Complex Analysis Homework 4

Transcription

1 Complex Analysis Homework 4 Isaac Defrain Steve Clanton David Holz April 9, 2009 Problem 1 Problem. Show that if f = u + iv is analytic then ( v) ( v) = 0. Explain this geometrically. Since f is analytic, CR is satisfied: u = x uˆx + y uŷ, v = x vˆx + y vŷ ( u) ( v) = x u x v + y u y v We may substitute partials with respect to y for partials with respect to x according to CR: ( u) ( v) = x u x v + ( x v)( x u) = 0 The gradient operator ( ) tells us the direction of the maximum rate of change in value of a function and the fact that ( u) ( v) = 0, means that u and v are orthogonal. So the directions of the maximum rate of change in value of u and v are orthogonal. Problem 2 Problem. Show that both the real and imaginary parts of an analytic function are harmonic, i.e., they both automatically satisfy Laplace s equation: φ = 0 f = u + iv is analytic so CR applies. u = 2 xu + 2 yu = x ( x u) + y ( y u) (By CR) = x y v y x v = x y v x y v = 0 u = 2 xv + 2 yv = x ( x v) + y ( y v) (By CR) = x ( y u) + y x u = y x u x y u = 0 The order of the derivatives does not make a difference since they all are continuous in R 2. Therefore, u = 0 and v = 0. This fact implies that also f = 0 since f = u + i v. 1

2 Problem 4 Problem. What is the most general function u = ax 2 + bxy + cy 2 that is the real part of an analytic function? Construct this analytic function, and express it in terms of z. Our function is analytic so CR applies again, and we know from Problem 2 that u = 0 xu 2 = x ( x u) = x (2ax + by) = 2a yu 2 = y ( y u) = y (bx + 2cy) = 2c thus u = 2a + 2c = 0 a = c x u = 2ax + by = y v Integrating with respect to y... v = 2axy by2 + g(x) y u = bx 2ay = x v Integrating with respect to y... v = 2axy bx2 + h(x) Equating these expressions for v, we get: g(x) = 1 2 bx2, h(y) = 1 2 by2 and the most general forms: u = ax 2 + bxy ay 2 v = 1 2 b(y2 x 2 ) + 2axy The most general analytic function which can be constructed is: f = a(x 2 y 2 ) + bxy + i( 1 2 b(y2 x 2 ) + 2axy) = (a i 2 b)x2 + (b + 2ia)xy + ( i 2 b a)y2 C = a i 2 b 2iC = b + 2ia = Cx 2 + 2iCxy Cy 2 = C(x + iy) 2 f(z) = (a i 2 b) 2

3 Problem 6 Problem. Solve the polar CR equations given that θ 0. Express your answer in terms of a familiar function, and interpret everything you have done geometrically. θ v = r r u = 0 θ u = r r v this case is trivial, but worth mentioning... If r = 0 then θ u = r r v = 0 If r u = 0 then θ v = 0 ip (r) v = v(r) = R(r)e u = u(θ) = Φ(θ)e iφ(θ) θ u = A u = Aθ + B, B C r r v = A r v = A A r v = dr = A ln r + C, r C C f = u + iv = Aθ + B + i( A ln r + C) = A(i ln r + θ) + (B + ic) = ia(ln r + iθ) + (B + ic) = f(z) = ia log z + D with D = (B + ic) = const 3

4 Problem 7 Problem. Use the Cartesian CR equations to show that the only analytic mapping that sends parallel lines to parallel lines is the linear mapping. Solution. In particular, our mapping will send horizontal lines to horizontal lines. We will consider the image of movement along a horizontal line (x + dx, y) (u + du, v) where (u, v) denotes the image of the point (x, y). Since dx and du are both horizontal, we know the total differential is du = x u dx + y u dx = a dx for some real number a. This equation tells us the partial derivatives are δu δu = a and δx δy = 0. Substitution from the CR equations gives us equations to determine v: Now, we can solve the differential equation: δv δv = a and δy δx = 0 u (x, y) = ax + g (y) δu δy = g (y) = 0 u (x, y) = ax + C 1 Likewise, we get v (x, y) = ay + C 2. Rewriting as a standard complex function we see f (z) = u (x, y) + iv (x, y) = ax + C 1 + iay + ic 2 = a (x + iy) + (C 1 + ic 2 ) = az + c a R and c C We can also remark that the same general solution solves a differential equation where a is an arbitrary complex number. Thus, we can generalize mapping horizontal lines to lines with any slope: f (z) = az + c a, c C 4

5 Problem 25 Problem. Sometimes the circle of convergence of a power series is so densely packed with singularities that it becomes a genuine barrier for the geometric mapping, beyond which it cannot be continued. This is called a natural boundary. An example of this is furnished by f (z) = z + z 2 + z 4 + z 8 + z 16 +, which converges inside the unit circle. Show that every point of z = 1 is either a singularity itself, or else has singularities arbitrarily near to it. Solution. First, we show that every 2 n -th root of unity is singular. For the base case of an inductive proof, we note that 1 is a singularity since f (z) = Next, we show that f (z) is a singularity whenever f ( z 2) is a singularity: f (z) = z + f ( z 2) f (z) z = f (z) 1 f (z) Now, we must show that every point on the unit circle that is not a 2 n -th root of unity is arbitrarily close to a 2 n -th root of unity. Note the 2 n -th roots partition the circle into arcs that are 2 1 n π < 2 3 n long. We know the distance between endpoints of an arc is less than the length of the arc. On a small arc, the distance between the endpoints will be greater than the distance between any ohter pair of points on the arc. Since the distance from a point on a small arc to its endpoints is less than the length of the arc, and since 2 3 n becomes arbitrarily small as n increases, we see that a point on the unit circle is either singular or arbitrarily close to a singularity. 5