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$\log \left(x^2+\sqrt{1-x^2}\right)$ branch point

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hbghlyj Posted at 2024-6-4 02:49:26 |Read mode
XbXEd[1].png
mathematica.stackexchange.com/a/169283/87896
download.gif
wolframalpha.com/input?i=branch+point+Log%5Bx … rt%5B1+-+x%5E2%5D%5D
Czhang271828 发表于 2023-4-14 13:13
数学一般用"分歧", 不用"分岐". 差点没找到.

$\log \left(x^2+\sqrt{1-x^2}\right)$有4个branch point
$$\left\{\pm1,\pm i \sqrt{\phi}\right\}\quad\phi=\frac{1}{2}(1+\sqrt{5})$$
我觉得这里$\pm1$是2次ramification point,而$\pm i \sqrt{\phi}$是logarithmic branch point,不知对不对?

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 Author| hbghlyj Posted at 2024-6-4 02:55:52
en.wikipedia.org/wiki/Branch_point#Riemann_surfaces
Let γ be a simple rectifiable loop in 
X around P. The ramification index of ƒ at P is
$\displaystyle e_{P}={\frac {1}{2\pi i}}\int _{\gamma }{\frac {f'(z)}{f(z)-f(P)}}\,dz.$

This integral is the number of times ƒ(γ) winds around the point Q. As above, P is a ramification point and Q is a branch point if eP > 1.

$\sqrt{1+\sqrt{1+x^2}}$的branch point是$\pm i,\tilde\infty$
https://www.wolframalpha.com/inp ... 5B1+%2B+x%5E2%5D%5D
download (1).gif
我覺得$\pm i,\tilde\infty$都是2次ramification point,不知对不对?


$\sqrt{1-\sqrt{1+x^2}}$的branch point是$0,\pm i,\tilde\infty$
https://www.wolframalpha.com/inp ... 5B1+%2B+x%5E2%5D%5D
download (2).gif
我覺得$0,\pm i,\tilde\infty$都是2次ramification point,不知对不对?

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