Singularities/Residues/ Applications 683where m ~ 2 and h(O) =/:-0. ThenandLetting z = Rei^8 it follows thatlh(e-i^8 /RI
R-+oo lim lzf(z)I = R-+oo lim Rm-1 =^0independently of 8, since lh(e-i^8 /R)I < %1h(O)I for R large enough. Hence
zf(z) =* 0 as R ---t oo, and it suffices to apply Lemma 9.3.
Corollary9.5 If f(z) = P(z)/Q(z), where P(z) and Q(z) are polynomials
prime to each other and of degrees n and m, respectively, with m - n ~ 2,
then
. J P(z)
J!!!1oo Q(z) dz= 0 (9.10-5)
'"'(Proof It follows from Corollary 9.4 since then f(z) = P(z)/Q(z) is regular
at oo and has at this point a zero of order ~ 2.
Lemma 9.4 Let 1: z = Rei^8 , a :::; 8 :::; (3, and suppose that:
- f(z) is continuous on 'Y for R :::; R-o
- zF(z) =* L as R---+ 0 for z E /, L being a constant.
Then
lim jJ(z)dz=iL(f3-a)
R->oo(9.10-6)Proof As in Lemma 9.3, assumption 2 implies that zf(z) = L + 'T/ with
l"ll < e, e > 0 arbitrary, provided that R < R,. Hence if R < min(R-o, R,),
we have