AN INVITATION TO THE STUDY OF UNIVALENT AND MULTIVALENT FUNCTIONS

We begin with the basic definition and some very simple examples from the theory of univalent functions. After a brief look at the literature, we survey the progress that has been made on certain problems in this field. The article ends with a few open questions.

i. THE HEART OF THE SUBJECT.
We are concerned with power series w f(z) b" z n b 0 + b z + b2 z2 + n I n=0 (i.i) in the complex variable z x + iy that are convergent in the unit disk E zl < i.Such a power series provides a mapping of E onto some domain D.
Two questions present themselves: (A) given the sequence of coefficients b0, bl, b2, what can we say about the geometric nature of D: and (B) given some geometric property of D what can we say about the sequence b 0, b I, b 2, An example of a nice geometric property is given in DEFINITION i.A function f(z) that is regular (holomorphic) in E is said to be univalent in E, if it assumes no value more than once in E. Such a function is also called simple or schlicht in E. When f(z) is univalent in E we say that the domain D f(E) is a simple (schlicht) domain.
Stated algebraically, f(z) is univalent in E if the equation w 0 f(z) has at most one solution in E for each complex w 0. If f(z) is univalent in E, then f' (z) # 0 in E. But one must be careful because the converse is false.
As trivial examples, we mention that f(z) -= z is univalent in E while 2 f2(z) z is not univalent in E. The function z + zn/n is univalent in E for each positive integer n.The function sin z is univalent in the larger disk z < /2.
divide by bl, and then write f(z) in the form (1.2) Geometrically this amounts to shrinking or expanding the domain D, and possibly rotating D. This does not disturb the unlvalence of the function.
When F(z) has the form (1. 2) we say that the function has been normalized.
Other normalizations are possible, but the set of conditions F(0) 0, F' (0) i, is the most usual and the one we will use here.
We now give a very important example of a normalized univalent function.
We start with the function l+z g( and from the properties of the general linear fractional transformation, it is easy to see that g(z) is univalent in E and g(E) is the half-plane 2 Re g > 0. If we square g(z) we see that the function h(z) g (z) is also univalent in E and maps E onto the entire complex plane minus the sllt along the negative real axis from 0 to -.To normalize E we subtract h(0) i and divide by h' (0) 4. Thus we arrive at This important function is called the Koebe function and it maps E onto the entire complex plane minus the slit along the negative real axis from-1/4 to -.On intuitive grounds it is a "largest" univalent function because we cannot adjoin to K(E) any open set without destroying univalence.A short computation (starting with i/(i-z) In=0 zn) gives the power series for K(z), This power series for the "largest" univalent function suggests immediately CONJECTURE i.If F(z) is univalent in E, and has the power series (1.2), then for n

2, 3,
This conjecture has been the subject of research for more than 60 years and still represents an open problem, although it has been settled in many special cases.A complete account of these results is far beyond the scope of this paper but the interested reader can pursue the matter in the books by Spencer and Schaeffer [27], Goluzln [8], Jenkins [14], Hayman [ii, 12], Pommerenke [25] and Schober [28].
The best result known today is due to D. Horowltz [13] who proved that anl __< 1.0657 n, (1.7) using a very deep method due to Carl FitzGerald [7].
The question raised by Conjecture i, suggests a large number of related problems, some of which are open while others have been solved completely.The Koebe function is the "heart of the subject" because it appears so often, that a theorem in this subject becomes very interesting if it does no___!tuse the Koebe function or some modification of it.We will mention some of these theorems and conjectures, but first we give 2. A SURVEY OF THE LITERATURE.
The specialist in the theory of univalent and multlvalent functions must be grateful to S. D. Bernardl, who has devoted much of his time to preparing an exhaustive bibliography of the subject [3].The first volume covers the period from 1907 to 1965 and lists 1694 papers.The second volume covers the period from 1966 to 1975 and lists 1563 papers, Each of these ,2 volumes has an exten- slve index which lists subtopics in this field and those papers that touch on each subtopic.Thus it is an easy matter for the specialist to determine the status of any problem up to the year 1975.
In addition, these two volumes give references to the many survey articles on the subject that have appeared during this period.Here we cite only three such; one by D. Campbell [5], a second by Anderson, Barth and Brannan I, pp.138-144], and third, the article by the author [9], It is the purpose of this paper to review this third survey and bring it up to date by reporting on the progress made on the problems mentioned there.QUESTION i. Find necessary and sufficient conditions on the sequence a2, a3, an, such that F(z) and every partial sum S (z) is unl- It is an easy matter to prove that if k;2 (3,2) then F(z) is univalent in Eo Indeed as Ao Hurwitz first observed, where 'k(Z2,Zl) z2k-i + z-2Zl + + zlk-l, < k for Zl, z2 e E, It follows that if the condition (3.2) is satisfied then the right side of (3.3)   is never zero for Zl, z 2 e E. Now suppose (3.2) is satisfied, then the same inequality is true for each S (z) and hence the condition (3.2) is a sufficient condition for F(z) and n every S n(z) to be univalent in E.
