384 Chapter^6
Theorem 6.31 Let f be defined on the closure of some open and bounded
set G C C, and suppose that:
1. f E 1J(G).
- J1(z) =j:. 0 everywhere in G.
- f(z) =j:. 0 at each z E G.
4. f is continuous on aa.
Then lf(z)I attains its minimum on aa.
Proof Similar to that of Theorem 6.29.
This property is known as the principle of the minimum modulus. It ap-
plies, in particular, to nonconstant analytic or conjugate analytic functions
(see Section 8.11).
EXERCISES 6.4
- Let J(z) = 2z + z + 4 and G = {z: JzJ::::; 1}. Find the points of G
where max Jf(z)J and min Jf(z)J occur.
2. Let f E 1J(G), where G = {z: lzJ < R}, and suppose that J1(z) =j:. 0
for all z E G. Prove that M(r) = maxJf(z)I, 0::::; r < R, is a strictly
increasing function of r.
3. Let f E 1J(G), where G is a bounded open set. Suppose that f is
continuous on 8G and that J 1 (z) =j:. 0 for all z E G. If f = u +iv,
prove that both the maximum and minimum values of u = u(x, y), as
well as those of v( x, y ), are attained on 8G.
4. Let f 1 , f2 E 1J(G), where G is a bounded open set. Suppose that f 1
and h are continuous on 8G and that J1Jz) =j:. Jt.(z) for all z E G.
Also, let u1 = Ref1, u2 = Refi. Show that if u1(x, y) = u2(x, y) for
all (x,y) E 8G, then u1(x,y) = u2(x,y) for all (x,y) E G.
- Let f E 1J(G), where G = {z: lzJ < r }, and suppose that J1(z) =j:. 0 for
all z E G, f continuous on aa, lf(O)I < k and IJ(z)I > k for Jzl = r.
Prove that f has at least one zero in G.
6.21 THE FUNDAMENTAL TOPOLOGICAL
PROPERTIES OF ANALYTIC FUNCTIONS
We recall (Definition 3.14) that a function f: A -7 B is called open (or
interior) if for every open subset G of A, f ( G) is an open subset of B.
This is what G. T. Whyburn [129] calls strongly open. The same author
calls f open if f(G) is open in f(A). This may also be called relatively
