CC BY
35Trudy Petrozavodskogo Gosudarstvennogo Universiteta
Seria “Matematika” Vipusk 12, 2005
Y^K 517.54
KOEBE DOMAINS FOR THE CLASS OF TYPICALLY REAL ODD FUNCTIONS
L. KOCZAN, P. ZAPRAWA
In this paper we discuss the generalized Koebe domains for the class Tand the set D C A = {z E C : \z\ < 1}, i.e. the sets of the form p|f eTM f (D). The main idea we work with is the method of the envelope. We determine the Koebe domains for H = {z E A : \z2 + 1\ > 2\z\} and for special sets fia, a < | .It appears that the set fi 4 is the largest subset of A for which one can compute the Koebe domain with the use of this method. It means that the set KT(2) (fi 4) U KT(A) is the largest subset of the still unknown set KT(2) (A) which we are able to derive.
Introduction
Let A denote the set of all functions which are analytic in the unit disk A = {z e C : \z\ < 1} and normalized by f (0) = f '(0) — 1=0. The notion of the Koebe domain was generalized in [5] as follows. For a given class A C A and for a given domain D C A, a set
f| f(D)
f eA
is called the Koebe domain for the class A and the set D. We denote this set by Ka(D).
It is easy to observe that if a compact class A has the property
f e A iff e-ivf (zeiv) e A for each p e R, (1)
© L. Koczan, P. Zaprawa, 2005
and D = Ar = {z e C : \z\ < r}, r e (0,1], then Ka(At) is a disk with the center in the origin. Moreover, if all functions belonging to A are univalent in Ar then the radius of this disk is equal to min{\f (z)\ : f e A, \z\ = r}.
The property (1) does not hold in classes consisting of functions with real coefficients, for example in the class of typically real functions
T = {f eA : ImtIm{(t) >f, \ e ■} . (2)
We use the notation:
S = {f e A : { is univalent in A},
Ar = {f e A : f has real coefficients},
((a,b)) - a line segment connecting a,b e C,
(A) - a closure of A, int(A) - an interior of A.
Koebe domains have a few simple properties.
Theorem A [5]. For a fixed compact class A C A the following properties of Ka(D) are true:
1. if A satisfies (1) and A C S then KA(Ar) = Am(r), where m(r) = min{\f (z)\ : f e A, z e dAr};
2. if A C Ar and D is symmetric with respect to the real axis then Ka(D) is symmetric with respect to the real axis;
3. if A C Ar, f e A —f (—z) e A, and D is symmetric with respect to both axes then Ka(D) is symmetric with respect to both axes;
4. if Di C D2 then Ka(Di) C Ka(D2);
5. if A1,A2 C A and A1 C A2 then KA2 (D) c KAi (D).
In [3] classes of typically real functions with n-fold symmetry were discussed, i.e.
T(") = {f e T : f (ez) = ef(z),z e A},
where e = e, n e N, n > 2. The main result of this paper is
Theorem B [3]. For k e N we have
T(2k-l) = {f : f (z) = g(z2k-1), g e T},
T(2k) = {f : f (z) = ^g^,g e T^}.
According to this theorem, the Koebe domains for T(n) can be determined by the relation
Kt(2k-i) (D) = {z : z2k-1 e Kt(D)},
Kt(2k) (D) = {z : zk e Kt(2) (D)}.
Results concerning the class T were obtained in [5]. In order to determine the Koebe domains also for T(n), while n is odd, we need to know these domains for the class T(2).
Koebe domains for T(2)
It is known (see for example [3]) that each function T(2) can be represented in the integral form by
f(z) = i
J0
*(1 + z2)
(1 + z2 )2 — 4z2t
dj(t),
where j e P[oA], i.e. j is a probability measure on [0,1]. While researching T(2) it is useful to work with functions
(3)
g(w)
w2 — 4t
dj ( t) ,
(4)
for which f (z) = g(z + 1).
Let T(e) denote a class of functions given by (4), i.e.
