb R^n $ .  Definitely this cannot work out in $\mathbb R$ , because the only differentiable function satisfying $det f'(p)=1 , \forall p$ and $f(a)=b$ is $f(x)=x+b-a$ and this function never gives identity unless $a =b$
29607
I am looking to a formula for the conformal map from the unit disc in the interior of an ellipse centered in $0$ and with semiaxes $a,b&gt;0$. I know that depends on elliptic function, but I didn't find any references.  Thanks in advance for any formula or reference.
78276
If $f$ is a harmonic function in a domain $D \subset \mathbb{C}$, and $g$ is a conformal mapping of a domain $D_0$ onto $D$, is $f \circ g$ harmonic in $D_0$?   I noticed this question while reading several pdf of lecture notes, and I'm having trouble figuring it out. Can anyone help?  Thank you so much!
88809
I refer to , which I understand says that the branch points of a nonconstant holomorphic map $F: X \to Y$ between Riemann surfaces $X$ and $Y$, both of which are not necessarily compact, form a discrete subset of range $Y$ of the map $F$. (Not sure if 'connected' is part of the definition of Riemann surfaces in other textbooks, but it is here.) Question 1: Do I understand correctly, and is this indeed true? What else I understand:  $A$ is a discrete subspace/subset (I just assume everything here is the usual subspace topology) of $B$ if and only if every singleton subset of $A$ is open in $A$ if and only if every point of $A$ is isolated.  This overflow question () seems to suggest this is not true but is true if $F$ is proper. However, I'm not sure the definitions of Riemann surface are the same (maybe some answers use definitions where Riemann surfaces are not necessarily connected or something).  2.1. I know $F$ is 'discrete' (meaning that its fibres are discrete; here, i refer to a ): see definition before Lemma 3.1, on p. 7), by Proposition II.3.12, and open, by open mapping theorem, which is Proposition II.3.8, but I'm not sure these help show $F$ is proper or are otherwise helpful. Ostensibly 'discrete' maps don't map discrete subsets to discrete subsets.   This stackexchange question () seems to suggest this is true.  I think this is true if $F$ is injective or at least something like locally injective. (Please don't make me type up all the stuff I did for this part.) However based on the question in (3) and based on the , which I think uses the term 'branch points' to mean the same thing as what Miranda means by 'ramification points' (and thus is different from Miranda's 'branch points'), it seems that no neighbourhood $U$ of a ramification point $p \in U \subseteq X$ of $F$ is such that the restriction $F|_U$ is injective.  Kinda related:    This page () says    Now, the ramification and branch points must form a discrete set  I don't think the page assumes anything like $F$ is proper.  The  text in (2.1) actually seems to assume proper based on the '3.7. Proper holomorphic maps' part, but said part refers to 'Lemma 3.17', so I think proper can be replaced with closed. Maybe $F$ is closed or something.  Kinda related (see Question 2 below):     says    If $F$ is proper then the image $\Delta = F(R)$ is discrete in $Y$.   This seems to have all the same assumptions as with Miranda. Ostensibly, either Miranda is wrong or there's some higher level machinery that allows us to omit $F$ proper.  Question 2: To possibly generalise this, what are the minimum requirements on $F, X$ and $Y$ to make $F$ map discrete subsets to discrete subsets? I mean, I'm not sure we really need 'holomorphic' here. I have a feeling this applies perhaps to maps that are just open/closed, continuous and discrete and spaces that are just Hausdorff and locally compact or something.  (again) Kinda related: