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单词 ProofOfMobiusCircleTransformationTheorem
释义

proof of Möbius circle transformation theorem


Case 1: f(z)=az+b.

Case 1a: The points on |z-C|=R can be written as z=C+Reiθ. They are mapped to the points w=aC+b+aReiθ which all lie on the circle |w-(aC+b)|=|a|R.

Case 1b: The line Re(eiθz)=k are mapped to the line Re(eiθwa)=k+Re(ba).

Case 2: f(z)=1z.

Case 2a: Consider a circle passing through the origin. This can be written as |z-C|=|C|. This circle is mapped to the line Re(Cw)=12 which does not pass through the origin. To show this, write z=C+|C|eiθ. w=1z=1C+|C|eiθ.

Re(Cw)=12(Cw+Cw¯)=12(CC+|C|eiθ+C¯C¯+|C|e-iθ)
=12(CC+|C|eiθ+C¯C¯+|C|e-iθeiθeiθC/|C|C/|C|)=12(CC+|C|eiθ+|C|eiθ|C|eiθ+C)=12

Case 2b: Consider the line which does not pass through the origin.This can be written as Re(az)=1 for a0. Thenaz+az¯=2 which is mapped to aw+a¯w¯=2. This is simplified asaw¯+a¯w=2ww¯ which becomes(w-a/2)(w¯-a¯/2)=aa¯/4 or|w-a2|=|a|2 which is a circle passing through the origin.

Case 2c: Consider a circle which does not pass through the origin. This can be written as |z-C|=R with |C|R. This circle is mapped to the circle

|w-C¯|C|2-R2|=R||C|2-R2|

which is another circle not passing through the origin. To show this, we will demonstrate that

C¯|C|2-R2+C-zRz¯zR|C|2-R2=1z

Note:|C-zRz¯z|=1.

C¯|C|2-R2+C-zRz¯zR|C|2-R2=zC¯-zz¯+z¯Cz(|C|2-R2)
=CC¯-(z-C)(z¯-C¯)z(|C|2-R2)=|C|2-R2z(|C|2-R2)=1z

Case 2d: Consider a line passing through the origin. This can be written as Re(eiθz)=0. This is mapped to the set Re(eiθw)=0, which can be rewritten as Re(eiθw¯)=0 or Re(we-iθ)=0 which is another line passing through the origin.

Case 3: An arbitrary Mobius transformationMathworldPlanetmath can be written as f(z)=az+bcz+d. If c=0, this falls into Case 1, so we will assume that c0. Let

f1(z)=cz+d  f2(z)=1z  f3(z)=bc-adcz+ac

Then f=f3f2f1. By Case 1, f1 and f3 map circles to circles and by Case 2, f2 maps circles to circles.

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更新时间:2025/5/5 2:36:53