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

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1 CHAPTER 9 Conformal Mapping and Bilinear Transformation BY Dr. Pulak Sahoo Assistant Professor Department of Mathematics University of Kalyani West Bengal, India sahoopulak1@gmail.com 1

2 Module-4: Bilinear Transformation and Inverse Points Inverse Points Let C be a circle having centre at z 0 and radius R. Two points P and Q are said to be inverse points with respect to the circle C if they are collinear with the centre, lie on the same side of it and the product of their distances from the centre is equal to R 2. Clearly Q is exterior to C if and only if P is interior to C. If Q is on C, then Q coincides with P (see Fig.1). Fig. 1: Note 1. z = 0 and z = are considered as a pair of inverse points. Remark 1. Let P (p) and Q(q) be the inverse points with respect to the circle C having centre at z 0 and radius R. Then p = z 0 + ρe iη and q = z 0 + R2 ρ eiη. 2

3 If z is any point on the circle C, then z = z 0 + Re iθ, 0 θ 2π. Hence z p z q = Re iθ ρe iη Re iθ R2 ρ eiη = ρ R. This is a new form of the equation of a circle. Theorem 1. Show that the set of all complex numbers satisfying z p z q = k, 0 < k 1 represents a circle (line) in the z-plane with respect to which p and q are inverse points. Proof. From the given condition and k 1, we have z p 2 = k 2 z q 2 (z p)(z p) = k 2 (z q)(z q) z 2 pz pz+ p 2 = k 2 ( z 2 qz qz+ q 2 ) z 2 p k2 q 1 k z p k2 q 2 1 k z + p 2 k 2 q 2 = k 2 ( z p ) ( k2 q z p ) k2 q = p k2 q 2 p 2 k 2 q 2 1 k 2 1 k 2 (1 k 2 ) 2 1 k 2 z p k2 q 1 k 2 2 = 1 (1 k 2 ) 2 [ p k2 q 2 (1 k 2 )( p 2 k 2 q 2 )] z p k2 q 1 k 2 = k p q 1 k 2. The above equation represents a circle having centre at z 0 is to be noted that p z 0 = k2 (q p) 1 k 2, q z 0 = q p 1 k 2, = p k2 q 1 k 2 which shows that (p z 0 )(q z 0 ) = r 2, the definition of inverse points. and radius k p q 1 k 2. It If k = 1, we have z p = z q, which shows that z is equidistant from both the points p and q, and hence lies on the perpendicular bisector of the line joining them. In this case, p and q are inverse points and q is the image of p in this line. This proves the theorem. Example 1. Suppose that L is the line passing through the points 1 and i. Are the points z 1 = 3i and z 2 = 2 + i inverse with respect to the line L? 3

4 Solution. The equation of the line L is y = x + 1. Again the equation of the line passing through the points z 1 = 3i and z 2 = 2 + i is y = x + 3. Obviously these two lines are perpendicular. Solving these two lines we obtain 1 + 2i as its point of intersection. Since (1 + 2i) 3i = (1 + 2i) (2 + i) = 2, z 1 and z 2 are inverse points with respect to the given line L. Theorem 2. Show that a bilinear transformation maps inverse points with respect to a circle or line onto inverse points with respect to the image circle and the image line. Proof. Let z p z q points (or symmetric points). Let = k be a circle (or a straight line for k = 1) with p and q as inverse be a bilinear transformation. Solving for z we get Then the circle is transformed into where w = az + b, ad bc 0 (1) z = dw + b cw a. dw+b cw a dw+b cw a p q = k w(cp + d) (ap + b) w(cq + d) (aq + b) = k w ap+b cp+d w aq+b cq+d α = ap + b cp + d, β = aq + b cq + d Thus the map of the circle z p z q = k cq + d cp + d w α w β = k, and k = k cq + d cp + d. = k under the bilinear transformation (1) is a circle or a straight line w α w β = k with respect to which α and β are inverse points or symmetric points which are respectively the images of p and q. This proves the theorem. 4

5 Example 2. Show that the bilinear transformation which carries the points z = i, 0, i into w = 0, 1,, respectively maps (i) the real axis Im z = 0 on w = 1, (ii) the upper half plane Im z > 0 on w < 1, (iii) the lower half plane Im z < 0 on w > 1. Solution. Let be the required bilinear transformation. Now w = az + b, ad bc 0 (2) f(i) = 0 ai + b ci + d = 0 ai + b = 0. (3) f(0) = i b d = 1 b + d = 0. (4) Also f( i) = ai + b ci + d = ci + d = 0. (5) Solving (2), (3), (4) and (5) we obtain w = f(z) = z i z + i, which is the required bilinear transformation. We now consider the following three cases separately. Case(i). Any point on the real axis can be taken as z = x. Then its image is w = x i x + i so that w = 1. Case(ii). Any point on the upper half plane can be taken as z = x + iy, y > 0. Then its image is w = x + i(y 1) x + i(y + 1). 5

