Math 460: Complex Analysis MWF 11am, Fulton Hall 425 Homework 5 Please write neatly, and in complete sentences when possible.

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1 Math 460: Complex Analysis MWF am, Fulton Hall 425 Homework 5 Please write neatly, and in complete sentences when possible. Do the following problems from the book: 2.6.8, 2.6., 2.6.3, 2.6.4, The next problems concern linear fractional transformations. A linear fractional transformation is a map of the form f(z = az + b cz + d, for a, b, c, d C, and ad bc 0. A linear fractional transformation with c = 0 is called affine. Problem C. ( Show that an affine transformation takes circles to circles and lines to lines. (2 Show that the composition of two linear fractional transformations is another linear fractional transformation. (3 Show that, for each linear fractional transformation f, there is a linear fractional transformation g so that g f(z = z for any z where both maps are defined. If f is not affine, what is the domain of definition of g? Deduce the image of C \ { d/c} under f. (4 Suppose f(z = az+b is a non-affine linear fractional transformation. Find an affine linear fractional transformation L (z cz+d so that f L (z is defined on C \ {0}. (5 If f is non-affine as above, find an affine transformation L 2 (z so that L 2 f(z has image C \ {0}. (6 Use L and L 2 above to show that f must send circles and lines to circles and lines. (Hint: In the non-affine case, consider L 2 f L, and use Problem A from Homework 4.

2 2 Solution Let γ indicate the circle through the vertices of the triangle, oriented counterclockwise. The difference in the contour integrals of /z over the triangle and over γ is given by the integral of /z over three closed loops, each formed by a triangular edge and a circular arc. Each such loop is contained in a convex open set on which /z is analytic, so Cauchy s Theorem guarantees that the integral of /z over each loop is 0. We conclude that the integral of /z over the triangle is equal to the integral of /z over γ. We ve seen that the latter is equal to the integral of /(z c over the circle, where c is the center of the circle traversed by γ, which we have computed to be 2πi. Solution The distance of + i to the origin is given by 2, which is less than 2, so that + i is contained inside the contour γ. The winding number around + i is given by /2πi times the integral of /(z i over γ. The integral of /(z i over the circle of radius 2, oriented counterclockwise, can be expressed as the sum of the integrals of /(z i over the four quarters of the circle (oriented counterclockwise, since the contributions from the radial arcs cancel. The three quarters of this circle that are distinct from γ give 0 when calculating the desired contour integral, by Cauchy s Theorem, since each is contained in a convex open set on which /(z i is analytic. Thus the winding number of γ around + i is equal to the winding number of the counterclockwise circle of radius 2 around + i. The latter is equal to the winding number of the circle around 0, i.e.. Solution The first integral can be written as z = z 2 = z = z+ z. The function /(z + is analytic on a convex open set containing z =, so Cauchy s Integral Formula applies, and we have z = z+ z = 2πi ind z =( ( z + z= = 2πi 2 = πi.

3 Similarly, we have z 2 = z+ = z+ = = 2πi z z + = 2πi ind z+ =( z= z 2 = πi. Solution Let γ denote the counterclockwise contour z = 3, and let γ and γ 2 indicate the counterclockwise contours z = and z + =. One may show that the integral of /(z 2 around γ is equal to the sum of the integrals of /(z 2 around γ and γ 2 : With appropriate choices of arcs, the difference γ γ γ 2 is equal to the sum of closed loops, each contained in a convex open set on which /(z 2 is analytic so that Cauchy s Theorem applies. Thus z =3 z 2 = by exercise z = z 2 + z+ = z 2 = πi πi = 0, Solution Let γ indicate the contour that traverses T counterclockwise. Note that we are given that f(z M and z = for z T. Thus f(z/z = f(z M for all z on γ. By Cauchy s Integral Formula, ind γ (0 f(0 = 2πi γ z. Note that the length of γ is given by l(γ = 2π, and we have also previously computed that ind γ (0 =. Taking absolute values, we find f(0 = f(z 2π z l(γ M = M. 2π γ Solution C(. Suppose f(z = az + b is an affine transformation. Then f consists of a rotation (z az followed by a translation (z z + b. Each of these maps take lines to lines and circles to circles, so f takes lines to lines and circles to circles. Solution C(2. Suppose f and f 2 are two linear fractional transformations given by f i (z = a iz + b i c i z + d i, with a i d b i c i 0 for i = or 2. We compute: 3

