A POLYA " SHIRE " THEOREM FOR FUNCTIONS WITH ALGEBRAIC SINGULARITIES

The classical "shire" theorem of P61ya is proved for functiuns with alge- braic poles, in the sense of L. V. Ahlfors. A function f(z) is said to have an alge- bralc pole at z 0 provided there is a representation f(z) lk=_Nak(z z 0) + A(z), where p and N are positive integers and A(z) is analytic at z O. For p i, the proof given reduces to an entirely new proof of the shire theorem. New quantitative results are given on how zeros of the successive derivatives migrate to the final set.

say that the shire of a pole I of f(z) will consist of all points z in the plane which are nearer to I than to any other pole of f(z).Then the final set of a meromorphic function consists of the boundaries of the shires of its poles.
If f(z) has but one pole, the final set is empty, and the single pole generates zero-free regions for the successive derivatives of f(z).The asymptotic size of such zero-free regions has been obtained in [4].Thus the poles of f(z) may be thought of as repellers of equal strength of the zeros of the derivatives of f(z).A refinement of the "equal strength" aspect of the Plya shire theorem is one of the features of this paper.
The general questions posed in P61ya's classic paper [i] concern the effect of the singularity structure of an analytic function on its final set.The present paper takes up the problem of determining the final sets of functions with algebraic singu- laritie8.We take a countable set A of points in the plane , and for each lea we let (%) denote the "principal branch line" (%) {z % + tlt < 0}, corresponding to I.
We assume in addition that each bounded region D intersects the branch lines of only finitely many leA.The functions f(z) we study are those which have "algebraic poles", in the sense of Ahlfors [5, p. 299], at the points of A. To say that f(z) has an algebraic pole at z 0 means that there are integers p > i and N > i such that f(z) has the representation k/p f(z) [
is analytic in D. By (z l) we mean the branch which is analytic for z(%) and real if Ira(z) Im(1) and Re(z) > Re(1).The factors (z-l) k/p(1) k > 0 cannot be thrown over into the analytic part AD(Z), as would happen if f(z) were meromorphic.Thus the singularities of f(z) in D arise from the algebraic poles and branch lines there, and may be subtracted off in the manner of poles for meromorphic functions.Since the unique representations (1.2) for f(z) are to hold for arbitrary D, then the coeffi- cients {c k()} k=0 must be entire coefficients; i.e., for each %EA, the series k0Ck(%)(z %)k/p(%) must converge for all complex numbers z.
Disregarding the branch lines, form the shires of the points %cA by the classical definition and let P(A) be the union of all the boundaries of the shires.Our princi- pal result asserts that in spite of algebraic singularities, P(A) is still the final set of f(z), with one exceptional case.Not surprisingly, difficulties arise when a horizontal line segment in P(A) determined by two points %l,%2eA coincides with the branch line of a third point %3eA.Should this occur, we will suppose that II and 2 are of unequal strength, in that either (Nl/Pl) # (N2/P2) or feN1 # ICN21 if (Nl/Pl) (N2/P2), where N 1 N(% I), Pl P(%I )' etc.
THEOREM i.If f(z) is defined and restricted as above, the final set of f(z) is Note that the theorem makes no mention of the branch lines apart from the exceptional case; the branch lines come into play only in the proof, and only in the excep- tional case.
We prove Theorem i in the next section, after which we make a series of Remarks on how the proof can be modified to cover various situations.For example, it is reasonable to ask how the theorem would be altered if we allow zeros of derivatives of all branches to contribute, instead of a single, fixed branch.
There is no hope of treating the case where (1.2) is replaced by a doubly in- finite series.Even if p i, essentially nothing is known about final sets for functions with isolated essential singularities.
Our proof of Theorem i covers the case where f(z) is meromorphic (p--i) and with- out restriction on the location or strength of poles.As such, it is a new proof of the Plya shire Theorem, independent of the two existing proofs in [6], [I], [3].More- over, the classical proofs do not extend to the case of algebraic singularities, and this is why it was necessary to devise a new proof.
The classical theory asserts that the location of the final set does not depend on the order of the poles or the size of their coefficients.Even so, our proof reveals the previously unobserved phenomenon that if adjacent poles have unequal strength, then all the zeros of sufficiently high order derivatives are pushed to one side of the common boundary of the two shires.
Various other features of the proof bearing on the migration of zeros to the flna.
set are mentioned in the Remarks.

