Problems 2 (due 22 January) 1. (a) Find the values of the constants a, b, c, d for which the polynomial u(x, y) = ax3 + bx2 y + cxy 2 + dy 3 is harmonic, i.e., ∂2u ∂2u + 2 = 0. ∂x2 ∂y (1) Solution: First, ∂2u ∂2u + 2 = (6a + 2c)x + (2b + 6d)y. ∂x2 ∂y So, in order to satisfy Eq. (1), we must have c = −3a, b = −3d, and u(x, y) = ax3 − 3dx2 y − 3axy 2 + dy 3 (b) Find v(x, y) such that f (z) = u + iv is analytic in the entire plane, where z = x + iy. Solution: The Cauchy-Riemann equations result in and ∂u ∂v = = 3ax2 − 6dxy − 3ay 2 ∂y ∂x (2) ∂v ∂u =− = 3dx2 + 6axy − 3dy 2 . ∂x ∂y (3) Integrating the first of these leads to v = 3ax2 y − 3dxy 2 − ay 3 + f (x), for some function f (x) of x alone. For this to be compatible with Eq. (3) requires that 3dx2 + 6axy − 3dy 2 = 6axy − 3dy 2 + f 0 (x), meaning that f 0 (x) = 3dx2 and f (x) = dx3 +k, where k is a constant. I.e., v = 3ax2 y − 3dxy 2 − ay 3 + dx3 + k, for some constant k. 1 2. Find all singularities of these functions: (a) tanh(z) = ez − e−z ez + e−z Solution: Since ez is entire, the singularities of this function are only at zeros of the denominator ez + e−z . Writing z = x + iy, singularities will occur when ex eiy = e−x eiπ−iy . Equality of these requires equality of the moduli: ex = e−x , or x = 0. Thus, any zeroes of the denominator above must lie on the y-axis with y = π − y, up to possible additional multiples of 2π. Thus, 1 +m m = 0, ±1, ±2, . . . . y=π 2 (b) 1 z(ez − 1) Solution: There is obviously a singularity at z = 0. The term (ez −1) also has additional zeros. Remember that ez “covers” C infinitely many times. Writing z = x + iy, solutions to (ez − 1) correspond to ex eiy = 1, which means both x = 0 and m = 0, ±1, ±2, . . . . y = 2mπ (c) z4 (z 2 + a2 )2 Solution: Since z 2 + a2 = (z + ia)(z − ia), the singularities are at ±ia. 2

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