M a t h e m a t i c a B a l k a n i c a New Series Vol. 9, 2005, Fasc. 3-4 Generating Starlike and Convex Univalent Functions Vanessa Bertoni, Dimitar K. Dimitrov 2 Presented by P.Kenderov We generate various starlike and convex univalent functions and polynomials through the location of their zeros and/or poles. It turns out that these starlike functions are extremal, in the whole class of univalent functions, for a functional which involves the Schwarzian derivative.. Introduction Let D = {z : z < } be the unit disc in the complex plane and AD) be the set of analytic functions in D. We denote by S the class of univalent functions with the usual normalization, S = {f AD) : f0) = 0, f 0) =, f fζ) whenever z ζ, z, ζ D}. Alexander [] was the first to introduce certain subclasses of univalent functions examining the geometric properties of the image fd) of D under f. The convex functions are those that map D onto a convex set. A function w = f is said to be starlike if, together with any of its points w, the image fd) contains the entire segment {tw : 0 t }. Thus we introduce the denotations S K = {f S : f is starlike in D}, = {f S : f is convex in D}. These classes obey very nice analytic characterizations, too. It can be proved that f S if and only if R z f ).) > 0 for every z D, f Research supported by the Brazilian foudation FAPESP under Grant 03/0469-4. 2 Research supported by the Brazilian foudations CNPq under Grant 300645/95-3 and FAPESP under Grant 03/0874-2.
406 V. Bertoni, D. K. Dimitrov and, similarly, f K if and only if.2) R + z f ) f > 0 for every z D. We refer to a recent paper of Gluchoff and Hartmann [5] for a historical overview of Alexander s seminal paper and to Duren s book [4] for comprehensive information about univalent functions. The characterizations.) and.2) suggest that the distribution of the zeros and/or of poles of f outside the unit disc should provide information about the geometric properties of fd) and, vice versa, once we know the shape of fd), we could be able to locate the zeros and the poles of f. This relation between the geometric properties and location of zeros and poles of univalent functions has been studied thoroughly. Recently Gluchoff and Hartmann [6] explored the idea in details, especially for the case of univalent polynomials. They also reviewed many of the contributions on this interplay. In this note we generate various starlike and convex univalent polynomials and functions examining the behaviour of their zeros. We are interested in the particular case when there exist points on the boundary of the unit disc for which the above inequalities.) and.2) reduce to equalities. In other words, all the starlike functions which will appear in the sequel satisfy.3) and the convex functions satisfy.4) R z f ) = 0 for some z D, f R + z f ) f = 0 for some z D. It is seen that.) is equivalent to the fact that the argument of fe iθ ) increases with θ [0, 2π). Then the vanishing of Rzf /f) on the boundary is equivalent to the presence of cusps on the curve fe iθ ). Similarly,.2) is the analytic equivalent of the fact that the argument of the tangent line to the same curve increases. Thus R + zf /f ) vanishes at z = e iθ only if the curvature of fe iθ ) vanishes, i.e., when this curve is not strictly convex. Let S n be the set of univalent polynomials of degree n from S. For any such polynomial p n, denote by z, z 2,...,z n its zeros that are located outside D, and by ξ, ξ 2,...,ξ n its critical points the zeros of p n). We are particularly motivated by a result of Brannan [2] who proved that the only starlike polynomials of degree n with real coefficients and with all its critical points ξ,...,ξ n on D are z ± z n /n. Since there is no univalent polynomial with a critical point in D, Brannan s example shows that these polynomials are