PROOFS AND REMARKS.
It is convenient to express the derivatives of the factors (z %)k/p in terms of the Gamma function, and in terms of certain factors Ck,n which were studied in the paper [4].For an arbitrary integer k, let us define where p > i, so that the tn-h derivative of (z %)k/p may be written as (see [7, p. 25.-,).We require certain information about the asymptotic behavior of the terms in (2.1).To begin with, if a and b are arbitrary real numbers, then [7, p. 257] Therefore, if N is a positive integer we have CI (p-l+n) (Np-.I) ,n F F 0 (n(1-N)/P), n =, CN,n r(p -I r (Np-l+n) using order-of-magnitude notation.
Let [x] denote the greatest integer not exceeding the real number x, and define max l_<k< (n-l) p] IC-k,n/Clnl n--1,2,3,...We shall need the result of Lemma 3.1 of [4] which asserts that there is a constant A, which depends only on p, such that Our first lemma is a result which shows that compact subsets of shires of points are zero-free for sufficiently high order derivatives of f(z).A similar result for meromorphic functions appears in Polya's original proof [1,3].LEMMA I. Let aeA and let f(z) be defined as in Theorem i.Then lid z al If (n)<-)/n!llln 1 uniformly on compact subsets of a-shire.
PROOF.Let K be a compact subset of a-shire and let R > 0 be such that K is con- tained in Iz a < (R/2).Let D be the disc Iz a < R and put A(D) {%AI(%) intersects D, # a}.Then for zD-%AU(D)(%) and z(a), f(z) can be represented as f (z)   .ak(z-a) k/p + .cj(%)(z-%) j/p(E) + (z) where (z) is analytic in D and where we have written a i ci(a) and N N(a).
We are going to show that the sequences F (z) Gn(Z) and H (z) all converge to 0 n n as n , unifory on D. Starting with F (z), we first break its defining sum into n three parts, with k ranging successively over -N + greatest integer function) and [(n-1)p] + 1 k < .The k t--)-h term of F (z) in the n first range has order of magnitude, for large n, r(-kp-l+n)/r(Np-l+n) 0(n-k+N)/p) o(i), n =, since k >--N + i.Since the sum over the first range has only finitely many terms, it follows that this part of F (z) is uniformly small for large n, zeD.where J is constant.By (2.2), the above does not exceed JAIa-N I-In-(N/p) k=l lakl Iz al(k+N)/p" Since k=l akl z a k/p is uniformly bounded over D, it follows that the compo- nent of F (z) coming from the sum over the range i _< k -< [(n l)p] also converges n uniformly to 0 on D. Looking at the third range, k -> [(n l)p] + i, we use (2.1) and (2.3) and bound the sum by J la_N I-I n(I-N)/P la kl nk Iz al (k+N)/p k=n .01a_NI-I n(i-N)/Plz-a IN/p J k=n (nk)( lakil/klz_all/p)k Since lak ll/k O, then lak ll/k Iz-al I/p <-y < (1/2) for all zeD, if k is sufficiently large.This means that the infinite sum in the above expression does not exceed k -i n+l k yn Y i -k=n for n sufficiently large.Since (/(l-y)) < i, then the component of F (z) taken from n the range k > [(n-l)p] + i is asymptotically smaller than n(l-N)/P(/(l-l)) n+l O, n .This completes the proof that F (z) 0, n uniformly in D n As for G (z), we need the fact that K is compact and in a-shire so that there n assuredly is a number T, 0 < T < i, such that z a < z l for all zeK and leA(D).We pick one leA(D) and break the corresponding sum over j into three parts exactly as above.A typical term in G (z) lying in the range -N(1) _< j _<-i has order n of magnitude "r n n-(J+N)/P(1) O, n ,.Hence this component of G (z) tends uni- n formly to 0 over K.The arguments for the ranges i <_ j _< [(n-l)p(1)] and [(n-l)p(X)] + I <_ j < run exactly as the corresponding ones for F (z) except that we now have n zeK and the additional factor n before each sum.We omit the details but conclude that G (z) 0, n , uniformly on K.
If zeK then (Iz -allo) <-7 < (i12), for some 0 < R and a constant 7, and so the 0-term in (2.6) is dominated by O(y/(l y))n.Then since n! F(n + i), we have Going back to (2.5) we now have (1 + o(1) and since [r(Np-l+n)Ir(l+n)] n (N/p)-I the desired result follows Lemma i shows that no point of any shire can be in the final set.The next lemma establishes the existence of zeros of derivatives near points on the boundaries of shires, and thus Theorem i will follow.
LEMA 2. Let $eP(A) and let f(z) be defined as in Theorem i.Then every neigh- borhood of $ contains zeros of f(n)(z) for n sufficiently large.
PROOF.First we note some reductions.If $ lies on the boundaries of a-shire, b-shire and other shires, then we can consider a point ' near which lies on the boundaries of a-shire and b-shire only.If we produce zeros in arbitrary neighbor- hoods of $', then zeros in neighborhoods of $ are implied by elementary arguments.
We note that in this reduction the point ' can also be taken not to lie on any branch line with the single exception that a and b have the same real part.In this case, re- call our hypothesis that a and b are of unequal strength.Our proof, then, splits in- to the two cases where either does or does not lie on a branch line, but in any case is on the common boundary of exactly two shires.
CASE I: E does not lie on a branch line.Suppose E lies on the common boundary of a-shire and b-shire, let R > 0 so that I al < (R/2) and b < (R/2), let be the union of the discs Iz a < R and Iz b < R, and put A()--{eAl(%) inter- sects , % # a, # b}.Then if ze-hA)(%), f(z) may be expressed as where A(z) is analytic in and p, ak, N, q, bk, M have obvious meaning.If we differentiate this representation, we get where akF (-kp-l+n) F (Np-I) (z-a) (k+N)/p F (z) [.
-1 n k---N+l a_NF (-kp F (Np-l+n) DkF (-kq-l+n) F (Np-I) (z-a) (N/p)+n c. (%) F (_j p (%) -l+n) F (Np-l) (z_a) (N/p)+n n %eA() j=-N(%) a_NF(-jp(%)-l)F(Np-l+n) (z-%) -(j/p())+n If we group Gn(Z) with I (z) and factor out the term [(z-a)/(z-b)]n we get the ex- n pression bkF (-kq-l+n) F (Np-I) a_Nr(Np where U (z) and V (z) denote the respective terms in the brackets.Let T be a compact n n neighborhood of , with in its interior and We calculate the asymptotic form of U (z) as n , zeT.Write U (z) in the form We proceed to treat the above infinite sum inside the brackets just as in Lemma i i.e., we break it up into summands with ranges M k -i, i k [(n-l)q] and [(n-l)q] + i k < .The kth term in the first range has dominant asymptotic form -(k+M)/q n 0, n , since k e -M + i.In the middle range, take note of (2.1) and (2.2) to see that the significant terms asymptotically are [bk[ 6n n (I-M)/q z b[k/q k=l < JA(q) n -M/q [bk[ [z b[k/q k=l where J is a constant and the terms of (2.1) now depend on q instead of p.This term tends to 0 with n in view of the fact that Ek=I [bk[ [z-b[ k/q is uniformly bounded on T. The sum in U (z) over the range [(n-1)q] + i k < behaves llke n .[bkl IC_k, /Cl,nl ICI, and the rest of the proof runs as in Lemma i.Consequently, we may say that U n(z) r(Np-1) (z.-.a)N/P + o(i) n a_Nr(Np-l+n F(Mq-I) (z-b) M/q where o(i) denotes a term which,converges uniformly to 0 in T.
Since T lies in [z b[ < (R/2), we may treat the term V (z) as we did the func- n tion AD(Z) in (2.6).That is, which means that V (z) 0 uniformly in T.Moreover, if we let W (z) U (z) + V (z), and the convergence is uniform in T.
The functions F (z) and H (z) in (2.7) have the same form as F (z) and G (z) in n n n n (2.5).It is easy to see that the arguments of Lemma i carry over to show that F (z) (2.10) All terms in (2.9) are analytic in T.
Let us now consider the covering properties of the mapping w (z-a)/(z-b) for z near .Recall that lies on the perpendicular bisector of the line segment from a to b.Moreover, we may assume without loss of generality that # (a+b)/2.The image i0 m e of under the map lies on the unit circle in the w plane.In fact, the image of the perpendicular bisector is the circle lwl i.Note that m # I and m # -I (the image of (a + b)/2) so that we can surround m with a sectorial neighborhcod S r < ]w] < r-1, 0 g < arg w < + g, where 0 < r < 1 and g > O, whose image omits the points w i and w -i.The inverse image of S under the mapping w= (z-a)/(z-b) is the region S in the z-plane which contains and is bounded by the four circles respectively by appropriate portions of lwl r, arg w e 6, lwl r -I and arg w + 6, and let YI' 2' Y3' 4 be the corresponding inverse images forming the bound- aries of S. Finally, suppose that and (i r) are sufficiently small that S T.