4 4 f f 2 (z = a a 2z+b 2 c 2 z+d 2 + b c a 2 z+b 2 c 2 z+d + d = a (a 2 z + b 2 + b (c 2 z + d 2 c (a 2 z + b 2 + d (c 2 z + d 2 = (a a 2 + b c 2 z + a b 2 + b d 2 (c a 2 + d c 2 z + c b 2 + d d 2. Let s keep track of the data of a linear fractional transformation f(z = (az + b/(cz + d via the matrix a b M(f :=, c d so that the condition ad bc 0 is equivalent to saying that the determinant det M(f i 0. Thus, to each linear fractional transformation f we can associate a matrix M(f with det M(f 0, and to each such matrix we can associate a linear fractional transformation. Note that M(f is actually not uniquely determined by f, since (az + b/(cz + d = (λaz + λb/(λcz + λ, for any λ 0. This won t matter for what we say below, but it is important to point out. (What might one do to associate a unique matrix M(f to each f? The above computation says that a matrix for the composition f f 2 is given by a a M(f f 2 = 2 + b c 2 a b 2 + b d 2, c a 2 + d c 2 c b 2 + d d 2 or in other words M(f M(f 2 is a matrix representation for f f 2. Since the determinant is multiplicative, det M(f f 2 = det M(f det M(f 2, and det M(f, det M(f 2 0 imply that det M(f f 2 0. Thus f f 2 is again a linear fractional transformation. Solution C(3. Given f(z = az+b, we have the matrix M(f. We cz+d seek g(z, a linear fractional transformation so that g f(z = z. Note that the identity map I(z = z = (z + 0/(0z + is evidently itself a linear fractional transformation, whose matrix M(I is the identity matrix M(I = ( 0 0.

5 By the above transformation rule, M(g M(f = M(I. Since det M(f 0, the matrix M(f invertible, and we find that M(g = M(f is a matrix representation for g. (Note that a linear fractional transformation defined in this way also satisfies f g(z = z. The matrix M(f can be calculated directly: M(g = Thus we have ad bc g(z = d b = c a d z + b c z + a ( d c = b cz + a. b a Evidently, this transformation is defined for z a/c. Finally, for any z C \ {a/c}, f(g(z = z, so that f maps C \ { d/c} to C \ {a/c}. Solution C(4. The non-affine map f(z is defined on C \ { d/c}, so in order to have f L (z be defined on C \ {0}, we need L (z to be a map that takes 0 to d/c. One such affine example is L (z = z d c. Solution C(5. The non-affine map f(z has image C\{a/c}, by part (3 above. In order to have L 2 f(z have image C \ {0}, we need L 2 to be a map that takes a/c to 0. One such affine example is L 2 (z = z a c. Solution C(6. We compute: ( L 2 f L (z = L 2 (f(z d a(z d c = L + b c 2 c(z d + d c acz + bc ad acz + bc ad = L 2 = L c 2 2 z cd + cd c 2 z a bc ad = L 2 + = a bc ad + a c c 2 z c c 2 z c bc ad =. c 2 z Finally, let L 3 (z = c2 z, so that L bc ad 3 L 2 f L (z =. z indicate the map s(z = /z, so that L 3 L 2 f L = s,. 5 Let s

6 and thus f = L 3 L 2 s L. In Problem A on homework assignment 4, we saw that s takes circles and lines to circles and lines, and by part ( above we know that affine transformations also take circles and line to circles and lines. Thus f takes circles and lines to circles and lines.