linear fractional transformation

复习提纲

1. stereographic projection (definition and the way to find a projecting point)
2. calculate square root for a given complex number
3. triangle inequality
4. differentiation of a holomorphic function. (definition, Cauchy-Riemann equation, method to calculate a derivative for a given function , find a harmonic conjugate for a given real part)
5. Maximal muduli theorem (proof is not required, just need know how to use it )
6. write a rational function into a sum of partial fractions
7. linear transformation (cross ratio, the way to find a center of a circle decided by three points, how to decide if four points are on a same circle, symmetric points, reflection with respect to a circle, determine a linear transformation which can realize some transformations between circles)

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• 疑问：哪些函数连续，连续引申出来的性质有哪些？
• 疑问：Prove fundamental theorem of algbra using maximum mudulii theorem
• 疑问：extended complex plane: C ∪ {∞} |z|>1/ε is the neiborhood of ∞

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• linear transformation:

  1. definition of order
  2. rational functions like:

      aZ + b/cZ + d with ad ≠ bc. (0.3) 

     a rational function is called linear transformation if (0.3) holds. One of the most important properties of linear transformations is the theorem shown as follows

     1. Theorem 0.4 Linear transformation maps circles to circles.
     2. Linear transformation is invertible.
     3. Compsition of two linear transformations are also linear transformations.
     4. Proposition 0.7: Given three distinct points in the Riemann sphere, denoted by z2, z3 and z4, there is a unique linear transformation which maps (z2, z3, z4) to (1, 0, ∞).

        proof：Clearly Sz = (z − z3/z-z4)/(z2 − z3 /z2 − z4) is a linear transformation which maps (z2, z3, z4) to (1, 0, ∞). If S1 and S2 are two linear transformations which map (z ,z ,z ) to (1,0,∞), then by Propositions 0.5 and 0.6, S1S2−1 is a linear transformation and moreover it maps (1, 0, ∞) to (1, 0, ∞). Assume S1S2−1(z)= az+b/cz+d. then clearly S1S2−1(∞) = a/c = ∞. This fact shows that c = 0. Therefore we can assume S1S2−1(z) =a'z+b'. When z=0,it holds that S1S2−1(0)=b =0. When z=1,it holds that S1S2−1(1)=a =1.All the above arguments show that S1S2−1(z) = z for all z a complex number. In other words, S1S2−1 is an identity map. 

     5. Definition 0.8: Sz = (z − z3/z-z4)/(z2 − z3 /z2 − z4) (0.4). We also define (z, z2, z3, z4) to be the right-hand side of (0.4). Conventionally (z, z2, z3, z4) is called cross-ratio of the four numbers z,z2,z3 and z4. one should know that the value of the cross-ratio (z, z2, z3, z4) is evaluated as follows: using z2, z3 and z4, we can find a linear transformation by Proposition 0.7. We denote this linear transformation by S. The cross-ration is obtained by evaluating S at z.

        1. the cross ratio have the following properties:
        2. Proposition 0.9: For any linear transformation T , (T z1, T z2, T z3, T z4) = (z1, z2, z3, z4) (对z1,z2,z3,z4做linear transformation,不改变cross ratio)

            proof: Letting Sz = (z,z2,z3,z4), then one can show that ST−1 is a map which sends (Tz2,Tz3,Tz4) to (1, 0, ∞). Therefore by Proposition 0.7, we know that ST −1(w) = (w, T z2, T z3, T z4) for any complex number w. Setting w = Tz1, the proof is finished. 

        3. Proposition 0.10: Im(z1, z2, z3, z4) = 0 if and only if the four points z1, z2, z3 and z4 lie on the same circle or straight line.

            proof:we sketch the proof. If the four points lie on a same circle, then we know that the angle ∠z3z2z4 equals to the angle ∠z3z1z4. Clearly ∠z3z2z4 is given by the argument of (z2 −z3)/(z2 −z4). ∠z3z1z4 is given by the argument of (z1 − z3)/(z1 − z4). Therefore we know that arg((z2 − z3)/(z2 − z4)) = arg((z1 − z3)/(z1 − z4)). This equivalently shows that Im(z1, z2, z3, z4) = 0. 

        4. Proof of Theorem 0.4

            Fixing z2, z3, z4 on a circle C, T z2, T z3, T z4 also determine a circle, say C′. Here T is a linear transformation. Choosing z an arbitrary point on C, then by Proposition 0.10, we have Im(z, z2, z3, z4) = 0. Using Proposition 0.9, it also holds Im(T z, T z2, T z3, T z4) = Im(z, z2, z3.z4) = 0. Still by Proposition 0.10, Tz should lie on the circle C′. The proof is done. 

  3. Definition 1: Given z2,z3,z4, we can determine a line or a circle, denoted by C, passing these three points. If z is arbitrarily given, then z∗ is called the symmetric point of z with respect to the circle C if z* satisfies the following equation

      (z*, z2, z3, z4) = conjugate of (z, z2, z3, z4). 