Let PS(1) denote the set of all functions F(z) of the form (1.2) for which F(z) and all the partial sums are univalent in E. Thus the condition (3.2) is a sufficient condition for F(z) to be in PS(1).However, the condl- tion is not necessary as we will soon see.
satisfies the condition (3.4).In fact every partial sum also satisfies (3.4)   and hence in (i z) e PS(1).Thus (3.4) is also a sufficient condition for F(z) to be in PS(1).Further Alexander proved that if a3z3 a5z5 2k+l G(z) ) is also a sufficient condition for a function G(z) to be in PS(1).But neither of these conditions is included in the other, nor in the condition (3.2).The determination of all functions in PS(1) is a difficult task.Still more difficult is the generalization to p-valent functions.DEFINITION

A function F(z)
a z n is said to be p-valent in a domain E if it assumes no value more than p times in E and there is some w 0 such that F(z) w 0 has exactly p solutions in E, when roots are counted in accordance %rlth their multiplicities.
Observe that the normalization a I i (used for univalent functions) has 2 been dropped.Indeed it is no longer available because f(z) Now let PS(p) be the set of all functions F(z) such that F(z) and all the partial sums S (z), with n > p have valence not exceeding p.The deter- mlnation of PS(p) seems to be very difficult, and very little is known beyond the trivial PS(1) C PS(2) C PS(3) ....In this direction Ozaki [24] has proved that if z k . n F(z) + anZ i _< k < p (_3.9) n=k+l and if k(p-k+l) >= 7. n(n+p-l) la n n=k+l (3 then F(z) and all partial sums Sn, with n _> k + I are at most p-valent in E. Thus Fz) e PSi).QUESTION 2. What can we say about F(z), given hy (1.2) when F(z) is biunivalent in E?
Of course we need DEFINITION 3. A function F(z) is said to be biunivalent in E if both F(z) and its inverse are univalent in E.
Let us examine this definition a little more closely.On a first reading, it sounds as though F(z) is 1-to-l, but l-to-i and regular is really univalent, so biunivalent must mean something different.Suppose that F(z) is regular in E and f' (0) that lles inside E in the w-plane and suppose that F F(F where F is w z z a suitable arc on Izl i.If F(z) is biunivalent, then it must be analytic -i on F and such that it can be continued across F so that F (w) is defined z z and regular throughout lwl < i.This additonal requirement is very stringent, and is very awkward to use in practice.
The concept of a biunivalent function was introduced by M. Lewin [19] who proved that if F(z), given by (1.2), is biunlvalent (belongs to the set BI), then ,,]a2[ < 1.51.At one time, I believed that the bound ]a < i was true n for every n, and that the extremal function should be z/(l-z) =i z However, in a very difficult paper, E. Netanyahu [23] ruined this conjecture by proving that in the set BI, max a2] 4/3.Jenson and Waadeland [15] proved that if f(z) e BI, then ]a3] < 2.51, 5ut this result is not sharp (a lower bound may be the correct one).
Prof. Waadeland noted (in a letter to the author) that the interesting example on page I0 of [15] is not in the set BI because it has a pole at w--16/25, and that another example due to Busklein [4] has the same defect although it was intended to show that in BI max a > n/4.At this time n is known about max a nl, max If'(z) l, max (arg f'(z)), etc, for the nothing set of functions BI.
QUESTION 3. Let A be the set of all functions of the form (z) where , 8 > 0, + i, and f(z) and g(z) are arbitrary functions from some given normalized set X. What can we say about (z)?
First suppose that X S, the set of normalized functions that are uni- then it is possible to find f and g in S such that (z) has infinite valence in E. However if , or 8 are in the interval (0,1/(l+e)), then the maximum valence of (z) is unknown.
DEFINITION 4. We let CV be the set of all normalized univalent functions F(z) for which F(E) is a convex region.We let ST denote the set of all normalized univalent functions F(z) for which F(E) is starlike with respect to the origin.A domain D is starlike with respect to the origin if the intersection of D with each ray (starting at the origin) is either the ray itself or a line segment.