Generating Univalent Functions 407 extremal in the sense that both the products ξ...ξ n and z...z n are the smallest possible. These facts follow also from a result of Dieudonné [3] who proved that the inequality a n /n holds for the last coefficient of every p n S n. It is easily seen then see Theorem below) that the polynomials P n = z ± z n /n 2 are convex and satisfy.4). We shall construct various polynomials and functions that are starlike or convex and their zeros have a specific location. It turns out that most of our examples obey various extremal properties. In the last section we show that the starlike functions we construct maximize, within the class of univalent functions, a functional which involves the Schwarzian derivative. 2. Preliminaries Let f AD) and suppose that f can be represented in the form 2.) f = z p q 2 with p = z/ζ k ), q = z/η k ), k= k= where and 2 may be finite or infinity. A necessary condition that f is analytic and univalent in D is that all its zeros ζ k and poles η k lie outside D. Moreover, in this special case R z f ) f = + R z p p ) R z q q In the next section we shall consider the case when q. In this case f reduces to a polynomial or to an entire function. In Section 3 we study the case p, i.e. functions that possess only poles. Using the properties of the Möbius map w = z/z z 0 ) we readily obtain ). Lemma. Let z 0 D. a) If z 0 >, then ) z z 0 R z z 0 + z 0 for every z D. Moreover, equality in the left-hand side inequality is attained if and only if z = z 0 / z 0 and the right-hand side inequality reduces to equality only at z = z 0 / z 0. b) If z 0 =, then Rz/z z 0 )) /2 for every z D and the equality is attained if and only if z =, z z 0. Thus, if all the zeros and poles of a univalent function f, defined by 2.), are real and positive, then the minimal value of both Rzp /p) and Rzq /q) is attained at z = and the maximal value of each of these
408 V. Bertoni, D. K. Dimitrov functions is attained at z =. Using Lemma we can draw a conclusion about the behaviour of transforms obtained from the Möbius one by rotation. Corollary. Let g = z m /r, with r >. Then m/ r) Rzg /g) m/ + r) where equality in the left-hand side inequality is attained only at z = e 2jπ/m and in the right-hand side one only at z = e 2j+)π/m, for j = 0,,...,m. We have the following close relation between convex and starlike functions. Theorem. Let F be analytic in D. Then F K if and only if zf S. This theorem shows that once a starlike function f is constructed, then one obtains immediately a corresponding convex function F by 2.2) only F = z 0 fζ) ζ dζ. 3. Starlike and convex polynomials and functions with zeros First we construct starlike and univalent polynomials. In what follows we denote by S n, S n and K n the set of univalent, starlike and convex algebraic polynomials of degree n. We begin with the construction of some polynomials whose nonzero roots lie on the positive real line. If we consider the particular case when all these zeros coincide, the only starlike polynomial is p n = z z/n) n and its corresponding convex polynomial is P n = z/n) n. The second example is when the positive zeros of p n are equidistant. If we set ζ k = +ck, k =,...,n, where c is a positive constant, then Lemma a) and the requirement.3) yield ζ k = +H n k, for k =,...,n, where H n = n j= /j is the harmonic number. Hence p n = z n k= ) z, + H n k and the corresponding convex polynomial is given by P n = z 0 p n ζ)/ζ)dζ
Generating Univalent Functions 409 0.5 0.5 - -0.5 0.5 - -0.5 0.5-0.5-0.5 - - The image of D and of the circumferences z = ρ, ρ = 0.6,0.7,0.8,0.9, under the starlike polynomial p n = z z n /n and the convex polynomial P n = z z n /n 2 for n = 7. Next we consider the case when f is an entire function. Let us try to find such a function with zeros ζ k, all being real and greater than one, and such that f is starlike, that is, satisfies.) but we require the existence of a point z D, such that.) reduces to an equality at z. As we have already observed, in this case the smallest value of Rzf /f) is attained at z = and then we must have =. ζ k k= If we choose ζ k = ck 2, with a positive constant c, then the sum /k 2 = π 2 /6 yields ζ k = + k 2 π 2 /6. Now the representation ) sin = z z2 k 2 π 2 k= implies that the starlike entire function with the desired zeros is 3.) 6 f = z sinh 6 sin 6z ) 6z ).
