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

# Ad w which is a linear map hence linear fractional

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- 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
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
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.

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