/TO —|
-e — Tj(u), - e C \[—e, e], j e p[',to]} ■
For z = 0 let z +1 = w, \w\ = g, arg w = p. The set {g(w) : g e T(e)} is a convex hull of the curve [0,1] 3 t ^ w2W4t (see for example [1]). It is easy to observe that if w is a point of real or imaginary axis then the set {g(w) : g e T(e)} coincides with a segment ((W, 4)) included in
this axis. Otherwise, the boundary of the set {g(w) : g e T(e)} consists of the arc of the circle
1
1
1
4g \cos p sin p
1
1
1
4g cos p sin p
(5)
i
w
0
w
having end points ^e iV and g2 ^2iJ-4 and not containing 0, and the line
segment ((|e-iV, q2 eJiJ-4)). We conclude from this fact that min{\g(w)\ :
g e T(e)
, arg }(-) = ft}, ft e [/, n] is achieved by functions 1w
ge(w) = (1 — e)-----------+ e—2-7 , w e C \ [—2, 2], e e [0, 1]. (6)
w w2 — 4
In terms of function f e T(2), the minimum of {\f (z)\ : f e T(2), arg f (z) = (3}, (3 e [0, 2], is achieved by functions
fE(z) = (1 — e)1 + z2 + e(( —hz2)l , z e A,e e [0,1], (7)
which correspond with functions (6).
We begin with finding the Koebe domain for T(2) and the lens H =
{z e A : \z2 + 1\ > 2\z\}. The set H, as it is known, has special properties. Firstly, H is the domain of univalence of T [2]. Secondly, it is the only domain of univalence of T(2) that is symmetric with respect to both axes of the complex plane.
We know from [5] that Kt(H) = A1.. By Theorem A point 4, Kt(2) (H) D Kt(H). The boundary of Kt(H) consists of images of some points belonging to the boundary of H under the functions
zz
f(z) =e(T—z2 + (1 — e)(TTzj2 , e e [0,1].
Since these functions are not in T(2) while e = 1, A 4 is a proper subset of Kt(2) (H). Furthermore, z = ^i and z = — ^i are the only common points of Kt(H) and Kt(2) (H).
Theorem 1. The set Kt(2) (H) is a bounded domain, whose boundary is the curve ^ ((— n,n]), where
1 13
^(p) = 2 cos3 P + i^(^ — sin2 p) sin p , p e (—n, n]. (8)
Proof. All functions of T(2) are univalent in H. Hence
Kt(2) (H)= H f*(H).
0<£<1
In order to determine the set Kt(2) (H) we will derive the envelope of the family of segments ((fo(z), fi (z))), while z ranges over the whole boundary of H. After that we will prove that this envelope is in fact the boundary of Kt(2) (H). Basing on Theorem A point 2 we can restrict determining the envelope to the first quadrant of the complex plane.
Let z e dH fl {z : Imz > 0}, which is equivalent to w = z + - =
2eip , p e (—n, 0). Observe that for these z
( ) w 1 gi(w) =
w2 — 4 4i sin p
and
go(w) = - = 1 e—ip. w2
Straight lines going through points go(w) and gi(w) for a fixed w = 2eip , p e (—n, 0) are of the form
.(f)»1 e-'- +1
f Є R, p Є (—n, 0). (9)
4 sin p 2
These lines are pairwise symmetric with respect to the imaginary axis (namely, for all t e R and p e (—n, 0) we have w—n—p(t) = —wp(t) ). Therefore, the envelope is symmetric with respect to the imaginary axis. This is a reason why we can restrict the set of variability of p to the interval ( — 2, 0). The line {w— 2 (t) : t e R} coincides with the imaginary axis.
The straight lines (9) can be written equivalently
x cot 2p — y — : = 0 , p e (— , 01 .
4 sin p 2
From the system
І 'У' 0/O _ Г,, ___ ________
4 sin —
x cot 2p — y — -j-J— = 0
r & 4 sin —
x = 0
sin2 2— 4 sin2 —
we derive the equation of the envelope of lines (9) in the first quadrant
1 з 1/3 2 \ /П\ /\
x =2cos y = — 2(2 — sin P)sin Є (— 2 ’0). ()
Let W(p) = 1 cos3 p — i 1 (2 — sin2 p) sin p , p Є ( — П, 0).
Observe that 1
arg W (p) < arg------------
4i sin p
and
arg W(p) > arg ^e—ip,
where the argument in the above is taken from (—n,n].
The first inequality is obvious. The second one is equivalent to
32
3 — sin2 p
------ tan p < tan p,
cos2 p
which is true for p e ( — 2, 0).
