Cauchy-Riemann Equations 193and v have continuous partial derivatives of every order. Recall that the
alternative notation for the partial derivative du/dx is ux, and when the
second-order partial derivatives are continuous, we have uxy = uyx. Then
from (4.2.2) we find uxx = vyx and
Similarly, vxx + vyy = 0, that is, both u and v are harmonic functions.
Two harmonic functions that satisfy (4.2.2) are called conjugate. We
conclude that a function f = f(z) is analytic in a domain G if and only if
the real and imaginary parts of f are conjugate harmonic functions. This
remarkable connection between analytic and harmonic functions leads to
numerous applications of complex analysis in problems such as the study
of two-dimensional electrostatic fields and two-dimensional flows of heat
and fluids. We discuss mathematical foundations of these applications in
Problem 5.3 on page 433. The lack of a three-dimensional analog of complex
numbers is one of the reasons why the corresponding problems in three
dimensions are much more difficult.
If u and v are conjugate harmonic functions, then, because of (4.2.2),
one of the functions uniquely determines the other up to an additive con-
stant. Similarly, either the real part u or the imaginary part v specify
the corresponding analytic function / = u + iv uniquely up to an addi-
tive constant. FOR EXAMPLE, if u(x,y) = xy, then, by the first equal-
ity in (4.2.2), ux = y = vy, so that v(x,y) = y^2 /2 + h(x) and hence
vx = h'(x). By the second equality in (4.2.2), vx = —uy = —x, that is,
h'{x) = —x and v(x, y) = (y^2 — x^2 )/2 + c, where c is a real number. Note
that vxx + vyy = —1 + 1 = 0, as it should; this is a useful computation to
ensure that the computations leading to the formula for v are correct. With
z = x + iy, the corresponding function f(z) is recovered from the equality
f(z) = u((z + z)/2, (z - z)/2i) + iv({z + 2)/2, (z - z)/2i),which in this case results in f(z) = —iz^2 /2 + ic, where c is a real number.
This answer is easy to check: f(z) = —iz^2 /2 + ic = xy + i(y^2 — x^2 )/2 =
u(x,y) + iv(x,y).
EXERCISE 4.2.9F Find all functions f = f(z) that are differentiable for all
z and have 9/(z) = x^2 — xy — y^2. Check that your answer is correct.
If we write z = r(cos6 + isin#) in polar coordinates, then f(z) =
u(r,9) +iv(r,6).
EXERCISE 4.2.10.B (a) Verify that in polar coordinates equations (4-2.2)