For example the functions z and z/(! z) are convex (in CV), since z/(l z) maps E onto the half-plane Re w > -1/2.The Koebe function z/(l z) 2 is not convex but it is starlike (in ST) because it maps E onto the entire complex plane minus the slit < w < -1/4.For the classes CV and ST, sharp coefficient bounds are known: If f(z) e CV then lanl l for every n We now return to question 3.
Styer and Wright [29] have proved that if e 8 i/2, then there are two functions f and g in ST for which (z) is infinite-valent.Such a conclusion may also hold for other values of e and 8, but at present the range of such pairs in unknown.
In [9] Goodman conjectured that if f and g are in CV, then (f + g)/2 is at most 2-valent.Styer and Wright [29] produced a pair of functions in CV for which (f + g)/2 is 3-valent and they venture the opinion that this sum "may very well be infinite-valent for some f and g in CV."Other questions about such sums have been raised and answered.For furthr details see [5, 9].
There are similar results for the geometric mean of normalized functions, (3,14) hut for brevity we omit the details [9, I0]0 We can replace the arithmetic or geometric mean in equations (3.11) and (3.14) hy other forms of composition.For this purpose we set and we define the Cvo operators (convolution, Yaltung) f,g and f,g by If f and g belong to the sets X and Y respectively, what can we say about H(z) or J(z)?
As an example consider CONJECTURE 2. If f and g E S (normalized univalent function in E) then f,g is also in S.
If this conjecture were true, then it would be easy to prove (in several different ways) that anl _< n for every function in S. Unfortunately, conjecture 2 is false, and indeed has been disproved on several different occasions (see [6]).One easy approach is to observe that Now Biernackf attempted to prove that if f(z) e S, then the integral in is also in S. Unfortunately his proof was erroneous, and a counterexample was found by Krzyz and Lewandowskl [17].This same counterexample disproves the conjecture 2, but also disproves the much weaker conjecture: If f e S and g e CV then fg is in S.
Suppose that f and g are in S. Since J(z), given by (3.18) is not necessarily univalent, the question naturally arises, what can be said about the valence of J(z)? Campbell and Slngh [6] gave the very surprising answer THEOREM I.There are two functions f S and G e ST such that fg has infinite valence in E.
We outline the proof.By a theorem due to Libera [20], if (z) e ST then g(z) 2 I (t dt (.20) is again in ST.We apply this theorem with (.z) z/(l z) 2, the Koebe function, and find that g(z) -= 2 We apply this with f(z) given by (3.22) and find that 2 ]" (3,24) This function has infinite valence in E, because at the points Zn 1 e-2n/b n i, 2, J(z n) always assumes the same value.
This theorem supplies one more counterexample for Conjecture 2, and as far as the author can see, it is the simplest of the many counterexamples now known [6, 9], Returning to question 4, we observe that different selections for X and Y and different properties for H(z) and J(z), yield a tremendous number of open problems.However the most interesting ones have been settled in a very important paper by Ruscheweyh and Sheil-Small [26].Here we merely state some of their results.
THEOREM 2. If f(z) and gz) are normalized convex in E, then H(z), given by (3.17), is also convex in E.
THEOREM 3. If f(z) and g(z) are normalized starlike in E, then J(z) given by (3,18) is also starlike in E.
THEOREM 4. If f(z) is normalized starlike in E and g(z) is normalized convex in E, then H(z), given by (.3.17) is starlike in E.
DEFINITION 5. Kaplan [16].A function f(z) of the form 63015) is said to be close-to-convex in E if there is a univalent starlike function (z) We let CC denote the set of all functions that are normalized and close-to- convex in E.
Geometrically, the condition (3.25) implies that the image of each circle zl r < i is a curve with the property that as 8 increases the angle of the tangent vector does not decrease by more than -= in any interval [81,82 ], Thus the curve can not make a "hairpin bend" backward to intersect itself.This means that each function in CC is univalent in E. Furthermore the class CC contains CV and ST.On the other hand there are univalent functions that are not in CC.Ruscheweyh and Shell-Small [26] extended Theorem 4 by proving: THEOREM 5.If f(z) is normalized close-to-convex in E and g(z) is rrmalized convex in E: then H(2), given by (3.17) is close-to-convex in E. THEOREM 6.If f(z) is normalized close-to-convex in E and g(z) is normalized starlike in E, then J(z) given by (3.18 is close-to-convex in E.
It is worthwhile to compare Theorem 1 and 6 and to observe that when "closeto-convex" is replaced by "univalent", the maximum valence of f**g Jumps from 1 to .
We now look at the bounds for the coefficients if f(z) is p-valent in E.
In his thesis, the author initiated for every n > p.
