576 Chapter^8
Since f(z) is not constant in G there is a first coefficient after ao, say
ak(k 2:: 1), which is not zero. Let
a 0 = A e ia , z - zo = rei.P (0 < r < R)
where a is chosen arbitrarily if a 0 = 0. We observe that A= JaoJ = Jf(zo)J
and B > 0. Then (8.11-1) can be written
(8.11-2)Next we choose the argument 1/J such that /3 + k,,P = a, i.e., such that
1/J = (a - /3)/k. With this choice (8.11-2) becomes
f(z) =(A+ Brk)efo + ak+Irk+Iei(k+l)(a-,8)/k + ...
and taking absolute values, we have
Jf(z)J 2:: JA + Brk - (Jak+iJrk+^1 + lak+2lrk+^2 +···)I
or
(8.11-3)Let g(r) = lak+ 1 lr + JaH2lr^2 + · · ·. Since r < R the power series on
the right converges (it may reduce to a polynomial, in particular, to the
constant 0). At any rate, g(O) = 0, and by the continuity of g there exists
r1 such that if r < r 1 , then
g(r) = Ja,.+ilr + lak+2lr
2
+ · · · <^1 / 2 B
Hence for the indicated choices of 1/J and r < r 1 , we have
if(z)I ;::: A+^1 / 2 Brk >A= lf(zo)I
If f(z) is conjugate analytic, then J(z) is analytic, and by the preceding
proof we have IJ(z)J > IJ(zo)J, or Jf(z)I > IJ(zo)I for some z in a deleted
neighborhood of zo E G.
Remarlks I. Theorem 8.29 shows that the modulus of an analytic (or
conjugate analytic) function does not have either a relative or an absolute
maximum in G.
II. A shorter topological proof of Theorem 8.29 based on the open
mapping theorem is given in Section 9.16.
Clearly, if f(z) = c for all z E G, then max Jf(z)I is attained everywhere
in G, and if max lf(z)I is attained at some point z 0 E G, then J(z) must
be a constant in G.
Theorem 8.30 Let f be a nonconstant analytic (or conjugate analytic)
function on a bounded region G, and suppose that f is continuous on
BG. Then Jf(z)I attains a maximum value somewhere on BG (maximum·
modulus principle).