1550251515-Classical_Complex_Analysis__Gonzalez_

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664 Chapter^9


has a Taylor series representation


f(z) = ao + aiz + · · · + anzn + · · ·

valid for every z EC. If f has a pole of order m (m;::: 1) at oo, then


g( z) = f ( ~) = ao + ai ( ~) + · · · + an zln + · · ·

must have a pole of order m at 0. This implies that am 'f. 0 and an = 0


for n > m. Hence it follows that

f(z) = ao + aiz + · · · + amzm

i.e., f(z) is a polynomial of degree m.
We have seen (Definitions 6.2) that the functions which are analytic in C
are called integral or entire functions. By the preceding theorems, the entire
functions having at most a pole at 00 are the polynomials, which reduce
to constants if they are regular at oo. Entire functions having an isolated
essential singularity at oo are called transcendental entire functions ..


Examples ez, sin z.


Definition 9.10 A function f is said to be meromorphic in an open set

A ifj is regular in A except possibly at a finite or infinite number of poles.
This definition is phrased so as to include the class of holomorphic (i.e.,
analytic) functions in the class of meromorphic functions.


Exampl1es The function f(z) = z/(z^2 +4) is meromorphic in C with poles

at z = ±2i. The function f(z) = tanz is meromorphic in C with poles at

z = ±(2k + l)7r/2 (k = O, 1,2, ... ). However, taiiz is not meromorphic in
C*, since oo is a cluster point for this function.


Theorem 9.8 The functions that are meromorphic in C* are rational
functions, and conversely.


Proof Let f be a meromorphic function in C. Then f has in C a finite

number of poles; otherwise, f would have a cluster point in C* (which is a

compact set). If f has no poles in C, then f will be either a constant or


a polynomial (other than a constant), depending on whether f is regular

or has a pole at oo.
Next suppose that f has some poles in C, and let ai, a 2 , ••• ak be the


finite poles off. If z = a 1 is a pole of order m off, by (9.2-1) we have, in


some deleted neighborhood 0 < lz - a 1 I < 8,


f(z) = A_m(z - ai)-m + · · · + A_1(z - ai)-^1 +Ao+ Ai(z - ai) + · · ·
= P1(z) + f1(z)
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