40 V. Bertoni, D. K. Dimitrov.5 0.5 0.5-2 -.5 - -0.5 0.5-0.5 - -0.5 0.5-0.5 - -.5 - The image of D and of the circumferences z = ρ, ρ = 0.5,0.6,0.7,0.8,0.9, under f and F, defined by 3.) and 3.2). Applying the integration formula 2.2), we obtain the corresponding convex entire function 2/3 F = sinh cosh ) 3.2) 6 sin 6z ). 6 Various further examples of starlike and convex functions, similar to 3.) and 3.2) may be constructed. For example, if we set ζ k = ck /2) 2, k =,..., then the sum /k /2) 2 = π 2 /2 yields ζ k = +k /2) 2 π 2 /2 and the corresponding starlike and convex functions are 3.3) 3.4) f = z cos 2z ) cosh, 2 F = cosh cos 2z ) + 2z )sin 2z ) 2 cosh 2 + 2 sinh ) 2. Similarly, if we set ζ k = k 2 for k = 2,..., then we the requirement Rf )/f)) = 0 yields ζ = 5. Thus we generate the following pair of starlike and convex functions: 3.5) f = z z/5 sinπ z z π z,
Generating Univalent Functions 4 F = 2 3.6) cos πz 2πSiπ + z))+ 5π 2 Siπ z)) 2Siπ)) ), where Si = z 0 sinζ/ζ) dζ is the sine integral function. Next we shall construct an entire function with zeros located at the points of intersection on m rays with infinitely many circles. Then the function must be of the form f zp = z k= z m /r k ), where r k >. By Corollary we conclude that ) + R z p + m, z D, p r k= k and the latter inequality reduces to equality if and only if z = exp2ijπ/m), for j = 0,...,m. Then we may choose the radii r k in a similar way as in the previous examples. We omit this technical detail. The resulting starlike entire function is 6/m f = z sinh sin 3.7) 6z m )/m 6/m 6z m )/m. only 4. Starlike and convex polynomials and functions with poles In this section we consider the case p. In other words, our functions will posses only poles η k. Recall that in this case R z f ) f = R z q q and, when η k >, the maximum of Rz/z η k )) is attained at z =. First we shall observe that it is impossible to transform the example where we generated the function 3.). Indeed, if + η k = ck 2, for k =, 2,..., with a positive constant c, then the sum /k 2 = π 2 /6 yields η k = + k 2 π 2 /6. Then η 0.645 which means that f must possess a pole in D. Consider now η k = k 2, for k = 2,... In this case we can calculate η using the equality and the result is + η + k=2 + η k =, η = 2 π coth π 4 + π cothπ. )
42 V. Bertoni, D. K. Dimitrov Thus, we generate the starlike function 4.) π z z) f = z ) z sinπ z, η If we set +η k = ck+), for k =, 2, 3, then the requirement.3) yields η k = 3k + )/2. So, the resulting pair of starlike and convex functions is 4.2) 4.3) f = z 6z/7) 4z/9) 3z/0), F = 630 83z 0)6z 7) 69 log 704z 9) 2 ). 0 2 5 5 0 5 2 3 4-5 - -0-2 The image of D and of the circumferences z = ρ, ρ = 0.5,0.6,0.7,0.8,0.9, under f and F, defined by 4.2) and 4.3). Our next example concerns a function with infinitely many poles located at the intersection points of m rays and infinitely many circumferences. In other words, the function will be of the form f = z q = z/ z m /ǫ k ), k=
Generating Univalent Functions 43 where ǫ k >. The conditions for starlikeness and Corollary yield R zq ) q m k= + ǫ k, z D, and the equality in the latter inequality is attained if only if z = expi2j + )π/m), j = 0,..., m. If we consider + ǫ k )/m = ck /2) 2, condition.3) implies ǫ k = mk /2) 2 π 2 /2 with m >. Thus we obtain the starlike function 4.4) cos 2/m f = z cos 2z m + )/m 5. Starlike and convex polynomials and functions with zeros and poles In this section, in accordance with the representation of f, neither p nor q. In the first example we consider real poles satisfying η k < and real zeros with ζ k >. Then we have zp ) zq ) + R R + p q ζ k= k k= η k, z D, with equality being attained only for z =. Let us fix the zeros ζ k = +π 2 k 2 /3. Then an appropriate choice for the poles, in order.3) to be satisfied, is η k = π 2 8)k + /2) 2, k =, 2..., and we obtain the starlike function f = z cosπ/ π 2 8) π 2 2)sinh 4z 3) + π 2 ) sin 3z ) 5.) 3 z cosh π ). z )/π 2 8) Next we construct a starlike function with one pole on the unit circumference and infinitely many zeros. Setting η =, from.3) we obtain ζ k = π 2 k /2) 2 +. This yields the starlike function 5.2) f = z cosh cos z. z + Another case is when both the zeros and the poles are located in m rays, but the poles are distributed on rotated rays in relation to the rays of the zeros.