6 T hus w = x + i(y 1) x + i(y + 1) = x 2 + y 2 2y + 1 x 2 + y 2 + 2y + 1 < 1. So the image of the upper half plane Im z > 0 is the region w < 1. Case(iii). In this case z = x + iy, y < 0. Then its image is T herefore w = w = x + i(y 1) x + i(y + 1). x + i(y 1) x + i(y + 1) = x 2 + y 2 2y + 1 x 2 + y 2 + 2y + 1 > 1. So the image of the lower half plane Im z < 0 is the region w > 1. Example 3. Find the bilinear transformation which maps the points 1, i, 1 in the z-plane into the points 0, 1, in the w-plane. Show also that by this transformation the area of the circle z = 1 is represented in the w-plane by the half plane above the real axis. Solution. Let w = az + b, ad bc 0 be the required bilinear transformation. Now Also Solving (6), (7) and (8) we obtain f(1) = 0 a + b c + d = 0 a + b = 0. (6) f(i) = 1 ai + b ci + d = 1 (a c)i + (b d) = 0. (7) f( 1) = a + b c + d = c d = 0. (8) b = a, c = d, a c = 1 i. Hence the required bilinear transformation is w = f(z) = az a cz + c = a c z 1 z + 1 = 1 z 1 i z

7 Let z be any point in the z-plane. Then its image is given by w = 1 i re iθ 1 re iθ + 1 (r cos θ 1) + ir sin θ = r sin θ + i(r cos θ + 1) 2r sin θ = r 2 + 2r cos θ i 1 r 2 r 2 + 2r cos θ + 1. We now consider the following three cases separately. Case(i). Let r < 1. Then Im w > 0. Therefore the image of a point lying inside the unit circle z = 1 lies in the upper half of the w-plane. Case(ii). Let r > 1. Then Im w < 0. Therefore the image of a point lying outside the unit circle z = 1 lies in the lower half of the w-plane. Case(iii). Let r = 1. Then Im w = 0, which shows that the map of a point lying on the unit circle z = 1 lies on the real axis of the w-plane. Combining the above three cases we can conclude that the area of the circle z = 1 in the z-plane is represented in the w-plane by the half plane above the real axis. Example 4. Show that the set of all bilinear transformations which maps unit disc onto itself is given by Solution. Let w = e iα 1 z 0 z, z 0 < 1, α R. w = az + b, ad bc 0 be the required bilinear transformation. Then clearly, c 0, otherwise the point at infinity in the two planes would correspond. Therefore, w = a c z + b/a z + d/c. (9) Now, w = 0 and w = are the inverse points with respect to w = 1 and these are the images of z = b/a and z = d/c respectively. We note that if z 0 is an inverse point of z = 1 then another inverse point is 1 z 0 Thus, because if z 0 = re iθ, then z 0 = 1 r eiθ = 1 re iθ = 1 z 0. b a = z 0 and d c = 1 z 0, z 0 < 1. 7

8 Hence (9) can be written as w = a c z 1 = az ( ) 0 z z0. (10) z 0 c 1 z 0 z The point z = 1 on the boundary of z = 1 must correspond to a point on the boundary of w = 1. Then from (10) we obtain Hence (10) becomes 1 = w = az 0 c az 0 c 1 z 0 = az 0 1 z 0 c = e iα, α R. w = e iα 1 z 0 z, z 0 < 1, α R. Example 5. Show that the set of all bilinear transformations which maps the upper half plane Im z > 0 onto w < 1 and the boundary Im z = 0 onto the boundary w = 1 is given by w = e iα, Im z 0 > 0, α R, where the point z 0 in the upper half plane is mapped onto the centre of the unit disc. Solution. Let w = az + b, ad bc 0 (11) be the required bilinear transformation. Then clearly, c 0, otherwise the point at infinity in the z-plane will not be mapped on the boundary w = 1. So (11) can be written as w = a c z + b/a z + d/c. (12) Now, w = 0 and w = are the inverse points with respect to w = 1 and these are the images of z = b/a and z = d/c respectively. We note that if z 0 is an inverse point with respect to the real axis then another inverse point is z 0. So Therefore (12) can be written as z 0 = b a and z 0 = d c. w = a c. (13) 8

9 The point z = 0 on the boundary corresponds to the point on the unit circle w = 1. So 1 = w = a c z 0 = a z 0 c a c = eiα, α R. Hence the transformation (13) assumes the form w = e iα, α R. (14) From (14), we see that the point z = z 0 corresponds to w = 0 and hence z 0 must be in the upper half plane, Im z 0 > 0. Thus we have w = e iα, Im z 0 > 0, α R. 9