Figure i
Let e >0.By (2.8) we can make the argument of W (z), as z ranges over a small T, n vary by less than by taking n sufficiently large; i.e., A T arg Wn(Z) < e for all large n.Now as z ranges over 71' w (z-a)/(z-b) traces out the circular arc F 1 on lwl r.Thus we can make [(z-a)/(z-b)] n wrap around lwl r as many times as desired by taking n large.Thus for suitably large n, the variation of the argument of W (z)[(z-a)/(z-b)]n for ze71 can be made large.Specifically require we n Ayl Wn(Z)[(z-a)/(z-b)]n > 4n + 4e, for all large n.On Y2 and Y4' (z-a)/(z-b) has .constantargument, so we may impose Ay2 arg W n(z)[(z-a)/(z-b)]n < and A4 arg Wn(Z)[(z-a)/(z-b)]n < simply by taking n large enough.

n
The sum over the second range is in view of (2 i) and the definition of n/Cl n[ [C I /C N [z al (k+N)/p k=l ,n ,n _< J a-Nl-I [(nl)p]..lakl 6n n k=l (1-N)/p ]z aI(k+N) Ip

n
Working with H (z), we first have n Gn(Z) + In(Z) Wn(Z)(z )nr (Np-I) (z-a) (Nlp)+n (n) only if f(n)(z) 0, and the foreBoing discussion yields the -1 which are the inverse images of the circle Iwl r, Iwi r and lines arg w + , arg w--e 6; see Figure i.Let F I, F2' 3' 4 denote the boundaries of S formed