Hence, the curve (10) is the envelope of the family of the straight lines (9) for p e ( — 2, 0), as well as the envelope of the family of the line segments ((g0(2eip), gl(2eip))) while p e ( — 2, 0).
Putting p = 0 into (10) we get the point 1 = fo(1) = go(2), and putting p = — 2 we obtain Ii = fl(i(\/2 — 1)) = gl(—2i).
It follows from arg[gl(2eip) — g0(2eip)] = —2p + 2 that this argument is a decreasing function of p e ( — 2, 0). Therefore, the bounded domain D for which the curve (10) and the intervals [0, 2] and [0, 4i] are its boundary is convex. Hence, each set {f (z) : f e T(2)} for z e dH f {z : Rez >
0, Imz > 0}, which is the same as {g(2eip) : g e T(£)} for p e [—2, 0] is disjoint from D (has exactly one common point with the closure of the curve (10)). It means that D C f (H f {z : Rez > 0, Imz > 0}) for each f e T(2).
Taking the interval (—n, n] instead of ( — 2, 0) in (10) we obtain a curve which is closed and symmetric with respect to both axes. Let us denote by E the set which has this curve as a boundary and which contains the origin.
From the above argument it follows that E C f (H) for each f e T(2). Since
E C n f (H) C n f£(H) = E,
feT(2) ££[0,1]
we have E = Kt(2) (H). □
Substituting cos p by ^2x in (10) one can write the equation of the boundary of Kt(2) (H) in the form
Now, we consider some special sets Qa for which we determine Koebe domains. After that we will be able to indicate the largest Koebe domain for Tand some set D which will be possible to determine applying the method of the envelope.
We need the following notation:
D
a
ra
l(z) = z + 1, z e A \ {0},
= {z e A : |(z + 1 )2 — a\ > 4 — a}, a <
= {w : |w2 — a\ > 4 — a, Rew > 0, Imw > 0}, a < |,
= {w : |w2 — a\ =4 — a, Rew > 0, Imw > 0}, a < |.
4 3 .
(11)
(12)
(13)
(14)
In particular, ^o = H and To is the arc of the circle |w| = 2 that is included in the first quadrant of the complex plane.
All domains Qa, a < 4 are symmetric with respect to both axes and l(Qa) fl {w : Rew > 0, Imw > 0} = Da. According to [3], l-1(D4) is the
quarter of the domain of local univalence for T(2) included in the fourth quadrant of the complex plane. It was proved in [4] that
Theorem C. Each function g e T(e) is univalent in D±.
Obviously, a < /3 < 3
^ Da C Dp. Hence, all functions g e T(e) are univalent in every set Da, a < |.
In order to determine KT(2) (Oa) we need the envelope of the family of line segments ((go(w),gi(w))) for w ranging over ra.
For w e ra
we have
w = \Ja + (4 — a)e1^ , V e (0, n),
(15)
where the branch of the square root is taken in such a way that %/T = 1. Denote
[2, 2 — a] a < 0,
Ia
[2 — a, 2] a e (0, 4]
and
*a(p) =
2 [p2
1 (2 2( + [p-
2]x
\J a(2 + p)(p — 2 + a)3
—ia[p2 + 2 (3a — 4)p + 1 a2 ^ a (2 — p)(p + 2 — a)
1
1
a
a
where p e Ia, a e (—<x>, 0) U (0, 3].
Theorem 2. The envelope of the straight lines going through g0(w) and gi(w), while w is of the form (15) and a e (—rc>, 0) U (0, |], coincides with the curve ^a (Ia).
Proof. Let w be of the form (15) and a e (—&>, 0) U (0, |]. Denote
\/a + (4 — a)elp = gel°, (17)
where the branch of the square root is chosen as in (15).
From this we observe that 0 e (0, n) and that the sign of g — 2 depends on a. Namely, for a e (0, |] we have g — 2 < 0 and for a e (—<x>, 0) we have g — 2 > 0.
