branch points of $\arcsin$

hbghlyj

en.wikipedia.org：
finite number of Riemann sheets for that branch point、infinite amount of Riemann sheets
Because of a branch point the function can take several values and we have

number of Riemann sheets = number of values of $f(z)$

A branch cut is an artificial line separating different Riemann sheets starting from the branch point. For the special case $f (z) = (z−z_0)^α$ if $α = M/N$ with $M,N$ coprime, then there are $N$ Riemann sheets, for the extension $f (z) = (z- z_1)^α(z - z_2)^β \dots$ we can choose multiple branch cuts, if the numbers $α, β,\dots$ are rationals the number of Riemann sheets is the lowest common denominator and if the powers of a subset of branch points adds up to an integer then we can join them by a finite branch cut. If instead of a product we have a sum the number of Riemann sheets is the product of the number for each term.
MSE–Branch point at infinity?
酷炫的用Three.js制作的黎曼曲面

• 0 is a branch point of the square root function.
• 0 is also a branch point of the natural logarithm.
• $i$ and $−i$ are branch points of the arctangent function $\arctan(z) = \frac1{2i}\log\left(i − z\over i + z\right)$.
• If the derivative $f'$ of a function $f$ has a simple pole at a point $a$, then $f$ has a logarithmic branch point at $a$. The converse is not true, since the function $f(z) = zα$ for irrational $α$ has a logarithmic branch point, and its derivative is singular without being a pole.
  

由欧拉公式得
\begin{equation}\arcsin (z)=-i \ln \left(i z+\left(1-z^{2}\right)^{\frac{1}{2}}\right)\label1\end{equation}

证明

$e^{i\arcsin(z)}=\cos(\arcsin(z))+i\sin(\arcsin(z))=\left(1-z^{2}\right)^{\frac{1}{2}}+iz$
$⟹i\arcsin(z)=\ln\left(\left(1-z^{2}\right)^{\frac{1}{2}}+iz\right)$
$⟹\arcsin (z)=-i \ln \left(i z+\left(1-z^{2}\right)^{\frac{1}{2}}\right)$

arcsin的branch point是±1和∞
证明
从\eqref{1}看出，由于分数指数1/2，在$z=±1$各有一个algebraic branch point。
由于 ln 在参数为零时有一个branch point，令\eqref{1}中 ln 的参数为零，
\begin{align*}
&i z+\left(1-z^2\right)^{\frac12}=0\\
\implies&{-i z}=\left(1-z^2\right)^{\frac12}\\
两边平方\implies&{-z^2}=1-z^2\;❌
\end{align*}
故\eqref{1}中 ln 的参数对于任何 $z$ 都不为零。
最后，代入 $z = 1/w$，$$i\arcsin (1/w)=\ln \left(i+\left(w^2-1\right)^{\frac{1}{2}}\right)-\ln w$$
当 $w\to0$，第一个 ln 的参数为零，所以 $\arcsin (1/w)$ 在 0 有一个logarithmic branch point。故 $\arcsin (z)$ 在 ∞ 有一个logarithmic branch point。

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Mathematica
In[]:= FunctionSingularities[ArcSin[z], z]
Out[]= z <= -1 || z >= 1
reference.wolfram.com/language/ref/ArcSin.html
ArcSin has both singularity and discontinuity in (-∞,-1] and [1,∞)

hbghlyj

或者使用$(\arcsin z)'=(1-z^2)^{-1/2}$的branch point是$\pm1$和∞，所以arcsin的branch point是$\pm1$和∞
Mathematica
In[]:= FunctionSingularities[(1 - z^2)^(-1/2), z]
Out[]= z^2 == 1 || z >= 1 || z <= -1
我有些不理解: 第一个 z^2==1 不就包含在 z >= 1 || z <= -1 里? 所以写 z >= 1 || z <= -1 就行吧.

hbghlyj

Singular point - Encyclopedia of Mathematics
For example, for the function $$
f(z)   =   
\frac{1}{(1+z ^{1/2} )(1 + z ^{1/6} )}
$$
the points $  a = 0,   \infty $ (for all curves) are algebraic branch points of order 5. As a point function, $  f(z) $ can be represented as a single-valued function only on the corresponding Riemann surface $  S $, consisting of 6 sheets over $  \overline{\mathbf C}  $ joined in a specific way above the points  $  0, \infty $. Moreover, three proper branches of  $  f(z) $ lie above the point $  a=1 $, which are single-valued on the three corresponding sheets of  $  S $; on one sheet of  $  S $ there is a pole of the second order, and on two sheets of $  S $ there are poles of the first order. In general, the introduction of the concept of a Riemann surface is particularly convenient and fruitful when studying the character of a singular point.