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9

One can give a very simple explicit counterexample (which I would imagine is a special case of the "complete treatment" mentioned in the other answer). Note first that if $\Re z<0$ then $\Re(1/z)<0$. Let $\Bbb D$ be the unit disk. If $z\in \Bbb D$ then $\Re(1/(z-1))<0$, hence $$\left|e^{\frac1{z-1}}\right|\le1.$$ And so there you are. Let ...


4

Let me explain how to create the most extreme counterexample: The function $$ f(z)=\sum_{n=0}^\infty z^{n!} $$ is analytic in the unit disk $\mathbb D$ and it can not be extended analytically to any region $\Omega$, with $\mathbb D\subsetneq \Omega$ (region=open & connected). This makes the unit circle $\partial\mathbb D$ the natural boundary of $f$. ...


3

As I've said on a comment above, for such a counterexample, it suffices to consider the domain as the unit disk, and take $f$ to be a generic function in $H^{\infty}$ whose boundary value is not analytic. Then we'll have the original function bounded, but no extension is possible, in such a way that we may take $g$ to be simply a large constant. A complete ...


2

If $f : \mathbb{C} \to D(0, 1)$ is a holomorphic function, then it is entire (its domain is $\mathbb{C}$), and it is bounded (its codomain is $D(0, 1) = \{w \in \mathbb{C} \mid |w| < 1\}$). By Liouville's theorem, $f$ must be constant. As constant functions are not one-to-one, there is no one-to-one holomorphic map $f : \mathbb{C} \to D(0, 1)$.


1

Usually this kind of equation can be recovered from the nullity of a well chosen determinant. If $y = T(x)$ then $cxy+dy-ax-b = 0$, so the graph of an homography is given by a linear combination between $1,x,y$ and $xy$. As a result, as long as everyone stays in $\Bbb C$, four pairs $(x_1,y_1) \ldots (x_4,y_4)$ are on the graph of a single homography if ...


1

This is really a comment but I do not yet have enough reputation to do so. The most important thing to note is that for any non-zero complex number $\lambda$, $$ \frac{a z + b}{cz + d} = \frac{\lambda a z + \lambda b}{\lambda c z + \lambda d}. $$ As such, you may assume without loss of generality that $ad - bc =1$. This then gives you four equations in four ...


1

$$\langle f,f \rangle=\iint_Df(z)\times\overline{f(z)}dxdy=\iint_D |f(z)|^2dxdy \leqslant 3\pi$$ Power series of $f(z)$ about $z=0$: $$\sum_{n=0}^{\infty}a_nz^n=\sum_{k=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^n$$ then you can easily prove (using polar coordinates, edit: or more easily by defining a base using $z^n$ then using the generalized form of Parseval's ...


1

Hint: Write $f$ as a power series, use polar coordinates on $\int_D |f|^2,$ and use the orthonality of the exponentials.


1

For any $\epsilon >0$ if $\vert z\vert <\epsilon $ then $\vert x \vert ,\vert y\vert < \epsilon $ and a routine calculation shows that $\vert f(z)-0\vert ^{2}=\left | \frac{x^{3}-y^{3}+i(x^{3}+y^{3})}{x^2+y^2} \right |^{2}=\frac{(x^{3}-y^{3})^{2}+(x^{3}+y^{3})^{2}}{(x^{2}+y^{2})^{2}}=2\frac{x^{6}+y^{6}}{(x^{2}+y^{2})^{2}}\leq 2\cdot ...


1

The difference is that in the first case some $\delta >0$ is fixed, and then a set $S_{\delta}= \{s\in\mathbb C:\Re(s)\ge 1+\delta\}$ is considered. This set is for every $\delta$ a strict subset of $S= \{s\in\mathbb C:\Re(s)>1\}$. For example, $S_{\delta}$ does not contain $1+\delta/2 $ or $1+(\delta/9) + 45i$ and so on. Of course the union of ...


1

Say that $f$ and $g$ equivalent, if for two (holomorphic) homeomorphisms $h_1,h_2:\mathbb CP^1\to\mathbb CP^1$ satisfied $h_2\circ f\circ h_1=g$. Maybe, in this sense?



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