在Abel's Theorem中的Stolz sector 切线斜率

hbghlyj

Counter example for Stolz sector in complex version of Abel's Theorem
abel.pdf
In search for a counterexample related to the Abel-Stolz theorem
Wolfram demonstrations
求证曲线$$\frac{| 1-z| }{1-| z|}=K$$在$z=1$的切线斜率为$±\sqrt{K^2-1}$.

K=1.3; Show[{Graphics[Circle[]], RegionPlot[Sqrt[(x - 1)^2 + y^2]/(1 - Sqrt[x^2 + y^2]) <=K&&x^2+y^2<=1, {x,-1,1}, {y,-1, 1}], Graphics[Line[{{1,0}-Normalize@{1,Sqrt[K^2-1]},{1,0},{1,0}-Normalize@{1,-Sqrt[K^2-1]}}]]}]

数值验证:

1. y/0.0001 /. NSolve[Sqrt[(x - 1)^2 + y^2]/(1 - Sqrt[x^2 + y^2]) == K /. x -> 0.9999, y][[1]]
2. 0.830592
3. Sqrt[K^2 - 1]
4. 0.830592

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hbghlyj

$$z=\frac{w-1}{w+1}$$代入$\frac{| 1-z| }{1-| z|}=K$得$$|w+1|-|w-1|=\frac2K$$这是一条以$±1$为焦点的双曲线的一支,它在$w=∞$的两条切线的夹角为$2\tan^{-1}\sqrt{K^2-1}$,根据Möbius变换的保角性质,原曲线在$z=1$的两条切线的夹角为$2\tan^{-1}\sqrt{K^2-1}$.证毕.

hbghlyj

$|w+1|-|w-1|=\frac2K$,这是一条以$±1$为焦点的双曲线的一支,它在$w=∞$的两条切线的夹角为$2\tan^{-1}\sqrt{K^2-1}$,

如图, $-1,1,w$分别对应点$A,B,C$, 则$AB=2$.
点$D$在$AC$上使$BC=DC$, 则$AD=\frac2K$.
当$|w|$很大时, $∠ADB=90°$, 则$\sec∠BAD=\frac{AB}{AD}=K⇒\tan∠BAD=\sqrt{\sec^2∠BAD-1}=\sqrt{K^2-1}$.