ad W which is a linear map Hence linear fractional transformations are

Ad w which is a linear map hence linear fractional

This preview shows page 61 - 64 out of 167 pages.

- ad ) W , which is a linear map Hence, linear fractional transformations are compositions of up to 3 known transforma- tions which map lines and circles to circles and lines. Hence, they do the same. Definition 39 (fixed point) If f is a map and f ( z ) = z , then z is called a fixed point of f . Example 34 Find all the fixed points of the linear fractional transformation w = az + v cz + d , where ad - bc = 0 . We are looking for z s.t. az + b cz + d = z Case 1: c = 0 a d z + b d = z a d - 1 z + b d = 0 If a d = 1 , then we have a unique solution z = - b d a d - 1 If a d = 1 and b = 0 , then 0 + b z = z . This is always true, i.e. z C , z is a fixed point and we have w = az + 0 0 z + d = a d z = z this is the identity transformation. If a d = 1 and b = 0 , then there is no solution, i.e. no fixed points. Case 2: c = 0 . az + b cz + d = z az + b = z ( cz + d ) = cz 2 + dz 0 = cz 2 + ( d - a ) z - b z = - ( d - a ) ± ( d - a ) 2 + 4 bc 2 c So w has up to 2 fixed points. The conclusion is that if f is a linear fractional transformation and f = Id , it has at most 2 fixed points. Corollary 6 If f is a linear fractional transformation and f ( z 1 ) = z 1 , f ( z 2 ) = z 2 , f ( z 3 ) = z 3 , with z 1 ,z 2 ,z 3 distinct, then f ( z ) = z , z C . Example 35 Given z 1 ,z 2 ,z 3 C distinct and w 1 ,w 2 ,w 3 C distinct, find all linear fractional transfor- mations T with T ( z i ) = w i 57
Image of page 61
Consider the following transformation T . We claim that it is the one we are looking for. w - w 1 w - w 2 w 1 - w 2 w 1 - w 3 = z - z 2 z - z 3 z 1 - z 2 z 1 - z 3 w - w 1 w - w 2 . z 1 - z 2 z 1 - z 3 = z - z 2 z - z 3 . w 1 - w 2 w 1 - w 3 ( w - w 2 )( z 1 - z 2 )( z - z 3 )( w 1 - w 3 ) = ( z - z 2 )( w 1 - w 2 )( w - w 3 )( z 1 - z 3 ) Plugging in z = z 3 into the LHS, we get 0, so the RHS must also be 0. But since z i are distinct, it must be true that if z 3 = 0 , then z 2 = 0 and z 1 = 0 and since w 1 = w 2 , the only term in the RHS that can be 0 is ( w - w 3 ) . So we conclude that w = w 3 , i.e. z = z 3 w = w 3 . Plugging in z = z 2 into the LHS, we get 0, so the RHS must also be 0. Since w 1 = w 3 and the z i are also distinct, the only term that we can get to be 0 from the RHS is ( w - w 2 ) . So we conclude that z = z 2 w = w 2 Plugging in z = z 1 in the LHS, we get 0. We get ( w - w 2 )( z 1 - z 2 )( z 1 - z 3 )( w 1 - w 3 ) = ( z 1 - z 2 )( w 1 - w 2 )( w - w 3 )( z 1 - z 3 ) = 0 ( w - w 2 )( w 1 - w 3 ) = ( w 1 - w 2 )( w - w 3 ) w - w 2 = w 1 - w 2 w 1 - w 3 ( w - w 3 ) This is linear, so it has only one solution. This is true for w = w 1 , which is the unique solution. So once again we get that z = z 1 w = w 1 Note: T is unique. Suppose that it wasn’t and that we could find another transformation S ( z i ) = w i . Consider S T - 1 ( w i ) = S ( T - 1 ( w i )) = S ( w i ) = w i . So S T - 1 has three fixed points and it is a linear fractional transformation S T - 1 = Id S = T . Example 36 Determine all linear fractional transformations f H D ( 0 , 1 ) , where H = { z C I ( z ) > 0 } and D ( 0 , 1 ) = { w C w < 1 } . Solution: We are looking for a transformation of the form f ( z ) = az + b cz + d with ad - bc = 0 . l H D (0,1) f(l) We are looking for linear fractional transformations of the form f ( z ) = az + b cz + d with ad - bc = 0 . First of all H is bounded by the real axis, call it l . Thus by the Inertia principle and since 58
Image of page 62
we are looking for a bijective mapping, as l.f.t. are continuous, we deduce that f ( l ) is a circle, which is the boundary of the disk D ( 0 , 1 ) , the unit circle.
Image of page 63
Image of page 64

You've reached the end of your free preview.

Want to read all 167 pages?

  • Winter '18
  • Mr Ali
  • lim, Complex number, Limit of a sequence, = Z

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes