Singularities/Residues/ Applications 651Since g(z) is supposed to be regular at z = O, we must haveg(z) = bo +biz+ b2z^2 + · · · (9.1-1)
{
valid for lzl < 8. Then we have the representationf(z) = g ( ~) ~ bo + biz-i + b2z-^2 + · · · (9.1-2)
valid for lzl > 1/8, and we define
f(oo) = g(O) = bo
From (9.1-2) it follows thatf'(z) = -biz-^2 - 2b 2 z-^3 - • • •
valid for lz I > 1/8. Letting
h(z) = f' ( ~) = -biz^2 - 2b 2 z^3 - .. •
we have, on applying (9.1-3),
f'(oo) = h(O) = 0(9.1-3)(9.1-4)(9.1-5)(9.1-6)Hence, if a function f is regular at z = oo its derivative is also regular at
z = oo but it vanishes at this point; i.e., f' has a zero at z = oo.
From (9.1-4) it follows that
lim f'(z) = f'(oo) = 0
z->oo
Some authors (see e.g., [15],I, p. 125) adopt the definition
f'(oo) = g'(O) =bi
However, with this definition of f'(oo) we have f'(oo) f limz_, 00 f'(z)
unless bi = 0.
Example Let
f(z) = az + b,
cz+d
ad - be = 1, c f 0
Then
( )
_ a+ bz
gz---d-'
c+ z
1
J'(z)= ---
(cz + d)^2 'z2
h(z)= --. (c + dz)^2
According to (9.1-3) and (9.1-6) we have
a
f(oo)=g(O)= -,. f'(oo)=h(O)=:O
c