Applying (17) we obtain
go(w) = - cos 0 — i- sin 0 gg
and
(g — |)cos 0 _ (g + 4 )sin 0
gi(w) = 2 , i6—o—™ — i
g2 + if — 8 cos 20 g2 + — 8 cos 20
The real equation of straight lines going through go(w) and gi(w) can be written in the form
[4 — g2(1 + 2cos20)]xtan0 + [4 + g2(1 — 2cos20)]y + 2gsin0 = 0. (18)
We conclude from (17) that
g2 = v^(4 — a)2 + 2a(4 — a) cos v + a2,
cot 20 = a + (4 — a)cos V. (19)
(4 — a) sin v
For convenience let
1
p
— \/(4 — a)2 + 2a(4 — a) cos v + a2. (20)
Hence, if v e [0, n] then p e Ia. From (19) we derive
g2 = 2P,
g2 cos 20 = a(p2 — 4 + 2a), (21)
g2 sin 20 = T^v/(4 — p2)(p2 — (2 — a)2),
and then _______________________
gcos 0 = hv7 a(p + 2)(p — 2 + a)), (22)
g sin 0 = tO^V a(2 — p)(p +2 — a)).
Applying (19) and (22) in (18) we obtain the equation equivalent to (18):
a(p + 2)-(4 — a — 2p)x + J a(2 p) (4 — a + 2p)y + a = 0. (23)
V p — 2 + a V p + 2 — a
The envelope of the family of these lines is obtained as the solution of the system
y—+0(4 — a — 2p)x + ^j a+-El(4 — a + 2p)y + a = 0
/ p—2+a 4p2 + 2(3a—4)p+ a2 x + / p+2 — 0 —4p2 + 2(3a—4)p—a2 0
Y a (p+2) (p—2+a)2 + y a(2—p) (p+2 —a)2 = .
In this way we get the curve given by (16). □
Remark. 1. In the limit case, taking into account lima^o ^a(Ia), we obtain the curve ^((— n, n]) defined in Theorem 1. One can calculate this limit putting s = p—0+a into the equation of ^a (in this case s e [0,1]).
2. The curve ^a(Ia) has one singularity for
po =__________________________________________________________________________
i2y/3( —112 + 80a — 9a2 + v/(3a — 4)(27a3 — 508a2 + 3280a — 4672)
(24)
while a < —12 (if a = —12 then po = 2). It can be concluded from
(x/(p))2 + (y/(p))2 =
p[24p4 + (9a2 — 80a + 112)p2 — 2(2 — a)(a2 — 16a + 16)]
8a(4 — a)(4 — p2 )(3p2 +4 — 2a)4
and the fact that po is the only zero of this expression in (2, 2 — a).
3. Using (22) one can obtain a new complex parametric equation of ra
w(p) = -j1 (Va(p + 2)(p — 2 + a)) + i^Ja(2 — p)(p + 2 — a))) ,p e Ia, |a|
(25)
which is useful in the following consideration.
First we need
Lemma 1. Let h(p) = arg[gi(w(p)) — go(w(p))], where go,gi are given by (6), the argument is taken from the interval (—2n, 0], and let w(p) be given by (25). Then the range of h is [— 3-, —n). Moreover, if a < 0 then h is decreasing in (2, 2 — a), and if a e (0, |] then h is increasing in (2 — a, 2).
Proof. Let w be defined by (15) and let k(v) = arg[gi(w) — go(w)]. The function k is decreasing for v e (0,n) because
k(v) = —[2 arg(a + (4 — a)eip) + arg(eip — 1)] .
Furthermore, by (20) for a e (0, |], p is a decreasing function of v Hence, there exists its inverse function v = v(p) and it is decreasing for p e [2 — a, 2]. Combining these facts, we conclude that h(p) = k(v(p)) is an increasing function for p e (2 — a, 2).
In the second case, for a < 0, it follows from (20) that p is an increasing function of v, and consequently, h(p) = k(v(p)) is a decreasing function for p e (2 - a, 2) . □
Theorem 3. The envelope of the line segments ((go(w),gi(w))), where go,gi are given by (6) and w is given by (15), is a convex curve of the form
1. 'Va ((2 — a, 2)) for a e (0, f] ,
2. ^a ((2,da)) for a e [—2, 0),
3. ^a ((ca,da)) for a e (—m, —2), where ^a is given by (16) and
Ca = y — — a)a, da = Y-(a2 — 8a + 8).
In this theorem and further on, the convexity of a curve means that the tangent line to this curve lies below the curve.
Proof. According to Theorem 2, ^a (Ia) , a e (—<x>, 0) U (0, 4] is the envelope of straight lines going through go(w) and gi(w), w e ra.
