312 Chapter 61. If f(z) = c (a constant) in A, f'(z) = 0 for all z EA.
2. If f'(zo) and g^1 (z 0 ) exist for z 0 E A, we have:
(a) (f + g)'(zo) = f'(zo) + g'(zo)
(b) (fg)'(zo) = f'(zo)g(zo) + f(zo)g'(zo)
(c) (f/g)'(zo) = [f'(zo)g(zo) - f(zo)g'(zo)]/g(zo)^2 provided that
g(zo) -:/- 0.3. If B is open, h: B ---+ C, F(z) = (ho f)(z), wo = f(zo), and f'(zo),
h' ( w 0 ) exist, it follows that F' ( z 0 ) exists andF'(z 0 ) = h'(w 0 )f'(z 0 ) (chain rule)
4. Let w = f(z) be analytic at z 0 , and let w 0 = f(zo). If f'(zo)-:/-0, the
equation w = f(z) is solvable with respect to z by an analytic function
z = F(w) = f-^1 (w) (the inverse off) in some neighborhood Ne(w 0 ).
Moreover, F'(wo) = 1/f'(zo).
Proofs Parts 1 and 2 follow from Theorem 3.2, as in calculus.
To prove Part 3, we let w = f(z), ( = h(w), 6w = f(zo + 6z)-f(zo),
6( = h(wo + 6w) - h(w 0 ), and write6( = (h'(w 0 ) + c:) 6w = h'(w 0 ) 6w + c; 6w (6.3-1)
where c: == c:(6w) = 6(/6w - h'(w 0 ) for 6w -:f. 0 and c:(O) = 0. Clearly,
c; ---+ 0 as 6w ---+ 0, so that c:( 6w) is continuous at the origin.
By dividing both sides of (6.3-1) by 6z -:f. 0, we get
6( = h'(wo) 6w +c: 6w
6z 6z 6zand we conclude that
F'(zo) = lim ~( = h'(wo)f'(zo)
.O.z-+0 uz
(6.3-2)The chain rule is often written in the form
d( d( dw
dz dw dzwith the understanding that dw /dz is to be evaluated at z 0 while d(/ dw
is to be evaluated at w 0 • As a consequence, if f is analytic in A and h is
analytic in B, the composite function ho f is analytic in A.
The proof of the first part of ( 4) will be made later in Theorem 9.25.As to the second part, the continuity of f at z 0 implies that corresponding
to the radius c; of Ne(wo) there is a o > 0 such that if l6zl = lz - zol < o
then J6wJ = Jw -wol = Jf(z) - f(zo)I < c:. Because of the existence of
the inverse function in Ne( wo) we have 6w =/=- 0 whenever 6z =/=- 0 (i.e.,