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• Foundations of Data Science 数据科学基础

## 数学代写|复分析作业代写Complex function代考|Linear Fractional ‘fransformations

The automorphisms (i.e., conformal self-mappings) of the unit disc $D$ are special cases of functions of the form
$$z \mapsto \frac{a z+b}{c z+d}, a, b, c, d \in \mathbb{C} .$$
It is worthwhile to consider functions of this form in generality. One restriction on this generality needs to be imposed, however; if $a d-b c=0$, then the numerator is a constant multiple of the denominator if the denominator is not identically zero. So if $a d-b c=0$, then the function is either constant or has zero denominator and is nowhere defined. Thus only the case $a d-b c \neq 0$ is worth considering in detail.
Definition 6.3.1. A function of the form
$$z \mapsto \frac{a z+b}{c z+d}, \quad a d-b c \neq 0,$$
is called a linear fractional transformation.
Note that $(a z+b) /(c z+d)$ is not necessarily defined for all $z \in \mathbb{C}$. Specifically, if $c \neq 0$, then it is undefined at $z=-d / c$. In case $c \neq 0$,
$$\lim _{z \rightarrow-d / c}\left|\frac{a z+b}{c z+d}\right|=+\infty \text {. }$$
This observation suggests that one might well, for linguistic convenience, adjoin formally a “point at $\infty$ ” (as in Section 4.7) to $\mathbb{C}$ and consider the value of $(a z+b) /(c z+d)$ to be $\infty$ when $z=-d / c(c \neq 0)$. It turns out to be convenient to think of a transformation $z \mapsto(a z+b) /(c z+d)$ as a mapping from $\mathbb{C} \cup{\infty}$ to $\mathbb{C} \cup{\infty}$ (again along the line of ideas introduced in Section 4.7). Specifically, we are led to the following definition:

Definition 6.3.2. A function $f: \mathbb{C} \cup{\infty} \rightarrow \mathbb{C} \cup{\infty}$ is a linear fractional transformation if there exist $a, b, c, d \in \mathbb{C}, a d-b c \neq 0$, such that either
(i) $c=0, f(\infty)=\infty$, and $f(z)=(a / d) z+(b / d)$ for all $z \in \mathbb{C}$, or
(ii) $c \neq 0, f(\infty)=a / c, f(-d / c)=\infty$, and $f(z)=(a z+b) /(c z+d)$ for all $z \in \mathbb{C}, z \neq-d / c$.

## 数学代写|复分析作业代写Complex function代考|Statement and Idea of Proof

The conformal equivalence of open sets in $\mathbb{C}$ is generally not very easy to compute or to verify, even in the special case of sets bounded by (generalized) circular arcs. For instance, it is the case (as will be verified soon) that the square ${z=x+i y:|x|<1,|y|<1}$ is conformally equivalent to the unit disc ${z:|z|<1}$. But it is quite difficult to write down explicitly the conformal equivalence. Even when a biholomorphic mapping is known (from other considerations) to exist, it is generally quite difficult to find it. Thus there is a priori interest in demonstrating the existence of the conformal equivalence of certain open sets by abstract, nonconstructive methods. In this section we will be concerned with the question of when an open set $U \subseteq \mathbb{C}$ is conformally equivalent to the unit disc.

Several restrictions must obviously be imposed on $U$. For instance, $U$ obviously cannot be $\mathbb{C}$ because every holomorphic function $f: U \rightarrow D$ would then be constant (Liouville’s theorem). Also, $U$ must be topologically equivalent (i.e., homeomorphic) to the unit disc $D$ since we are in fact demanding much more in asking that $U$ be conformally equivalent. For instance, $D$ could not be conformally equivalent to ${z: 1<|z|<2}$ since it is not topologically (i.e., homeomorphically) equivalent. [Although at the moment we have no precise proof that the two are not homeomorphic, this assertion is at least intuitively plausible.]

Surprisingly, the two restrictions just indicated (that $U$ not be $\mathbb{C}$ and that $U$ be topologically equivalent to $D$ ) are not only necessary but they are sufficient to guarantee that $U$ be conformally equivalent to the disc $D$.
To give this notion a more precise formulation, we first define formally what we want to mean by topological (homeomorphic) equivalence (reiterating the definition in Section 6.3, prior to Theorem 6.3.4):

Definition 6.4.1. Two open sets $U$ and $V$ in $\mathbb{C}$ are homeomorphic if there is a one-to-one, onto, continuous function $f: U \rightarrow V$ with $f^{-1}: V \rightarrow U$ also continuous. Such a function $f$ is called a homeomorphism from $U$ to $V$.

# 复分析代写

## 数学代写|复分析作业代写Complex function代考|Linear Fractional ‘fransformations

$$z \mapsto \frac{a z+b}{c z+d}, a, b, c, d \in \mathbb{C} .$$

$$z \mapsto \frac{a z+b}{c z+d}, \quad a d-b c \neq 0,$$

$$\lim _{z \rightarrow-d / c}\left|\frac{a z+b}{c z+d}\right|=+\infty .$$

(ii) $c \neq 0, f(\infty)=a / c, f(-d / c)=\infty$ ，和 $f(z)=(a z+b) /(c z+d)$ 对所有人 $z \in \mathbb{C}, z \neq-d / c$.

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## MATLAB代写

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