This curve (whole or only a part of it) is also the envelope of line segments ((go(w), gi(w))), but only for thesep which satisfy the inequality
arggi(w(p)) < arg^a(p) < arggo(w(p)) , p e Ia
(26)
For w € ra (w is of the form (15)) we have
arggi(w) = arg[(g2 - 4) cos 9 — i(g2 + 4) sin 9], ( )
arg go(w) = arg[cos 9 — i sin 9] = -9.
Let a € (0, 3] and let be given by (16).
It follows from (27) that argg1(w) € (— n, — ^) and argg0(w) € ( — 2,0). Moreover, arg^a(p) € ( — n, 0) for p € (2 — a, 2). Since the left hand side of (26) is fulfilled, it is sufficient to discuss only the right hand side inequality. We rewrite it as follows
p2 + 2(3a — 4)p + 1 a2 1(2 — p)(p + 2 — a)3 ^ 1(2 — p)(p + 2 — a)
p2 — 2(3a — 4)p + 4a2y (2 + p)(p — 2 + a)3 _ y (2 + p)(p — 2 + a)
which holds for a € (0, 3] and p € [2 — a, 2].
Therefore, the inequality(26) is true for a € (0, 3] and w € ra. We conclude from this that the curve ^a(/a) is really the envelope of line segments ((go(w),gi(w))) for a € (0, 3].
Let now a € (—w, 0) and w € Ta.
From (27) we obtain arg g1(w) € ( — f, 0) and arg g0 (w) € ( — f, 0). The left hand side of (26) is equivalent to
and equivalently
and further on
p2 + 2 (3a — 4)p + 4 a2 p2 — 2 (3a — 4)p + 4 a2
(3a — 4)p + 4a2 y (2+ p)(p — 2 + a)
(2 — p)(p + 2 — a)3 <
3 —
and then
[p2 + 2(3a — 4)p + 1 a2][—p2 + ap — 2(2 — a)]
> [p2 — 2(3a — 4)p +4a2][p2 + ap — 2(2 — a)].
After simple calculations it takes form
(4 — a)p[p2 — ^(a2 — 8a + 8)] > 0. (28)
The inequality (28), and in consequence, the inequality arg gi(w) — arg ^a(p) holds only for p € [2, 1 (a2 — 8a + 8)] because of 2 <
\J2 (a2 — 8a + 8) < 2 — a.
The right hand side of (26) turns to
p2 + 2(3a — 4)p + 1 a2 /(2 — p)(p + 2 — a)3 ^ /(2 — p)(p + 2 — a)
p2 — 2 (3a — 4)p + 1 a2 y (2 + p)(p — 2 + a)3 — \j (2 + p)(p — 2 + a)
(29)
Hence
[p2 + 2(3a — 4)p + 1 a2](p + 2 — a) — [p2 — ^(3a — 4)p + 1 a2](p — 2 + a), and then
p2 + -a(2 — a) > 0. (30)
It is easy to check that if a € (—2,0) and p € [2, 2 — a], then (30) holds.
It means that (29) is fulfilled. If a € (—<x>, —2) then (30) is satisfied only
for p € [\J — 1 a(2 — a), 2 — a].
Our next goal is to prove the convexity of the above derived envelope of the line segments.
In view of Remark 2 the envelope of the straight lines going through go(w)
and g1(w) has no singularities for a € [—12,0) U (0, |]. If a < —12 then
this envelope has the only singularity corresponding to po given by(24), but po < ca. Indeed,
— 112 + 80a — 9a2 + v/(3a — 4)(27a3 — 508a2 + 3280a — 4672) <
—24(2 — a)a
and then
(a — 2)(7a + 4)(a — 4) > 0 ,
which is true for a < —12.
Therefore, the envelope of the line segments ((go(w),g!(w))) has no singularities, and, by Lemma 1, is convex. □
Let a < 0 and
p € [2, 2 — a]. For $a and Ya we have
((2, 2 — a)) = {g!(w) : w € Ta} and Ya ((2, 2 — a)) = {go(w) : w € Ta}.
We need four lemmas to prove Theorem 4.
Lemma 2. For a < —2 and p € (2,ca) we have arg[$a(p) — Ya(p)] — argY^(p) > 0, where the arguments of $a(p) — Ya(p) and Y'a(p) are taken from the interval ( — , — §) .