44 V. Bertoni, D. K. Dimitrov More precisely, we consider functions p and q of the form 5.3) 5.4) Then we have p = q = ) zm, k= k= r k + zm ǫ k ). zp ) zq ) + R R + m m p q r k= k k= + ǫ k, z D, with equality only for z = expi2jπ/m). Then the right-hand side of the above inequality vanishes for r k = mπ 2 k 2 /3 + and ǫ k = mπ 2 k /2) 2. Thus we obtain the starlike function 5.5) f = z 3/mcos /m sinh 3/m sin 3z m )/m 3z m )/m cosh z m )/m. Another interesting example is obtained if we require that the starlike function f possesses only one pole on the unit circumference and two real zeros r, r 2 >, namely we consider q = z/e iθ ), with 0 θ < 2π, and p = z/r ) z/r 2 ). Then + R zp ) zq ) R p q 2 + +, z D, r r 2 where, by Lemma, this inequality reduces to equality if and only if z = provided θ 0. We choose θ = π/4. If we set r k = ck for k =, 2, then the right-hand side vanishes only when r k = 3k +. This leads to following pair of starlike and convex functions: 5.6) 5.7) z/4) z/7) f = z e i π 4 z) 22i 27 + 29i) ) ) 2 log + e i3π 4 z) 2i + e iπ 4 22 + z) z F =. 56 Our last example concerns a function with the same pole on the unit circumference D and 2m zeros located at the intersections of m rays and two
Generating Univalent Functions 45 5 4 3.5 2 0.5-2 3 4 5 6 - -2-0.5 0.5.5-0.5 The image of the circumferences z = ρ, ρ = 0.6,0.7,0.8,0.9, under f and F, defined by 5.6) and 5.7). circumferences with radii r and r 2. Then we have again q = exp iπ/4)z but now 2 ) p = zm. Then + R zp ) R p k= r k zq ) 2 q 2 + m k= r k, z D. Obviously inequality is attained at z =. If r k )/m = ck 2, then r k = 5k 2 m/2 +, for k =, 2, which yields the starlike function 5.8) f = z zm / + 5m/2)) z m / + 0m)) e i. π 4 z) 6. Some extremal properties involving starlike functions In this section we discuss an interesting extremal property involving the starlike functions that were generated in this paper. Recall that the Schwarzian
46 V. Bertoni, D. K. Dimitrov The image of the circumferences z = ρ, ρ = 0.6,0.7,0.8,0.9, under f the starlike function 5.8) with m = 5. derivative, or, simply, the schwarzian of a function w at z is denoted by {w, z} and is defined by w ) {w, z} = w w ) 2 2 w. In 949 Zeev Nehari [7] proved the following interesting necessary condition for a function f to be univalent in terms of its schwarzian: Theorem 2. If f S, then 6.) z 2 ) 2 {f, z} 6, and 6 is the smallest possible constant to set on the right-hand side of 6.). That 6 is the best possible constant follows from the fact that for the Koebe function k = z/ z) 2 z 2 ) 2 {k, z} = 6 z 2 ) 2 z 2 2, and the right-hand side of the latter equal 6 for every real z. For Brannan s polynomials p n = z z n /n we obtain z 2 ) 2 {p n, z} = n )zn 3 z 2 ) 2 2n 2) + n + )z n ) 6.2) 2 z n )
