# 数学代写|复分析作业代写Complex function代考|A Puzzling Phenomenon

## 数学代写|复分析作业代写Complex function代考|A Puzzling Phenomenon

Throughout Chapter 2 we witnessed a very strange phenomenon. Whenever we generalized a familiar real function to a corresponding complex function, the mapping sent infinitesimal squares to infinitesimal squares. At present this is a purely empirical observation based on using a computer to draw pictures of the mappings. In this chapter we begin to explore the theoretical underpinnings of the phenomenon.

Let’s go back and take a closer look at a simple mapping like $z \mapsto w=z^2$. As we already know, this maps the origin-centred circle $|z|=r$ into the circle $|w|=r^2$, and it maps the ray $\arg (z)=\theta$ into the ray $\arg (w)=2 \theta$. An obvious consequence of this is that the right angle of intersection between such circles and rays in the $z$-plane is preserved by the mapping, which is to say that their images in the $w$-plane also meet at right angles. As illustrated in [4.1], a grid of infinitesimal squares formed from such circles and rays must therefore be mapped to an image grid composed of infinitesimal rectangles. However, this does not explain why these image rectangles must again be squares.

As we will explain shortly, the fact that infinitesimal squares are preserved is just one consequence of the fact that $z \mapsto w=z^2$ is conformal everywhere except at the two critical points $z=0$ and $z=\infty$, where angles are doubled. In particular, any pair of orthogonal curves is mapped to another pair of orthogonal curves. In order to give another example of this, we first dismember our mapping into its real and imaginary parts. Writing $z=x+i y$ and $w=u+i v$, we obtain
$$u+i v=w=z^2=(x+i y)^2=\left(x^2-y^2\right)+i 2 x y .$$
Thus the new coordinates are given in terms of the old ones by
\begin{aligned} & u=x^2-y^2, \ & v=2 x y . \end{aligned}
We now forget (temporarily!) that we are in $\mathbb{C}$, and think of (4.1) as simply representing a mapping of $\mathbb{R}^2$ to $\mathbb{R}^2$. If we let our point $(x, y)$ slide along any of the rectangular hyperbolas with equation $2 x y=$ const., then we see from (4.1) that its image $(u, v)$ will move on a horizontal line $v=$ const. Likewise, the preimages of the vertical lines $u=$ const. will be another family of rectangular hyperbolas with equations $\left(x^2-y^2\right)=$ const. Since their images are orthogonal, the claimed conformality of $z \mapsto z^2$ implies that these two kinds of hyperbolas should themselves be orthogonal.

## 数学代写|复分析作业代写Complex function代考|Local Description of Mappings in the Plane

Referring to [4.3], it’s clear that to find out whether any given mapping is conformal or not will require only a local investigation of what is happening very near to the intersection point q. To make this clearer still, recognize that if we wish to measure $\phi$, or indeed even define it, we need to draw the tangents [dotted] to both curves and then measure the angle between them. We could draw a very good approximation to one of these tangents simply by joining $q$ to any nearby point $p$ on the curve. Of course the nearer $p$ is to $q$, the better will the chord $q p$ approximate the actual tangent. Since we are only concerned here with directions and angles (rather than positions) we may dispense with the tangent itself, and instead use the infinitesimal vector $\overrightarrow{\mathrm{qp}}$ that points along it. Likewise, after we have performed the mapping, we are not interested in the positions of the image points $Q$ and $P$ themselves; rather, we want the infinitesimal connecting vector $\overrightarrow{Q P}$ that describes the direction of the new tangent at $Q$. We will call this infinitesimal vector $\overrightarrow{\mathrm{QP}}$ the image of the vector $\overrightarrow{\mathrm{qP}}$. However natural this terminology may seem, note that this really is a new sense of the word “image”.

Let us now summarize our strategy. Given formulae such as (4.1), which describe the mapping of the points to their image points, we wish to discover the induced mapping of infinitesimal vectors emanating from a point $q$ to their image vectors emanating from the image point $Q$. In principle, we could then apply the latter mapping to $\overrightarrow{q P}$ and to $\overrightarrow{q S}$, yielding their images $\overrightarrow{Q P}$ and $\overrightarrow{Q S}$, and hence the angle of intersection of the image curves through $Q$.

# 复分析代考

## 数学代写|复分析作业代写Complex function代考|A Puzzling Phenomenon

$$u+i v=w=z^2=(x+i y)^2=\left(x^2-y^2\right)+i 2 x y .$$

$$u=x^2-y^2, \quad v=2 x y .$$

## 数学代写|复分析作业代写Complex function代考|Local Description of Mappings in the Plane

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