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# 物理代写|电磁学代写Electromagnetism代考|PHYS122 Analytic Functions and Conformal Mappings

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## 物理代写|电磁学代写Electromagnetism代考|Analytic Functions and Conformal Mappings

A point in an $(x, y)$ plane can be addressed “uniquely” (i.e., unique in both directions) by a complex number
$$z=x+i y .$$
Notice that $z$ here has no relation to the third Cartesian coordinate, which is anyway irrelevant for plane problems. The complex number is a two-dimensional quantity, whose properties are quite similar to those of a two-dimensional vector. This, therefore, allows to identify $z$ with the vectors $\langle x, y\rangle$ (see Fig. 3.29).
$$z=x+i y \Leftrightarrow\langle x, y\rangle=\mathbf{r} .$$
In any case, the complex number $z$ and the vector $\mathbf{r}$ address the same point. It is also possible to use Vector Calculus on complex numbers, where those operations simplify in some sense. The scalar product, for instance,
$$\mathbf{r}_1 \bullet \mathbf{r}_2=\left\langle x_1, y_1\right\rangle \bullet\left\langle x_2, y_2\right\rangle=x_1 x_2+y_1 y_2,$$
and the vector product, or more precisely, its only component that does not vanish in the plane case is
$$\mathbf{r}_1 \times \mathbf{r}_2=\left\langle x_1, y_1\right\rangle \times\left\langle x_2, y_2\right\rangle=x_1 y_2-y_1 x_2 .$$

## 物理代写|电磁学代写Electromagnetism代考|The Complex Potential

When comparing the statements of the last two sections, one realizes that real part and imaginary part of an analytic function behave exactly like the potential and flux function of an electrostatic field. The reader is encouraged to compare eqs. (3.362) and (3.380), as well as (3.363), (3.364) and (3.386), (3.387). One may conclude that every analytic function can be regarded in an electrostatic way. Its real part $u$ can be identified with the potential $\varphi$ and its imaginary part $v$ with the flux function $\psi$ of the related field. In light of this point of view, the analytic function $w(z)$ is called complex potential.
$(3.391)$
The accompanying field is

$$\left.\begin{array}{l} E_x=-\frac{\partial u}{\partial x}=-\frac{\partial v}{\partial y} \ E_y=-\frac{\partial u}{\partial y}=+\frac{\partial v}{\partial x} \end{array}\right} .$$
Naturally, with $w(z)$ so is iw(z) an analytic function. The potential here is
$\begin{array}{|cc|}\tilde{w}(z)=i w(z)= & -v(x, y)+i u(x, y) \ \vdots & \vdots \ & \text { potential } \quad \text { flux function }\end{array}$
and the accompanying field is now
$$\left.\begin{array}{l} \tilde{E}_x=\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y} \ \tilde{E}_y=\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x} \end{array}\right}$$

## 物理代写|电磁学代写|电磁学代考|解析函数和共形映射

$$z=x+i y$$

$$z=x+i y\Leftrightarrow\langle x, y\rangle=\mathbf{r} 。$$

$$\ǞǞ2=left\langle x_1, y_1\right\rangle Ǟleft\langle x_2, y_2\right\rangle=x_1 x_2+y_1 y_2。$$

$$\mathbf{r}_1 \times \mathbf{r}_2=left\langle x_1, y_1\right\rangle \times\left\langle x_2, y_2\right\rangle=x_1 y_2-y_1 x_2 。$$

## 物理代写|电磁学代写电磁学代考|复数电势

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