数学代写|复分析作业代写Complex function代考|Möbius Transformations: Basic Results

数学代写|复分析作业代写Complex function代考|Non-Uniqueness of the Coefficients

To specify a particular Möbius transformation $M(z)=\frac{a z+b}{c z+d}$ it seems that we need to specify the four complex numbers $a, b$, c, and d, which we call the coefficients of the Möbius transformation. In geometric terms, this would mean that to specify a particular Möbius transformation we would need to know the images of any four distinct points. This is wrong.
If $k$ is an arbitrary (non-zero) complex number then
$$\frac{a z+b}{c z+d}=M(z)=\frac{k a z+k b}{k c z+k d} .$$
In other words, multiplying the coefficients by k yields one and the same mapping, and so only the ratios of the coefficients matter. Since three complex numbers are sufficient to pin down the mapping- $(a / b),(b / c),(c / d)$, for example-we conjecture (and later prove) that
There exists a unique Möbius transformation sending any three points
$(3.25)$ to any other three points.
In the course of gradually establishing this one result we shall be led to further important properties of Möbius transformations.

If you read the last section of Chapter 1 , then (3.25) may be ringing a bell: the similarity transformations needed to do Euclidean geometry are also determined by their effect on three points. Indeed, we saw in that chapter that such similarities can be expressed as complex functions of the form $f(z)=a z+b$, and so they actually are Möbius transformations, albeit of a particularly simple kind. However, for such a similarity to exist, the image points must form a triangle that is similar to the triangle formed by the original points. But in the case of Möbius transformations there is no such restriction, and this opens the way to more flexible, non-Euclidean geometries in which Möbius transformations play the role of the “motions”. This is the subject of Chapter 6.

数学代写|复分析作业代写Complex function代考|The Group Property

In addition to preserving circles, angles, and symmetry, the mapping
$$z \mapsto w=M(z)=\frac{a z+b}{c z+d} \quad(a d-b c) \neq 0$$ is also one-to-one and onto. This means that if we are given any point $w$ in the $w$ plane, there is one (and only one) point $z$ in the $z$-plane that is mapped to $w$. We can show this by explicitly finding the inverse transformation $w \mapsto z=M^{-1}(w)$. Solving the above equation for $z$ in terms of $w$, we find [exercise] that $M^{-1}$ is also a Möbius transformation:
$$M^{-1}(z)=\frac{d z-b}{-c z+a}$$
Note that if $M$ is normalized, then this formula for $M^{-1}$ is automatically normalized as well.

If we look at the induced mapping on the Riemann sphere, then we find that a Möbius transformation actually establishes a one-to-one correspondence between points of the complete $z$-sphere and points of the complete $w$-sphere, including their points at infinity. Indeed you may easily convince yourself that
$$M(\infty)=(a / c) \text { and } M(-d / c)=\infty$$
Using (3.26), you may check for yourself that $M^{-1}(a / c)=\infty$ and $M^{-1}(\infty)=$ $-(\mathrm{d} / \mathrm{c})$

Next, consider the composition $M \equiv\left(M_2 \circ M_1\right)$ of two Möbius transformations,
$$M_2(z)=\frac{a_2 z+b_2}{c_2 z+d_2} \quad \text { and } \quad M_1(z)=\frac{a_1 z+b_1}{c_1 z+d_1}$$
A simple calculation [exercise] shows that $M$ is also a Möbius transformation:
$$M(z)=\left(M_2 \circ M_1\right)(z)=\frac{\left(a_2 a_1+b_2 c_1\right) z+\left(a_2 b_1+b_2 d_1\right)}{\left(c_2 a_1+d_2 c_1\right) z+\left(c_2 b_1+d_2 d_1\right)}$$
It is clear geometrically that if $M_1$ and $M_2$ are non-singular, then so is $M$. This is certainly not obvious algebraically, but later in this section we shall introduce a new algebraic approach that does make it obvious.

复分析代考

数学代写|复分析作业代写Complex function代考|Non-Uniqueness of the Coefficients

$$\frac{a z+b}{c z+d}=M(z)=\frac{k a z+k b}{k c z+k d}$$

$(3.25)$ 任何其他三点。

数学代写|复分析作业代写Complex function代考|The Group Property

$$z \mapsto w=M(z)=\frac{a z+b}{c z+d} \quad(a d-b c) \neq 0$$

$$M^{-1}(z)=\frac{d z-b}{-c z+a}$$

$$M(\infty)=(a / c) \text { and } M(-d / c)=\infty$$

$$M_2(z)=\frac{a_2 z+b_2}{c_2 z+d_2} \quad \text { and } \quad M_1(z)=\frac{a_1 z+b_1}{c_1 z+d_1}$$一个简单的计算[练习]表明 $M$ 也是莫比乌斯变换:
$$M(z)=\left(M_2 \circ M_1\right)(z)=\frac{\left(a_2 a_1+b_2 c_1\right) z+\left(a_2 b_1+b_2 d_1\right)}{\left(c_2 a_1+d_2 c_1\right) z+\left(c_2 b_1+d_2 d_1\right)}$$

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