CHAPTER 9. Conformal Mapping and Bilinear Transformation. Dr. Pulak Sahoo

Transcription

1 CHAPTER 9 Conformal Mapping and Bilinear Transformation BY Dr. Pulak Sahoo Assistant Professor Department of Mathematics University of Kalyani West Bengal, India sahoopulak@gmail.com

2 Module-3: Bilinear Transformation - Normal Form Fixed Points or Invariant Points The points which coincide with its transformation are called fixed or invariant points of the transformation. If z is a fixed point of the transformation T, then T (z) z. The identity transformation I(z) z has every z C { } as its fixed points. The fixed points of the transformation w z 2 are solutions of z 2 z, i.e. z 0,. Theorem. Prove that in general there are two values of z for which w z but there is only one if (a d) 2 + 4bc 0. Show also that if there are distinct invariant points p and q, then the transformation can be written in the form w p z p w q k z q and that if there is only one invariant point p, the transformation can be written in the form where k is a constant. Proof. Let be a bilinear transformation. For fixed points we have () w p k + z p, (2) w az + b, ad bc 0 (3) w z i.e. cz 2 (a d)z b 0. (4) 2

3 In general (4) gives two values of z and hence two fixed points. When (a d) 2 + 4bc 0, there is only one fixed point. Let p and q be two fixed points of the transformation (3). Then from (4) we obtain cp 2 ap b dp and cq 2 aq b dq. Now w p az + b (a cp)z + (b dp) p (a cp)(z p). Similarly, w q (a cq)(z q). T herefore, w p w q (a cp)(z p) (a cq)(z q) k z p, z q where k a cp a cq. If there is only one fixed point then p q p + q 2 a d, by (4). 2c T herefore, w p cz + a 2cp (a cp)(z p) (a cp)(z p) c(z p) + (a cp) k + (a cp)(z p) z p, where k c. This proves the theorem. a cp Remark. Equations () and (2) are known as the normal form or canonical form of a bilinear transformation. Remark 2. A bilinear transformation w f(z) with more than two fixed points in the extended complex plane must be the identity transformation. Note. (i) A bilinear transformation having only one fixed point is called a parabolic transformation and if the fixed point is p then the transformation is of the form w p k + z p, 3

4 where k is a constant. (ii) A bilinear transformation having two fixed points p and q can be written in the form w p z p w q k z q where k is a constant. If k, k, then it is called elliptic transformation and if k > 0 is real, then it is called hyperbolic transformation. (iii) A bilinear transformation which is neither parabolic nor elliptic nor hyperbolic is called loxodromic. That is it has two fixed points and satisfies the condition k ae iα, α 0, a. Example. Find all the bilinear transformations which have fixed points and. Solution. Let w f(z) az + b, ad bc 0 be the required bilinear transformation. Now f() a + b c + d and f( ) a b d c. Solving we get a d and b c. Therefore, is the required transformation. w f(z) az + b bz + a, a2 b 2 0 Example 2. Find the normal form of the bilinear transformation w 3iz+ z+i. Solution. For fixed points of the transformation we have z 3iz + z + i i.e. z 2 2iz 0 i.e. (z i) 2 0 i.e. z i. So the given transformation has only one fixed point and hence its normal form is where k is a constant. w i z i + k, 4

5 Example 3. Find the bilinear transformation with fixed points and carrying the point z i onto w i. Solution. Since the transformation has two distinct fixed points, its normal form is w + z + w k, z where k is some constant. Since it maps z i onto w i we have i + i + i k. i Solving we get k. Hence the required transformation is w + z + w z i.e. w z. We now develop the specific bilinear transformation which maps three distinct points in the extended z-plane onto three distinct points in the extended w-plane. For that we first introduce the concept of cross ratio. Cross Ratio For three distinct complex numbers z, z 2, z 3 in C { }, the cross ratio of four points z, z, z 2, z 3 is defined as (z, z, z 2, z 3 ) (z z )(z 2 z 3 ) (z z 2 )(z 3 z). If one of the numbers is replaced by infinity, say z 3, then (z z )(z 2 /z 3 ) (z, z, z 2, ) lim z 3 (z z 2 )( z/z 3 ) z z. z z 2 Note 2. The cross ratio will change if the order of the factors is changed. Since four letters z, z 2, z 3, z 4 can be arranged in 4! 24 ways, there will be 24 cross ratios but as a matter of fact there will be only six distinct cross ratios. The six distinct cross ratios are (z, z 2, z 3, z 4 ), (z, z 2, z 4, z 3 ), (z, z 3, z 2, z 4 ), (z, z 3, z 4, z 2 ), (z, z 4, z 2, z 3 ), (z, z 4, z 3, z 2 ). Theorem 2. A bilinear transformation leaves a cross ratio invariant. 5

