Math 412 Group Work #7 Spring 2009
SOLUTIONS
Problem 1. Let
f(z) = x2+y2+i(2xy).
(a): Use the Cauchy-Riemann equations to determine where fis differen-
tiable.
SOLUTION: We have
ux(x, y) = 2x, uy(x, y) = 2y, vx(x, y) = 2y, vy(x, y ) = 2x.
For all (x, y)∈R2, we have ux=vy. However, uy=−vxonly holds if 2y=−2y; that
is, uy=−vxif and only if y= 0. Thus, the Cauchy-Riemann equations only hold for
points (x, 0) along the real-axis of the complex plane. Thus, fis only differentiable
along the real axis: {z∈C: Im(z) = 0}.
(b): Evaluate the derivative at the points z0where f0(z0)exists.
SOLUTION: We have
f0(z) = f0(x+iy) = ux(x, y) + ivx(x, y ) = 2x+i(2y) = 2x,
since y= 0 along the real axis.
(c): Determine all singular points of the function f.
SOLUTION: The only points of differentiability of flie along the real axis, and no
such point possesses a neighborhood centered at it so that all points in the neighbor-
hood are differentiable. Thus, there are no points of analyticity of f, which implies
that all points of Care singular points of f.
Problem 2. Suppose that f:C→Cis an entire, real-valued function.
Show that fmust be a constant function.
SOLUTION: Since f(x+iy) = u(x, y) is real-valued (i.e. v(x, y) = 0), we know that
vx(x, y) = vy(x, y) = 0 for all (x, y )∈R2. Thus, since ux(x, y) = vy(x, y ) = 0 and
uy(x, y) = −vx(x, y) = 0, we conclude that u(x, y) = c(c∈C) must be a constant
function. Thus, f(x+iy) = u(x, y ) = cis a constant function.
Problem 3. Suppose that f(x+iy) = u(x, y ) + iv(x, y). For what relations
among the real constants a, b, c, d with
u(x, y) = ax3+bx2y+cxy2+dy3
