656 Chapter^9
Conversely, if f(z) can be expressed in the form (9.2-4) where h(z) is
analytic at z = a, and h( a) -f:. 0, then the point a is a pole of order m of
f. In fact, by the Taylor expansion theorem we have
h(z) = h(a) + h'(a)(z - a)+ ... + h(m)(a) (z -ar + ...
m!valid for lz - al < r, and we obtain
h(a) h'(a) h(m)(a)
f(z)= (z-a)m + (z-a)m-1 +···+ m! +···valid for 0 < lz - al< r. Since h(a) "f:. 0, z =a is a pole of order m off.
Similarly, if f has a pole of order m at oo, we have
f(z)= ···+ B_l +Bo+B1z+···+Bmzm
z=z m(B m+--+···+-+--+··· Bm-1 Bo B-1 )
z zm zm+l
or
(9.2-5)where h(l/z) = 1/J(z) = Bm + Bm-1Z + ·· · + Bozm +···is analytic at
z = 0 and 1/J(O) = Bm -f:. 0. Conversely, if f(z) can be expressed in the
form (9.2-5), where h(l/z) = 1/J(z) is analytic at z = 0and1/J(O) -f:. 0, then
oo is a pole of order m of f.
9.3 Behavior at a Pole
Theorem 9.2 The poles of f(z) are ~p.e zeros of 1/ f(z) with the same
multiplicity, and conversely.
Proof For a "f:. oo it follows at once from (9.2-4). If a = oo, then (9.2-5)
holds with 1/J(z) = h(l/z) analytic at z = 0and1/J(O) "f:. 0. We have
1 1
F(z) = f(z) = zmh(z)To show that F(z) has a zero of order m at oo, considerG(z) = F (;) = h(~~z) = ;(:) = zmw(z)
where w(z) = 111/J(z). Since 1/J(z) is analytic at z = 0 and 1/J(O) "f:. O, the
function w( z) has the same property. Hence z = 0 is a zero of order m of
G(z). This implies that z = oo is a zero of order m of F(z).
The converse follows in the same manner.