6 Proof. Let w az + b, ad bc 0 (5) be a bilinear transformation. Let w, w 2, w 3, w 4 be the images of the points z, z 2, z 3, z 4 respectively under the transformation (5). To prove the theorem we have to show that (w, w 2, w 3, w 4 ) (z, z 2, z 3, z 4 ). We have (w, w 2, w 3, w 4 ) (w w 2 )(w 3 w 4 ) (w 2 w 3 )(w 4 w ). Now w w 2 az + b cz + d az 2 + b cz 2 + d (ad bc)(z z 2 ) (cz + d)(cz 2 + d). Similarly, Hence we obtain w 2 w 3 (ad bc)(z 2 z 3 ) (cz 2 + d)(cz 3 + d), w 3 w 4 (ad bc)(z 3 z 4 ) (cz 3 + d)(cz 4 + d), w 4 w (ad bc)(z 4 z ) (cz 4 + d)(cz + d). This proves the theorem. (w, w 2, w 3, w 4 ) (z z 2 )(z 3 z 4 ) (z 2 z 3 )(z 4 z ) (z, z 2, z 3, z 4 ). Theorem 3. Given three distinct points, z, z 2, and z 3 in the extended z-plane and three distinct points w, w 2, and w 3 in the extended w-plane, there exist a unique bilinear transformation w f(z) such that f(z k ) w k for k, 2, 3. Proof. We assume that none of the six points is. Let w f(z) az + b, ad bc 0 be the bilinear transformation. We wish to solve for a, b, c, and d in terms of z, z 2, z 3, w, w 2, and w 3. For k, 2, 3 we have w w k az + b az k + b cz k + d (ad bc)(z z k) ()(cz k + d). (6) 6

7 So from (6) we obtain w w w w 3 ( cz3 + d cz + d ) ( z z z z 3 ). (7) Replacing z by z 2 and w by w 2 in (7) we obtain w 2 w 3 cz + d z2 z 3. (8) w 2 w cz 3 + d z 2 z Multiplying (7) and (8) we have (w w )(w 2 w 3 ) (w w 3 )(w 2 w ) (z z )(z 2 z 3 ) (z z 3 )(z 2 z ). (9) Solving for w in terms of z and the six points gives the desired transformation. If one of the points, say z 3 case, we would have, (9) would be modified by taking the limit as z 3. In this (w w )(w 2 w 3 ) (w w 3 )(w 2 w ) z z z 2 z. Now we suppose that f(z) and g(z) are both bilinear transformations that agree at three or more points in the extended complex plane. Therefore, w k f(z k ) g(z k ) for k, 2, 3. Thus for k, 2, 3 we have (f g)(z k ) f (g(z k )) f (w k ) z k and so by Remark 2, f g I. This gives f g which proves the uniqueness part of the theorem. Example 4. Find a bilinear transformation that maps the points z w i,,, respectively. i, 2, 2 onto Solution. We know that the transformation which maps the points z, z 2, and z 3 in the extended z-plane onto the points w, w 2, and w 3 in the extended w-plane is Here (w w )(w 2 w 3 ) (w w 3 )(w 2 w ) (z z )(z 2 z 3 ) (z z 3 )(z 2 z ). z i, z 2 2, z 3 2, w i, w 2, and w 3. 7

8 Therefore we obtain the required transformation as (w i)( + (w + )( i) (z i)(2 + 2) (z + 2)(2 i) i.e. w i (w + )( i) 2(z i) (z + 2)(2 i) i.e. izw + 6w 3z + 2i i.e. w 3z + 2i iz + 6. Example 5. Let f(z) be a bilinear transformation such that f( ), f(i) i and f( i) i. Find the image of the unit disc {z C : z < } under f(z). Solution. Let w az+b, ad bc 0 be the required bilinear transformation. Now cz+d f( ) a/c a c. (0) Also Solving (0), () and (2) we obtain f(i) i ai + b ci + d i (a d)i + (b + c) 0. () f( i) i ai + b ci + d i (a d)i (b + c) 0. (2) a c d b. Thus w f(z) z z+ be the required transformation. Putting z reiθ we see that w reiθ re iθ + (r cos θ ) + ir sin θ (r cos θ + ) + ir sin θ r 2 r 2 + 2r cos θ + + i 2r r 2 + 2r cos θ +. The map of a point lying on z lies on the imaginary axis in the w-plane because for r, Re w 0. If r <, then Re w < 0, i.e. the image of a point lying inside the unit circle z lies in the left half of the w-plane. 8