Generating Univalent Functions 47 By L Hospital rule the right-hand side of the latter converges to 6 when z tends to. In other words, we have lim z z 2 ) 2 {p n, z} = 6. Surprisingly, the limit 6 is attained for the majority of the starlike functions that we generated. We shall calculate the functional z 2 ) 2 {f, z} for some of them. Consider this functional for the starlike function 3.). For real values of z we obtain z 2 ) 2 {f, z} = 3z + ) 2 52 344z + 63z 2 + 90z 3 60z 4 + 40 88z + 65z 2 30z 3 + 2z 4 )cos2 6z ) ) +2 6z )4 + 2z z 2 + 4z 3 )sin2 6z )) / 6 6z )z cos 6z ) + z 2)sin 6 + z)) 2. Again, using L Hospital rule, by lengthly but straightforward calculations we obtain lim z z2 ) 2 {f, z} = 6. Let us consider now the starlike function 4.2), restricting ourselves to real values of z. We obtain z 2 ) 2 6z ) 2 6z 4 560z + 945) 6.3) {f, z} = 8z 2 35z + 35) 2. The maximum of this ratio, when z [, ], is attained at z =. Moreover, we have lim z z2 ) 2 {f, z} = 6. For the starlike function 5.2) the product 6.) is z 2 ) 2 {f, z} = + z) 2 72 24z 28z 2 + 6z 3 + 2z 4 + 5z 5 + 24 + 48z 22z 2 + 5z 3 + 4z 4 + z 5 )cos2 z ) +4 z 30 4z + 5z 2 + 2z 3 )sin2 z )))/ 6 2 z cos2 z ) + z + z)sin2 z )) 2 ) and again we obtain lim z z2 ) 2 {f, z} = 6.
48 V. Bertoni, D. K. Dimitrov We have verified that the latter limit relation holds also for the starlike functions 3.5) and 3.7). For the function 4.4) we obtain the limit relation Bibliography lim z 2 ) 2 {f, z} = 6 when z e iπ/m with z <. [] J. W. A l e x a nder. Functions which map the interior of the unit disc upon simple regions, Ann. of Math. 7, 95-6, 2 22. [2] D. A. B r a nnan. On univalent polynomials, Glasgow Math. J., 970, 02 07. [3] J. D i eudonné. Recherches sur quelques problèmes relatifs aux polynomes et aux fonctions bornées d une variable complexe, Ann. Sci. École Norm. Sup. 48, 93, 247 258. [4] P. D uren. Univalent Functions, Springer-Verlag, New York, 983. [5] A. G l uchoff a nd F. H a r tmann. On a much underestimated paper of Alexander, Arch. Hist. Exact Sci. 55, 2000, 4. [6] A. G l uchoff a n d F. H a r t m a nn. Zero sets of polynomials univalent in the unit disc, manuscipt. [7] Z. N e hari. The Scwarzian derivative and schlicht functions, Bull. Amer. Math. Soc. 55, 949, 545 55. Vanessa Bertoni Departamento de Ciências de Computacão e Estatística, ICMC, Universidade de São Paulo 3566-970 São Carlos, SP Brazil e-mail: vbert@icmc.usp.br Dimitar K. Dimitrov Departamento de Ciências de Computação e Estatística IBILCE, Universidade Estadual Paulista 5054-000 São José do Rio Preto, SP Brazil e-mail: dimitrov@ibilce.unesp.br Received 30..2004