1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/ Applications

where

P1(z) = A-m(z - ai)-m + · · · + A-1(z - ai)-^1

fi(z) =Ao+ Ai(z - ai) + A2(z -ai)^2 + · · ·

665

We note that Pl ( z) (the principal part of f at a 1 ) is regular for every

z # a 1 (including z = oo ), and has at a 1 a pole of order m, while Ji ( z) =
f(z) - P1(z) has the same poles as f except at a 1 where f 1 (z) is regular.

For instance, a 2 is a pole of Ji since

lim lf1(z)I 2:: lim llf(z)l - IP1(z)ll

z-+a2 z-+a2

=I z-+a2 lim lf(z)I - z-+alim IP1(z)ll = oo

2

Then, if P2(z), ... , Pk(z) are the principal parts off at the poles a2, ... ,
ak, the function

fk(z) = f(z) -P1(z)- P2(z)-· · · - Pk(z)

is regular in C and has at most a pole at oo. Hence by Theorems 9.6 and 9.7,
fk(z) is a polynomial (which may reduce to a constant). Therefore, we have

f(z) = P1(z) + P2(z) + · · · + Pk(z) + fk(z)

which shows that f(z) is a rational function.
To prove the converse, let

J(z) = P(z) = aozn + a1zn-l +···+an

Q(z) bozm + b1zm-l + · · · + bm

(9.6-1)

where a 0 -=f; 0, b 0 -=f; 0, and the fraction P(z)/Q(z) is supposed to be in
simplest terms: i.e., P(z) and Q(z) have no common factors of the form
z -c. Let (3 be a zero of order q of the denominator. Then we have

Q(z) = (z -(3)qQ1(z)

where Q 1 (z) is a polynomial of degree m-q and Q 1 ((3) # 0. It follows that

P(z) h(z)

f(z) = (z - f3)qQ1(z) = (z -(J)q

where h(z) = P(z)/Q 1 (z) is regular at z = (3, and h((J) # Oj so that
(3 is a pole of order q of f(z) (Section 9.2). Hence a rational function
P(z)/Q(z), when expressed in simplest terms, has poles at the zeros of Q(z)
and with the same multiplicity. This justifies the definition introduced in
Section 5.16.
As to the behavior off at oo, we have seen in Section 5.16 that if m 2:: n,

oo is a regular point off, and if m < n then oo is a pole of order n - m of