# 数学代写|复分析作业代写Complex function代考|Stereographic Formulae

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## 数学代写|复分析作业代写Complex function代考|Stereographic Formulae

In this subsection we derive explicit formulae connecting the coordinates of a point $z$ in $\mathbb{C}$ and its stereographic projection $\widehat{z}$ on $\Sigma$. These formulae will prove useful in investigating non-Euclidean geometry, but if you don’t plan to study Chapter 6 then you should feel free to skip this subsection.

To begin with, let us describe $z$ with Cartesian coordinates: $z=x+i y$. Similarly, let $(X, Y, Z)$ be the Cartesian coordinates of $\widehat{z}$ on $\Sigma$; here the $X$ – and $Y$-axes are chosen to coincide with the $x$ – and $y$-axes of $\mathbb{C}$, so that the positive $Z$-axis passes through $\mathrm{N}$. To make yourself comfortable with these coordinates, check the following facts: the equation of $\Sigma$ is $X^2+Y^2+Z^2=1$, the coordinates of $N$ are $(0,0,1)$, and similarly $S=(0,0,-1), 1=(1,0,0), i=(0,1,0)$, etc.

Now let us find the formula for the stereographic projection $z=x+$ iy of the point $\widehat{z}$ on $\Sigma$ in terms of the coordinates $(X, Y, Z)$ of $\widehat{z}$. Let $z^{\prime}=X+i Y$ be the foot of the perpendicular from $\widehat{z}$ to $\mathbb{C}$. Clearly, the desired point $z$ is in the same direction as $z^{\prime}$, so
$$z=\frac{|z|}{\left|z^{\prime}\right|} z^{\prime}$$
Now look at [3.23a], which shows the vertical cross section of $\Sigma$ and $\mathbb{C}$ taken through $\mathrm{N}$ and $\widehat{z}$; note that this vertical plane necessarily also contains $z^{\prime}$ and $z$. From the similarity of the illustrated right triangles with hypotenuses $\mathrm{N} \widehat{\mathrm{z}}$ and $\mathrm{Nz}$, we immediately deduce [exercise] that
$$\frac{|z|}{\left|z^{\prime}\right|}=\frac{1}{1-Z^{\prime}}$$
and so we obtain our first stereographic formula:
$$x+i y=\frac{X+i \gamma}{1-Z}$$
Let us now invert this formula to find the coordinates of $\hat{z}$ in terms of those of z. Since [exercise]
$$|z|^2=\frac{1+Z}{1-Z}$$
we obtain [exercise]
$$X+i Y=\frac{2 z}{1+|z|^2}=\frac{2 x+i 2 y}{1+x^2+y^2}, \quad \text { and } \quad Z=\frac{|z|^2-1}{|z|^2+1}$$

## 数学代写|复分析作业代写Complex function代考|Preservation of Circles, Angles, and Symmetry

From (3.3) we know that a general Möbius transformation $M(z)=\frac{a z+b}{c z+d}$ can be decomposed into the following sequence of more elementary transformations: a translation, complex inversion, a rotation, an expansion, and a second translation. Since each of these transformations preserves circles, angles, and symmetry, we immediately deduce the following fundamental results:

• Möbius transformations map circles to circles. ${ }^9$
• Möbius transformations are conformal.
• If two points are symmetric with respect to a circle, then their images under a Möbius transformation are symmetric with respect to the image circle. This is called the “Symmetry Principle”.

We know that a circle $\mathrm{C}$ will map to a circle-of course lines are now included as “circles”-but what will happen to the disc bounded by C? First we give a useful way of thinking about this disc. Imagine yourself walking round $\mathrm{C}$ moving counterclockwise; your motion gives $\mathrm{C}$ what is a called a positive sense or orientation. Of the two regions into which this positively oriented circle divides the plane, the disc may now be identified as the one lying to your left.

Now consider the effect of the four transformations in (3.3) on the disc and on the positively oriented circle bounding it. Translations, rotations, and expansions all preserve the orientation of $C$ and map the interior of $C$ to the interior of the image $\widetilde{\mathrm{C}}$ of $\mathrm{C}$. However, the effect of complex inversion on $\mathrm{C}$ depends on whether or not $\mathrm{C}$ contains the origin. If $\mathrm{C}$ does not contain the origin, then $\widetilde{C}$ has the same orientation as $\mathrm{C}$, and the interior of $\mathrm{C}$ is mapped to the interior of $\widetilde{\mathrm{C}}$. This is easily understood by looking at [3.24].

If $\mathrm{C}$ does contain the origin then $\widetilde{\mathrm{C}}$ has the opposite orientation and the interior of $\mathrm{C}$ is mapped to the exterior of $\widetilde{\mathrm{C}}$. If $\mathrm{C}$ passes through the origin then its interior is mapped to the half-plane lying to the left of the oriented line $\widetilde{\mathrm{C}}$. See [3.25].

# 复分析代写

## 数学代写|复分析作业代写Complex function代考|Stereographic Formulae

$$z=\frac{|z|}{\left|z^{\prime}\right|} z^{\prime}$$

$$\frac{|z|}{\left|z^{\prime}\right|}=\frac{1}{1-Z^{\prime}}$$

$$x+i y=\frac{X+i \gamma}{1-Z}$$

$$|z|^2=\frac{1+Z}{1-Z}$$

$$X+i Y=\frac{2 z}{1+|z|^2}=\frac{2 x+i 2 y}{1+x^2+y^2}, \quad \text { and } \quad Z$$

## 数学代写|复分析作业代写Complex function代考|Preservation of Circles, Angles, and Symmetry

• 莫比乌斯变换将圆映射到圆。 ${ }^9$
• 莫比乌斯变换是共形的。
• 如果两点关于圆对称，则它们在莫比乌斯变换下 的像关于像圆对称。这被称为“对称原理”。
我们知道一个圆C将映射到一个圆圈一一当然线现在被 包含为“圆圈”一一但是以 C 为界的圆盘会发生什么? 首 先，我们给出一种思考这张光盘的有用方法。想象自己 四处走动 $\mathrm{C}$ 逆时针移动; 你的议案给 $\mathrm{C}$ 什么叫做稆极的 感觉或方向。在这个正向圆将平面划分为两个区域中， 现在可以将圆盘识别为位于您左侧的那个。
现在考虑 (3.3) 中的四个变换对圆盘和包围它的正向圆的 影响。平移、旋转和扩展都保持方向 $C$ 并绘制内部图 $C$ 到图像的内部 $\widetilde{\mathrm{C}}$ 的C. 然而，复数反演对C取决于有没有 C包含原点。如果C不包含原点，那么 $\widetilde{C}$ 具有相同的方向 $\mathrm{C}$, 和内部 $\mathrm{C}$ 映射到的内部 $\widetilde{\mathrm{C}}$. 通过查看 [3.24] 可以很容 易地理解这一点。
如果C那么确实包含原点 $\widetilde{\mathrm{C}}$ 有相反的方向和内部C被映射 到外部 $\widetilde{C}$. 如果 $\mathrm{C}$ 通过原点然后它的内部被映射到位于定 向线左侧的半平面 $\widetilde{\mathrm{C}}$. 见[3.25]。

## 有限元方法代写

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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