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682 CHAPTER 19 Complex Numbers and Functions
The theorem states that the Cauchy-Riemann equations are necessary for f = u + iv to
be differentiable at a point. If u, v, and their first partial derivatives are continuous, then the
Cauchy-Riemann equations are also sufficient for f to be differentiable.
Using the Cauchy-Riemann equations, we can establish the following results.
THEOREM 19.5
Let f be differentiable on an open disk D.Let f = u + iv, and suppose that u and v are con-
tinuous with continuous first and second partial derivatives, and satisfy the Cauchy-Riemann
equations on D. Then
1. If f (z) = 0on D, then f is a constant function on D.
2. If | f (z)| is constant on D,sois f (z).
Proof Conclusion (1) is easy to prove. For each z in D,
∂u ∂v
f (z) = 0 = + i
∂x ∂x
implies that ∂u/∂x = ∂v/∂x = 0on D. By the Cauchy-Riemann equations, ∂u/∂y = ∂v/∂y = 0
on D also, so u and v are constant functions and therefore so is f .
Conclusion (2) is more involved. Suppose | f (z)|= k for all z in D. Then
2
2
2
| f (z)| = u(x, y) + v(x, y) = k 2 (19.2)
for (x, y) in D.If k = 0, then f (z) = 0 for all z in D.If k = 0, differentiate equation (19.2) with
respect to x to get
∂u ∂v
u + v = 0 (19.3)
∂x ∂x
and with respect to y to get
∂u ∂v
u + v = 0. (19.4)
∂y ∂y
Use the Cauchy-Riemann equations to write equations (19.3) and (19.4) as
∂u ∂u
u − v = 0 (19.5)
∂x ∂y
and
∂u ∂u
u + v = 0. (19.6)
∂y ∂x
Multiply equation (19.5) by u and equation (19.6) by v and add the resulting equations to get
∂u 2 ∂u
2
2
(u + v ) = k = 0.
∂x ∂x
Therefore, ∂u/∂x = 0on D, and by the Cauchy-Riemann equations, ∂v/∂y = 0 also. A similar
manipulation shows that ∂u/∂y = ∂v/∂x = 0, so u(x, y) and v(x, y) are constant, hence f (z) is
constant on D.
There is an intimate connection between differentiable complex functions and harmonic
functions. Recall that a real-valued function u(x, y) of two real variables is harmonic on a set of
points in the x, y-plane if u satisfies Laplace’s equation
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