x和y互为有理函数,则为分式线性函数

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A,B,C,D是常数

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首先, 有理函数的次数定义为(既约)分子和分母的次数的最大值, 见Rational function #Degree:

There are several non equivalent definitions of the degree of a rational function.
Most commonly, the degree of a rational function is the maximum of the degrees of its constituent polynomials $P$ and $Q$, when the fraction is reduced to lowest terms. If the degree of $f$ is $d$, then the equation
$$f(z) = w$$
has $d$ distinct solutions in $z$ except for certain values of $w$, called critical values, where two or more solutions coincide or where some solution is rejected at infinity (that is, when the degree of the equation decrease after having cleared the denominator).

In the case of complex coefficients, a rational function with degree one is a Möbius transformation.

1楼的问题等价于: 有理函数的逆也是有理的,则是分式线性函数. 见Which kinds of rational functions of one variable have an inverse relation that contains a branch that is a rational function?

If $R$ is a rational function and $S$ a branch of $R^{-1}$ in an open set $U \subset \Bbb C$ then $S(R(z)) = z$ in $U$. If $S$ is also a rational function then it follows that $S(R(z)) = z$ globally (as meromorphic functions).

It follows that $S$ is injective and therefore has degree one. Then $R$ has degree one as well.

So the only rational functions with a (local) rational branch of the inverse are rational functions of degree one (which are the Möbius transformations).

最后, 一个分式线性函数$y=\frac{ax+b}{cx+d}$的逆总是一个分式线性函数$y=\frac{dx-b}{-cx+a}$.

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补充,其中的一步的证明:
$\Bbb C$上的有理函数为单射, 则为分式线性函数
Prove that injective rational function on complex numbers are mobius transformations

All we need is the fundamental theorem of algebra and a little trick. Write $p,q$ in lowest terms. Consider $p(x)-tq(x)$. Except at possibly one value of $t$, this is a polynomial of degree $\max(\deg p,\deg q)$ and except at finitely many more values of $t$, it has distinct roots (these values correspond exactly to the roots of the discriminant, a nonzero polynomial in $t$). Pick a $t=t_0$ which isn't one of the above finite values to avoid. Then $p(x)-t_0q(x)$ has $\max(\deg p,\deg q)$ distinct roots, which correspond to solutions of $p/q=t_0$. If $f$ is to be injective, then $\max(\deg p,\deg q) \leq 1$ and you have your result.

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courses.maths.ox.ac.uk/pluginfile.php/31198/m … mp;zoom=auto,-79,548
Remark 10.11. Note that the non-trivial part of the proof of the above theorem is the fact that $g$ is continuous! In fact the condition that $f^{\prime}(z) \neq 0$ follows from the fact that $f$ is bijective - this can be seen using the degree of $f$ : if $f^{\prime}\left(z_0\right)=0$ and $f$ is nonconstant, we must have $f(z)-f\left(z_0\right)=\left(z-z_0\right)^k g(z)$ where $g\left(z_0\right) \neq 0$ and $k \geq 1$. Since we can chose a holomorphic branch of $g^{1 / k}$ near $z_0$ it follows that $f(z)$ is locally $k$-to-1 near $z_0$, which contradicts the injectivity of $f$. For details see the Appendices. Notice that this is in contrast with the case of a single real variable, as the example $f(x)=x^3$ shows. Once again, complex analysis is "nicer" than real analysis!
若$f'(z_0)=\cdots=f^{(k-1)}(z_0)=0$, 则$f(z)$是局部的$k$对1函数

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hbghlyj 发表于 2022-3-24 17:26
补充,其中的一步的证明:
$\Bbb C$上的有理函数为单射, 则为分式线性函数

Why is every holomorphic bijection of the Riemann sphere a Möbius transformation?
Entire one-to-one functions are linear