orientation of a given circle

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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)

• 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 ∞

orientation of a given circle

1. Using z2,z3 and z4, we can determine a unique circle passing across these three points. We denote this circle by C. If z is on the circle, then we have Im(z, z2, z3, z4) = 0.

2. C automatically separate the complex plane into two parts. One part contains all z where Im(z, z2, z3, z4) < 0. We call this part the algebraic left-hand side of the circle C with respect to the triple (z2, z3, z4). Another part contains all z where Im(z,z2,z3,z4) > 0. We call this part the algebraic right-hand side of C with respect to the triple (z2, z3, z4).

3. The definition in 2 for the left and right-hand side of C corresponding to the triple (z2,z3,z4) is an algebraic way to describe the side for a given circle. Here is a geometric way to understand it. If the triple (z2,z3,z4) is given, then we can decide a unique direction on the circle C so that by following this direction we can go from z2 to z3 and then to z4 in order. One can easily see that the direction that we can have is just counterclockwise or clockwise direction. But once (z2, z3, z4) is given in order, then the counterclockwise or clockwise direction is uniquely fixed so that along this direction we go from z2 to z3 and then to z4 in order. Clearly if you are moving along counterclockwise direction, then the interior part of circle C is on your left. This corresponds to the left-side defined in 2 where Im(z, z2, z3, z4) < 0. Meanwhile the exterior part of C is on your right. This is the right side defined in 2 where Im(z,z2,z3,z4) > 0. But if you are moving clockwisely, then the situation is different. Now you can see that exterior part of C is on your left which corresponds to the region where Im(z,z2,z3,z4) < 0, while the interior part is on your right which corresponds to the region where Im(z, z2, z3, z4) > 0.

• 总结：Proposition 0.1: Given a triple (z2,z3,z4) on C, we can find a direction on C so that by following this direction, we go from z2 to z3 and then to z4 in order. The geometric right-hand side of C coincide with the algebraic right-hand side of C. The geometric left-hand side of C coincides with the algebraic left-hand side of C.

1. With the above proposition and the fact that cross ratio is invariant under linear transformations, we can show that Proposition 0.2: Linear transformations map left-hand (right-hand) side to left-hand (right-hand) side.

   • Remark 0.3: Proposition 0.2 should be understood as follows. given (z2,z3,z4) a triple on a circle C, we can decide a direction on C. Given an arbitrary linear transformation T, the triple (z2,z3,z4) is sent to (Tz2,Tz3,Tz4) which decide a direction for the imaging circle of C. Therefore Proposition 0.2 tells us that the left side of C with respect to the direction given by (z2,z3,z4) coincides with the left side of the imaging circle of C with respect to the direction given by (Tz2,Tz3,Tz4).

     proof: If C is determined by z2 , z3 and z4 and the direction of the circle C is given by the triple (z2, z3, z4), then the imaging circle is determined by T z2, T z3 and T z3. Here T is a linear transformation. Moreover if we go from z2 to z3 and then to z4 in order, then in the imaging circle we can induce a direction which let us go from Tz2 to Tz3 and then to Tz4 in order. (z2,z3,z4) decide a direction for C. (Tz2,Tz3,Tz4) decide a direction for the image of C. If z is on the left of C, then Im(z,z2,z3,z4) < 0. Therefore Im(Tz,Tz2,Tz3,Tz4) = Im(z,z2,z3,z4) < 0. This tells us that Tz is on the left of the imaging circle of C whose direction is given by the triple (Tz2,Tz3,Tz4). The proof is finished since the right-side case can be similarly treated.