1550251515-Classical_Complex_Analysis__Gonzalez_

(jair2018) #1
Sequences, Series, and Special Functions 577

Proof Similar to that ~f Theorem 6.29.

Remarks I. Some authors call Theorem 8.29 the weak form of the maxi-
mum modulus principle, and Theorem 8.30 the strong form of the principle.
This remark applies also, mutatis mutandi, to the following two theorems.
II. The region G in Theorems 8.29 and 8.30 may be simply or mul-
tiply connected. Furthermore, since the weak form of the maximum
modulus principle is a local property, it holds also for multiple-valued an-

alytic functions, provided that lf(z)I is single-valued in the region under

consideration.

III. If the region G in Theorem 8.30 is unbounded, the maximum mod-


ulus principle is no longer valid in general; i.e., it may or may not hold

depending on the case. For instance, if G = {z: lzl > 1} and f(z) = 1/z,

the maximum of lf(z)I is attained on 8G = {z: lzl = 1}. However, if


f(z) = ez, then lf(z)I =ex, and if G = {z: Rez > O}, lf(z)I is unbounded

on G while lf(z)I = 1 on 8G = {z: Rez = O}. Here oo is not regarded
as a boundary point of G.

Theorem 8.31 Let f be a nonconstant analytic (or conjugate analytic)

function in a region G, and suppose that f(z) 'f. 0 for every z E G. Then

lf(z)I does not attain a minimum value anywhere in G.

Proof Since f(z) 'f. 0 everywhere in G, the function g(z). = 1/ f(z) is


analytic (respectively, conjugate analytic) in G. By Theorem 8.29 lg(z)I =

1/lf(z)I does not attain a maximum value anywhere in G. Hence it follows

that lf(z)I does not assume a minimum value anywhere in that region.

Note If f(z 0 ) = 0 for some point z 0 E G, then lf(z)I will attain a


minimum value at z 0 and the conclusion of the theorem is not valid.

Theorem 8.32 Let f be a nonconstant analytic (or conjugate analytic)

function on a bounded region G. Suppose that f(z) 'f. 0 for every z E G

and that f is continuous on 8G. Then lf(z)I attains a minimum value

somewhere on {)G (minimum modulus principle).

Proof Similar to that of Theorem 6.31.

Example Given f(z) = z + 2 to find max lf(z)I and min lf(z)I on lzl :5 1.
Since the given function is conjugate analytic on lzl :::; 1, both max Jf(z)J
and min Jf(z)I will occur on the boundary of the unit disk. For z = eit,
0 :5 t :5 271", we have


Jf(z)J = yf(cost + 2)^2 + (-sint)^2
= v'5 + 4cost
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