The reader will find in [9] some historical notes and an account of the progress on this conjecture up to the year 1968.When p i, inequality (3.26)   yields anl <_ n[all and this is equivalent to Conjecture i. Certainly, we can hardly expect to prove (3.26) in full generality, as long as the inequality lanl <= nlal{ for univalent functions is still unsettled.However, there are some recent advances that are worth recording.
We generalized Definition 5 of close-to-convex univalent functions so that the new class CC) includes p-valent functions.Without going into too much detail, a function (z) belongs to the class ST) (starlike p-valent) if f(z) is regular in E and if there is some r 0 < i such that for each r in Jr0,1) the image of the circle zl r under f(z) is a curve that is starlike with respect to the origin (arg f(rei8) is monotonic increasing in D) and the curve makes p complete turns around the origin.(z) for r I < [z < i (we omit some minor refinements here) [21].
The main point .here is that ST(p) C CC(p) (any starlike function is that a question selected at random may be answered already either explicitly or implicitly in some work previously published.With this note of warning to the reader we pose a few problems, that w believe are still open.
1. Suppose that f(z) is in S and that f(E) is bounded by a rectifi- able curve of length L. For flxed L* > 2, let S(L*) denote the subset of those functions for which L L*.We can ask many questions about S(L*).For example, in this set find sup If(z) I, find max a nl for each n.If f(z omits y in E, find min lI, etc. 2 Let A denote the area of f(E) For each fixed A* let S(A*) de- note the subset of those functions in S for which A < A*.We can ask the same type questions for this set that we asked about the set S(L*).Here, we have the help of the well-known formula for the area of the image of the disk but there is no way to find an analogous formula for L (4.3) the length of the boundary of f(Er).
3. Let s and 8 be fixed and suppose that f(z) omits the two values Zl.ffi rle and z 2 r2e What can we say about mln (r I + r 2) for f (z)   in S? More generally if (s,t) is any fixed function what can we say about the minimum or maximum value of (rl,r2) for f(z) in S? As a simple example, it is known that if 8 s + , then Now in (4.7) and (4.8) replace 0 on the right side by C where C is some fixed positive constant.We can no longer assert that f(z) must be uni- valent.Can we give some upper bound on the valence of f(z) as a function of the constant C?
6. Let M be some collection of functions each regular in E (not neces- sarily univalent).The Koebe domain D for the set M is the largest domain D that is contained in f(E) for every function in the set M. For example, if M S then the Koebe domain is the disk lwl < 1/4.
Koebe domains have been determined for a large number of different sets M, but we mention here two sets M for which the Koebe domain is unknown.We first recall that a function f(z) in A is said to be typically-real if for all z in E we have (Im z) (Ira f(z)) > 0. (4.9) This merely states that the upper half of the unit disk must go into the upper half-plane under f(z), and the lower half of the unit disk must go into the lower half-plane.The interval (-i,i) goes into the real axis.We let TR denote the set of all typically-real functions with the normalization (4.1).
The Koebe domain for the set TR is the domain bounded by the curve in polar coordinates sin e 0 < e < One open problem (suggested by E. Merkes) is to find the Koebe domain for the set of odd typically-real functions, with f' (0) I.
If D denotes the domain of univalence in any of the above families Todorov [31, 32] proved that one can add all of the boundary points of D (with one exception) to D without destroying the univalence of f(z).
Suppose that in (4.11), we assume that < a I < a 2 < a 3 < an < (4.12) and that all are real.Let K. be the closed disk (4.13) Todorov [31] proved that, f(z) given by 4.11 is univalent in Kj {ajl for each j i, As interesting as the above results may be, the really difficult problem is still untouched.Find D the domain of p-valence, for each p > i for the P family of functions of the form (4.12) for some reasonable set of conditions on the constants and a k.
For this problem, I can not even suggest a conjecture for D P Finally we observe that Todorov [32] has investigated the applications of sums of the form (4.11) to magnetic fields.In this work, some of the residues may be negative.

F
Figure i function, If f(z) e ST, then [a < n for every n > i and the Koebe function is an n extremal function.It is clear from the definitions of CV and ST that CV C ST C S.(3.13) the properties of the exponential function and the logarithm function, A.W. GOODbL%N it can be shown that (i-z) -l+bi C-l+bl) in (l-z) e is univalent in E, The other items in (3.22) merely normalize the function,

CONJECTURE 3 .
If f(z) [n-I a z is p-valent In E, then n lakl <3.261 [ant-< k=l pk)'.(p-k).' (n-p-l).' (n2-k2) Now a function f(z) is said to belong to the class CC(p) (close-to- convex of order p) if there is a function (z) in ST(p) and an r I reflection in the real axis.