I need to show that a) every Möbius transformation that maps the disk onto itself can be written as $g(z)=e^{i\theta}\frac{z-z_0}{\bar{z_0}z-1}$, and b) that this function maps the unit disk onto the unit disk. I'm currently trying to make sense of the last part. Taking modules I get the following result: $|z|^2+|z_0|^2&lt;1+|z||z_0|$, and I have been stuck there since. How can I prove this inequality given that $|z|,|z_0|&lt;1$?
12873
This is the question I'm stumbling with:   When $|\alpha| &lt; 1$ and $|\beta| &lt; 1$, show that:  $$\left|\cfrac{\alpha - \beta}{1-\bar{\alpha}\beta}\right| &lt; 1$$    The chapter that contains this question contains (among others) the triangle inequalities:   $$\left||z_1| - |z_2|\right| \le |z_1 + z_2| \le |z_1| + |z_2| $$  I've tried to use the triangle inequalities to increase the dividend and/or decrease the divisor:  $$\left|\cfrac{\alpha - \beta}{1-\bar{\alpha}\beta}\right| &lt; \cfrac{|\alpha| +|\beta|}{\left|1-|\bar{\alpha}\beta|\right|}$$  But it's not clear if or why that would be smaller than one.  I've also tried to multiply the equation by the conjugated divisor $\cfrac{1-\alpha\bar{\beta}}{1-\alpha\bar{\beta}}$, which gives a real divisor, but the equation does not appear solvable.  Any hint would be much appreciated.
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.
110270
Problem: Let $f(z)$ be a meromorphic function on the complex torus $\mathbb C/\Lambda$ that as a function on $\mathbb C$ has no zeros and no poles on $\partial P$, the boundary of the fundamental parallelogram $P$. Show that  \begin{equation}                 \frac{1}{2 \pi i}\int_{\partial P}z\frac {f'(z)} {f(z)} dz \in \Lambda \, . \end{equation}  Thoughts: By the Residue Theorem  \begin{equation} \frac{1}{2 \pi i}\int_{\partial P}z\frac {f'(z)} {f(z)} dz = \sum_{z_0 \in \text{ Int }P} v_{z_0}(f)z_0. \end{equation}  I don't know why the latter is on the lattice.  Thanks!
121454
I am thinking about the following problem:     Let $f(z)=2z^2-1.$Then what is  the maximum value of $|f(z)|$ on the unit disc $D=\{z \in \mathbb C:|z|\leq 1\}$ ?   I guess i have to use the maximum modulus principle.But  i also notice that $|f(z)|=|2z^2-1|\leq 2|z|^2+1\leq 2+1=3.$  So, is $3$  the maximum value of $|f(z)|?$ Can someone point me 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 $?
44315
Given a Möbius transformation that maps one pair of concentric circles to another pair of concentric circles, why is the ratio of the radii preserved through the map?   I thought about how Möbius transformations are compositions of rotations, scaling, inversion, and translation, and that intuitively, these types of maps shouldn't change the ratio of radii between two circles.   Would it be correct to just say that if $\frac{r_1}{r_2}$ is the ratio of radii between the two circles, then   1) The radii are invariant under translation, $z \mapsto z+a$, so $\frac{r_1}{r_2}$ stays the same  2) Under scaling by a factor $z \mapsto az$, $\frac{ar_1}{ar_2} = \frac{r_1}{r_2}$   3) Under inversion, $z \mapsto \frac{1}{z}$, $\frac{1/r_1}{1/r_2} = \frac{r_2}{r_1}$  Or is there a different/better way to think about this problem?
32979
Let $f(z) = z + g(z)$ where $g$ is holomorphic. Suppose that $|\operatorname{Im} g(z)| &lt; 1$ for $z ∈ [−1 − i, 1 − i]∪[−1 + i, 1 + i]$ and $|\operatorname{Re} g(z)| &lt; 1$ for $z ∈ [−1 − i, −1 + i] ∪ [1 − i, 1 + i]$.   Show that $f$ has exactly one zero on the square $Q =$ {$x + iy ∈ \Bbb C : |x| &lt; 1, |y| &lt; 1$}.  My attempt:  I let $h(z) = z$. Then, I want to compare $|g(z)|$ and $|h(z)|$ because if $|g(z)| &lt; |h(z)|$ then by Rouché's theorem, $h$ and $h+g$ have the same number of zeros, and $h$ has in fact one zero. But then $h+g = f$ and thus $f$ would also have the same number of zeros as $h+g$ which has one zero.  This is what I could come up with: $|g(z)| = |u(z) + iv(z)|$. Then for $z \in Q$, we have $|g(z)| \leq |u(z)| + |v(z)| &lt; 1 + 1 = 2$ (since $|\operatorname{Im} g(z)| &lt; 1$ for $z ∈ [−1 − i, 1 − i]∪[−1 + i, 1 + i]$ and $|\operatorname{Re} g(z)| &lt; 1$ for $z ∈ [−1 − i, −1 + i] ∪ [1 − i, 1 + i]$)  But I don't know how to continue from here. Any help please?
32927
Let $D^n$ be the closed unit disk in $\mathbb{R}^n$. I'm looking for a homeomorphism of $D^n$ to itself which is the identity on the boundary, and takes the origin to a given point $z$ in its interior. I can visualize such a function as "shifting around the mass" within $D^n$, but I can't seem to write one down.
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!
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.)
90763
Let $z_0$ be a point in an open set $U \subset \mathbb{C}$.  Let $f,g:U \to \Delta(0,1)$ where $g$ is a bijection such that $g(z_0)=0$ and $g'(z_0)\neq 0$. Suppose that $f(z_0)=0$, then $|f'(z_0)|\leq |g'(z_0)|$  I am not even sure how to start this.  I was assuming Schwarz lemma.
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.
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.
11603
Let $f(z)=\sum_{n=0}^{\infty}a_nz^n$ be a holomophic function in the unit disc. Show that if $$\sum_{n=2}^{\infty} n\vert a_n \vert &lt;\vert a_1\vert$$ and $a_1\not = 0$, then $f$ is injective.   I am able to show that $f'$ never vanishes by showing that $\vert f'-a_1\vert &lt;\vert a_1\vert$, but that doesn't imply that $f$ is injective. For example, $e^z$ is not inejective but its derivative never vanishes. What is required to prove that $f$ is injective? How do you show $f$ is injective on any set?
12548
By considering the conjugate of its parametric form, evaluate $$\frac{1}{2\pi i}\int_{\gamma(0;1)}\frac{\overline{f(z)}}{z-a}dz$$ when $|a|&lt;1$ and $|a|&gt;1$, where $f$ is holomorphic in in the disk $(0;R),  R&gt;1$.  Typically when doing these kinds of integration and parametrization, $|z|=n$ is given, but it's different in this case (or is it not?). Can someone help me out?
122888
This is a problem from my complex analysis textbook.  The hint is to consider $g(z)=(z-iR)^2(z+iR)^2 f(z)$ and to show that $|g(z)| \leq 8R^2$.  This is what i have tried: Consider $Re z \geq 0$, then $|(z-iR)^2 f(z)|\leq \left(\frac{|z-iR|}{Re z}\right)^2\leq 2$   But then I don't know what to do next. Any help I'd appreciate :)
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.