Differentiation 357provided that l~zl is taken small enough, say l~zl < 6, in which case
llfzl - lfzll - (11111+11121) > 0, and (6.13-1) implies that l~wl > 0. Hence
w 0 has just one inverse image in N 0 (z 0 ), namely, the point z 0 itself.
RemarksI. If f is analytic in A'. The condition If z I f= If z I implies that f' ( z) f= 0
in A'. If f is conjugate analytic in A' the condition implies that J' ( z) f=
0.
II. Theorem 6.22 does not hold in the large. For instance, if f(z) = ez,then f'(z) = ez f= 0 for all z. Yet the points z 0 and z 0 + 2k7ri (k
an integer) all map into wo = ezo. However, in a sufficiently small
neighborhood of zo the mapping is one-to-one.III. Since J(u,v/x,y) = lfzl^2 -lfzl^2 , the condition lfzl f= lfzl implies that
J f= 0 in A'. Thus our theorem is equivalent to the existence of a local
inverse to the mapping defined by u = u(x, y), v = v(x, y), where u andv are the components off. However, here we have shown the validity of
the result without assuming the continuity of u.,, uy, v.,, and Vy.
IV. The condition lfzl f= lfzl in A' is also equivalent to fMz) f= 0 in A'.Theorem 6.23 Let f be defined on A, where A= N 0 (zo), 6 > 0, and
suppose that:
1. f E '.D(A).
- lfzl f= lfzl for each z E A.
3. f is continuous on 8A = {z: lz - zol = 6}.
4. f is one-to-one on A.
Then f(A) contains a neighborhood of w 0 = f(zo); i.e., for some e > 0
we have Ne(w 0 ) C f(A). In other terms, f is locally open at zo.
Proof Consider the real-valued function
g(z) = lf(z) - f(zo)Ifor z E 8A. Since f is one-to-one on A, it follows that g(z) > 0 for each
z E 8A. Also, g is continuous on 8A, for f is continuous on 8A. Since 8A
is a compact set, g(z) must assume an absolute minimum mat some point
of 8A, and we have m > 0 because g( z) > 0 for each z E 8A.
Now, let e = %m and consider the neighborhood Ne(wo). We wish to
prove that Ne(w 0 ) C f(A), or that w' E Ne(wo) implies w' E f(A).
To accomplish this, we introduce another real-valued function
h(z) = lf(z) - w'Ifor z E .A, with w' kept fixed. Then h is continuous on the compact set A
and attains an absolute minimumμ somewhere in A. In fact,μ is attained