1550251515-Classical_Complex_Analysis__Gonzalez_

652 Chapter9

However, g'(z) = -1/(c+dz)^2 , so that g'(O) = -l/c^2.

Definition 9.3 A function f that is regular at oo is said to have a zero
of order m, or of multiplicity m (m 2:'.: 1 an integer) at oo iff i = 0 is a zero

of order m of g(z) = f(l/z). It follows from (9.1-5) that the derivative of

a regular function at oo has at least a zero of order 2 at oo.

Example The function f(z) = (z + 1)/(z + 2) is regular at z = oo since
g(z) = (1+z)/(1+2z) is regular at z = 0, and f(oo) = g(O) = 1. Also,

1 z^2

f'(z) = (z + 2)2' h(z) = (1+2z)2

so that f'( oo) = h(O) = O, the point oo being a zero of order 2 off'.

Definition 9.4 A point where a function f is regular is said to be a regular

point of f. If f is regular at every point of a set E then f is said to be

a regular function on this set.

Theorem 9.1 If f is regular at oo and this point is a zero of order m

(m 2:'.: 1) off, then oo is a zero of order m + 1 off'.

Proof If oo is a zero of order m of f we have

t(~) =g(z)=zmG(z)

where G(z) is analytic at z = 0 and G(O) f= 0. Hence,

f' (~) ( ~;) = mzm-^1 G(z) + zmG'(z)

= zm-^1 [mG(z) + zG'(z)]

= zm-1,,p(z)

where ,,P(z) = mG(z) + zG'(z) is analytic at z = 0. Since

J' ( ~) = -zmH,,p(z)

and 'lj;(O) = mG(O)-=/= 0, it follows that oo is a zero of order m + 1 off'.

Remark In Theorem 8.20 we have shown that if a finite number a is a

zero of order m 2:'.: 2 of an analytic function f, then a is a zero of order

m - 1 off'.

Example For f(z) = 1/z^2 , z = oo is a zero of order 2, and for f'(z) =
-2/ z^3 , z = oo is a zero of order 3.