     1. 如果z也在圆上，根据proposition 0.10 和 proposition 0.7那么应该有z*=z。
     2. Remark 1: if (z2′ , z3′ , z4′ ) and (z2, z3, z4) determine an identical circle C, then the two symmetric points given by Definition 1 are equal.
     3. Since cross ratio is invariant under linear transformations, it holds

         (Tz*,Tz2,Tz3,Tz4) = conjugate of (Tz,Tz2,Tz3,Tz4). 

        By the Definition 1, Tz∗ must be the symmetric point of Tz. Therefore, we conclude that
     4. Proposition 1:linear transformation maps symmetric pair to symmetric pair. More pre- cisely if (z, z∗) is a symmetric pair with respect to the circle determined by z2, z3 and z4, then (Tz,Tz*) is a symmetric pair with respect to the circle determined by Tz2,Tz3 and Tz4.
     5. In fact, the concept of symmetric point is not new to us. In the following arguments, we still use the notations in Definition 1. If we assume C is a straight line, then we know that ∞ must be on C. Therefore we can assume z3 = ∞. by Definition 1, we know that

         (z∗ − z4)/(z2-z4)=conjugate of (z-z4)/(z2-z4) (0.1) 

        If z4 =0 and z2 =1,then C is just the x-axis. From(0.1),we see that z* =conjugate of(z). They are symmetric with respect to the x-axis. For arbitrary z2 and z4, we can also show that z and z* are symmetric with respect to the line given by z2 and z4. In fact if we take absolute values on both sides of (0.1), we get |z − z4| = |z* − z4|. By Remark 1, z4 can be arbitrary point on C, therefore z∗ can only be z or the symmetric point of z with respect to C. If z = z∗, then by (0.1) we know that Im((z − z4)/(z2 − z4)) = 0. this shows that z is located on the line C. In other words, if z is not on C, z∗ must be different from z. That is z* must be the symmetric point of z with respect to C. If C is a circle with center a and radius R, then we have

         conjugate of (z,z2,z3,z4)=conjugate of (z−a,z2 −a,z3 −a,z4 −a) 

        from the proposition 0.9. Noticing that z2, z3 and z4 are located on C, therefore, we have |zj −a|^2 =R2, j=2,3,4. Applying the above equalities to (0.4), we deduce that

         conjugate of (z,z2,z3,z4)=(R^2/conjugate(z-a)+a,z2,z3,z4). 

     6. Proposition 2 : If z* is the symmetric point of z with respect to the circle centering a and having radius R, then

         z* = a+ R^2/conjugate(z-a) 

        1. The symmetric point of a point on the circle C with respect to C is itself
        2. If C is a circle centering at a, then the symmetric point of a with respect to C is ∞.
        3. Example 3: Given a circle C, the map from z to z∗ is called reflection. Reflect the imaginary line with respect to the circle |z − 2| = 1.

            solution:Let w be a point on the reflection. then its symmetric point with >>> respect to |z − 2| = 1 must be on the imaginary line. By Proposition 2, we know that 2+1/[conjugate(w)-2] must be pure imaginary. Assume w = w1 + iw2, it is clear from the above equality that 2(w1 −2)^2 +2(w2)^2 +w1 −2 = 0. It is a circle centering at (7/4, 0) with radius 1/4. 

           Example 4: Given the unit circle |z| = 1 and a linear transformation Sz = z/(z + 2).Find out the image of the unit circle under the given linear transformation.

            solution: Pick up one point on the unit circle, say 1. Its image under the action of the linear transformation is 1/3(找特殊点). By Example 2, (0, ∞) is a symmetric pair with respect to the unit circle. Then by Proposition 1, (0, 1) is symmetric with respect to the image circle. By Proposition 2, if a is the center of the imaging circle. R is its radius. Then we have a1 −(a1)^2 +R^2 =0, a2 =0. (0.5) Here we assume a = a1 + ia2.(找symmetric pairs:linear transformation 后还是sysmmetric) Moreover 1/3 is on the image circle, therefore |a − 1/3|2 = R2.Connecting this equation with the (0.5), we get a1 = −1/3 and R = 2/3.(代值) 

           Example 5: Find linear transformation which carries |z| = 2 to |z + 1| = 1, the point −2 to the origin, the origin to i.

            Since (0,∞) is a symmetric pair of |z| = 2, it holds that (i,T∞) is symmet- ric pair of |z + 1| = 1. Here T is the linear transformation we are searching. By Proposition 2, we can easily show that T∞ = (−1+i)/2. Since -2---0,0---i,∞---(-1+i)/2, T can be explicitly written out as follows: T z = (z + 2)/((−1 − i)z − 2i). 

           3点即可确定一个linear transformation:

            f(z)=az+b/cz+d ---->f(z)=a'z+b'/c'z+1 

转载于:https://my.oschina.net/smartman/blog/330433