Proof. Let a < —2. Firstly, we are going to prove that w(ca) is the only point, given by w( p) , p € (2, 2 — a) , for which the tangent line to the curve
Ya(Ia) coincides with the straight line going through go(w) and g1(w). Let us discuss the equation
(31)
\Ja(p + 2)(p — 2 + a) — i\Ja(2 — p)(p + 2 — a) , (32)
Let Ea be a bounded domain whose boundary is of the form:
• [0, 1) U *a ([2 — a, 2]) U i(0, '84_320“ ^ for a € (0, 4],
• [0, 1) U *a ([2, da]) U ((da, 2 — a]) U (—i(0, )) for a €
[—2, 0),
• [0, 2 )UYa ([2, Ca))U*a ([Ca, da])U$a ((da, 2 — a])U ( — i(0, ))
for a € (—<x>, —2).
Theorem 4. For each w € ra and a € (—&>, 0) U (0, |].
(33)
(34)
Re [$a(p) — Ya(p)] • ImYa(p) = Im [$a(p) — Ya(p)] • ReYa(p) .
This equation, by (31) and (32), is equivalent to
[—ap + 4(2 — a)][4p2 — 2ap — 2(4 — a)(2 — a)] =
[ap + 4(2 — a)][4p2 + 2ap + 2(4 — a)(2 — a)] and hence p2 = — 1 a(2 — a).
Thus the expression arg [$a(p) — Ya(p)] — arg Ya(p) does not change the sign for all p € (2, ca).
Moreover,
^ ( ) Y ( )] t I 2p — (4 — a) (p + 2)(p + 2 — a)
arg [*a(p) - Ya(p)] = ---arcta^,p +(4 - ay\j (p - 2)(2 - a - p)
and for p € (2, 4) we have ReYa(p) < 0 and ImYa(p) < 0. Thus
ar~r (p)= n arctan| ap + 4(2 — a) l(p + 2)(2 — a — pm
argYa(p) = -n - “rcta^ap - 4(2 - a) y (p - 2)(p + 2 - a)) '
For p € (2, 4) the inequality
arg [$a(p) — Ya(p)] — arg Y'a(p) > 0 (35)
is equivalent to
ap + 4(2 — a) (p + 2)(2 — a — p) > 2p — (4 — a) (p + 2)(p + 2 — a)
ap — 4(2 — a) y (p — 2)(p + 2 — a) > 2p +(4 — a) y (p — 2)(2 — a — p) ,
and, in consequence, to p2 < — 2a(2 — a). Therefore, (35) holds for p €
(2,4) n (2, Ca).
The function arg [$a(p) — Ya(p)] — arg Y'a(p) is continuous for p € (2, 2 — a), positive for p € (2, 4) n (2, ca) and its only zero is ca. Thus,
(35) holds for p € (2, ca). □
Proofs of next three lemmas will be omitted.
Lemma 3. For a < 0 the function arg Ya(p) is decreasing in (2, 2 — a).
Lemma 4. The function argY'a(p) is
1. increasing in (2, 2 — a) for a € [—4,0),
2. decreasing in (2, ba) and increasing in (ba, 2 — a) for a < —4,
where ba = \J6(2 — a).
Lemma 5. For a < 0 the function arg&'a(p) is increasing in (2, 2 — a).
Proof of Theorem 4.
We will prove this theorem only for a < —4. For other a the proof is easier, because we need only some elements of the argument presented below. Let w = w(p) be given by (25). Denote by Bp a closed and convex set whose boundary consists of a ray lp with the end point in go(w(p)) and going through 2go(w(p)), a ray kp with the end point in g1(w(p)) and going through 2g1(w(p)), and a segment {{go(w(p)), g1(w(p)))).
The set {g(w(p)) : g € T(e)}, p € (2, 2 — a) is a segment of the disk whose boundary is given by (5). Moreover, 0 € {g(w(p)) : g € T(e)}. From it and from the fact that w = 0 belongs to the circle (5) we conclude that for p € (2, 2 — a)
{g(w(p)) : g € T(e)}CB .
V
Let p € (2, ba].
From Lemmas 1 - 4 it yields that Bp C
{u € C : arg [g1(w(p)) — go(w(p))] < arg [u — go(w(p))] < arggo(w(p))} C{u € C : arg [g1(w(ba)) — go(w(ba))] < arg [u — go(w(p))] < arggo(w(p))}
C{u € C : arg d go(w(p))\p=ba < arg [u — go(w(p))] < arggo(w(p))}.
