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684 CHAPTER 19 Complex Numbers and Functions

∂u ∂u
= g(z) = − i .
∂x ∂y
Therefore,
∂U ∂u ∂U ∂u
= and =
∂x ∂x ∂y ∂y
for (x, y) in D. This means that U(x, y) − u(x, y) is constant on D, so for some real number K,
U(x, y) = u(x, y) + K.
Now define f (z) = G(z) − K. Then f is differentiable on D, and
f (z) = G(z) − K = U(x, y) + iV (x, y) − K = u(x, y) + iv(x, y).
We may therefore choose v(x, y) = V (x, y), proving the theorem.

Given a harmonic function defined on a domain, we are rarely interested in actually pro-
ducing a harmonic conjugate. However, knowing that a harmonic conjugate exists enables us to
go from harmonic u to a differentiable complex function f = u + iv, bringing complex function
methods to bear on some problems. We will exploit this in solving Dirichlet problems by confor-
mal mappings in Chapter 23. We will also use complex integration to derive important properties
of harmonic functions in Chapter 20.

SECTION 19.2 PROBLEMS

In each of Problems 1 through 12, find u and v so that 7. f (z) = z/Re(z)
f = u + iv, determine all points (x, y) at which the 3
8. f (z) = z − 8z + 2
Cauchy-Riemann equations hold, and determine all z at
which f is differentiable. Familiar facts about continuity of 9. f (z) = (z) 2
real-valued functions of two real variables can be assumed. 10. f (z) = iz +|z|
1. f (z) = z − i 11. f (z) =−4z + 1/z
2
2. f (z) = z − iz 12. f (z) = (z − i)/(z + i)
13. Let z n = a n + ib n be a sequence of complex num-
3. f (z) =|z|
bers. We say that this sequence converges to w =
4. f (z) = (2z + 1)/z
c + id if the real sequences a n → c and b n → d.Show
2
5. f (z) = i|z| that, if f (z) is continuous at z 0 and z n is a sequence
6. f (z) = z + Im(z) converging to z 0 , f (z n ) converges to f (z 0 ).

19.3 The Exponential and Trigonometric Functions

z
The complex exponential function e is defined for all z = x + iy by
x
x
z
e = e cos(y) + ie sin(y).
iy
Notice that f (iy) = e = cos(y) + i sin(y) is just Euler’s equation. The functions u(x, y) =
x
x
e cos(y) and v(x, y) = e sin(y) are continuous with continuous first partial derivatives, which
z
satisfy the Cauchy-Riemann equations. Therefore, e is differentiable for all z. Furthermore,
using equation (19.1),

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October 15, 2010 18:5 THM/NEIL Page-684 27410_19_ch19_p667-694