1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/Applications 751

has a zero other than a in iz - al :::; 6'. Thus if z 1 E N 6 1(a) is such that

f ( z1) = b, then f' ( z1) -:f O, and z1 is a simple zero of f ( z) - b.

Theorem 9.24 (Open Mapping Theorem). Let f be a nonconstant ana-

lytic function in a region R. If A is an open subset of R, then f(A) is an

open subset of C. That is, the mapping defined by a nonconstant analytic

function is open (Definition 3.14). In particular, f(R) is open and also a

region, since the continuous image of a connected set is connected.

Proof Let Wo be a point in f(A), and let f(zo) = w 0 , where.z 0 EA. Since

A is open there exists a neighborhood N 0 (z 0 ), 6 > 0, with 6 chosen as in

Theorem 9.23, such that N 0 (z 0 ) C A. By the same theorem, there exists

e > 0 such that for every w E Ne(w 0 ) there is. some z E N 0 (z 0 ) such that

w = f(z). Hence Ne(wo) C f(A), which shows that f(A) is open.

Note Theorem 9.24 provides a brief and elegant proof of the weak form
of the maximum modulus principle (Theorem 8.29). In fact, since f(R)

is open, wo = J(zo) lies in a neighborhood Ne(wo) C J(R). In Ne(wo)

there are points W1 such that lw 11 > lw 0 1, say, w 1 =· (1 + e')w 0 • For a

suitable f the point w 1 is of the form w 1 = f(z 1 ) for some z 1 ER. Hence

lf(z1)I > IJ(zo)I.

Similarly, if u = Ref, there are points w 1 E Ne( w 0 ) (those to the right

of the line u = uo) such that u1 > uo, and points w2 E Ne ( wo) (those to

the left of u = u 0 ) such that u 2 < u 0 • Hence u attains neither a maximum

nor a minimum at (xo, Yo).

Theorem 9.25 (Inverse Function Theorem). Let f be analytic in some

open set A, let z 0 E A, and suppose that f'(z 0 ) -:f 0. Then there exists a

neighborhood U of Zo and a neighborhood V of w 0 = f(zo) such that f:

U ~ Vis bijective (one-to-one and onto), so that the inverse (single-valued)
function f-^1 exists. Furthermore, f-^1 is analytic in V.

y v

0 x^0 u

Fig. 9.34