1550251515-Classical_Complex_Analysis__Gonzalez_

382 Chapter^6

Theorem 6.28 Let f be defined in some open set G C C and suppose

that:

1. f E 1J(G).

2. J1(z) ":/:- 0 everywhere G.
   Then lf(z)I does not attain a maximum value anywhere in G.

Proof It follows from hypothesis (2) that f is nonconstant in G. Let z E G

be any given point, and choose l~zl > 0 small enough so that z + ~z E

N 0 (z) C: G. From formula (6.12-2) we have

f(z + ~z) = f(z) + fz~z + fz~z + 1]1~z + 1]2~z (6.20-1)

Let f(z) = M ei^8 , fz = Aeioi, fz = BeiP, ~z = rei'l/J, where any one of the

arguments(}, a, (3 is chosen arbitrarily ifthe corresponding modulus is zero.

Of course, because of (2), A":/:-B, so that A and Bare not simultaneously

zero. Substituting in (6.20-1), we obtain

f(z + ~z) = Mei^8 + Arei(<;'H) + Brei(/3-7/J) + r(771ei'l/J + 772e-i'l/J) (6.20-2)

First, suppose that A> B at z, and choose 'ljJ such that a+ 'ljJ = 0. Then

(6.20-2) becomes

f(z + ~z) = (M + Ar)ei^9 + Brei(/3+01-U) + r(771ei'l/J + 772e-i'l/J)

and we obtain

lf(z + ~z)I ~ M +Ar - (Br+ rl111I + rl1121)

= M + r( A - B - 1111 I - 11121) (6.20-3)

by choosing r small enough so that l111l+l112I < A-B, or A-B-11111-11121 >

0. This is possible, since 771 ~ 0 and 772 ~ 0 as r ~ 0. Then from (6.20-3)
   we conclude that

lf(z + ~z)I > M = lf(z)I

Now suppose that A< Batz. Choosing 'ljJ such that (3-'ljJ =(},we get

f(z + ~z) = (M + Br)ei^8 + Arei(01+/3-^9 ) + r(771ei'l/J + 772e-i'l/J)

so that

lf(z + ~z)I ~ M +Br - (Ar+ rl771 I+ rl112I)

= M + r(B -A-11111-11121)

again by selecting r small enough so as to have B - A - 1111 I - 1112 I > 0.

We obtain, as before,

lf(z + ~z)I > M = lf(z)I