Let p € (ba, ca). From Lemmas 1 - 4 we have Bp C
{u € C : arg [g1(w(p)) — go(w(p))] < arg [u — go(w(p))] < arggo(w(p))} C{u € C : arg d go(w(p)) < arg [u — go(w(p))] < arggo(w(p))} .
It means that for p € (2, ca) there is BpnEa = 0. Therefore, {g(w(p)) : g € T(e)} nEa = 0 and {g(w(p)) : g € T(e)} n\\(£a) = {}'Q(^})}.
Let p € (ca, da). From Lemma 1, from the inequalities arg go(w(ca)) < arggo(w(p)) and argg1(w(p)) < argg1(w(da)) and from the fact that the segment {{go(w(p)), g1(w(p)))) is tangent to dEa (or equivalently to *([ca, da])) we obtain Bp nEa = 0, and thus {g(w(p)) : g € T(e')}nEa =
0. Furthermore, the only common point of {g(w(p)) : g € T(e)} and cl(Ea) is a point of tangency.
Let p € (da, 2 — a). From Lemma 1, Lemma 4 and Lemma 5 we have
Bp C {u € C : argg1(w(p)) < arg [u — g1(w(p))] <
< arg [go(w(p)) — g1(w(p))]}
C {u € C : arg g1(w(p)) < arg [u — g1(w(p))] <
< arg [go(w(da)) — g1(w(da))]}
= {u € C : argg1(w(p)) < arg [u — g1 (w(p))] < arg — g1(w(p))\p=da } .
Hence Bp n Ea = 0 and {g(w(p)) : g € T(e)} n Ea = 0. Moreover, {g(w(p)) : g € T(^} n\t(Ea) = {}to(3(v))}.
Let Aa = l-1(Da), i.e. Aa = Qa n {z € A : Rez > 0, Imz < 0}.
Corolary. Kt(2) (Aa) = Ea.
Proof. For each f € T(2) and z € A there are Imz = 0 ^ Imf (z) = 0 and Rez = 0 ^ Ref (z) = 0. Hence
Ea n {f (z) : f € T(2), z € dAa, z = 1, RezImz = 0} = 0 .
This and Theorem 4 leads to
Ea n{f (z) : f € T(2), z € dAa, z = 1} = 0 .
Moreover, if z = 1 is a regular point of f € T(2) then f (1) > 2 (because for x € (0,1) there is f (x) > 1+xx2). It means that Ea C f (Aa) for each
f € T(2). Therefore, Ea C Kt(2) (Aa). By the definition of the Koebe domain, Kt(2) (Aa) C f|ee[oj1] fe(Aa).
The univalence of fe in Aa (by Theorem C) and (33) leads to n fe (Aa ) =
ee[o,1]
Ea. From the above argument Ea C Kt(2) (Aa) C Ea. □
Theorem 5. The set Kt(2) (^a), a € (—&>, 0) U (0, 4] is a bounded domain, symmetric with respect to both axes of the complex plane. The
boundary of this domain in the fourth quadrant coincides with:
1. ^a ([2 — a, 2]) for a € (0, 3] ,
2. ^a ([2, da]) U $a ((da, 2 — a]) for a € [—2, 0),
3. Ya ([2, ca)) U ^a ([ca,da]) U $a ((da, 2 — a])for a € (—tt, —2) , where ^a, $a, Ya are given by (15), (29), (30) respectively, and ca =
\J— 2(2 — a)a , da = \J 1 (a2 — 8a + 8).
Proof. Let us denote Ga = int((Ea U Ea U (—Ea) U (—Ea)).
I. Let a € (0, 4]. For each f € T(2) the set f (Qa) is symmetric with respect to both axes of the complex plane. Therefore, by Corollary 1,
Ea U Ea U (—Ea) U (—Ea) C f (^a) ,
so
Ga \ {z € C : RezImz = 0} C f (^a).
If 0 < x < 1 then f (x) > 1+x2 and if —1 < x < 0 then f (x) < 1+xx2. Hence for each f € T(2)
f (( —1,1)) ^ ( — ^, 2) = Ga n {z € C : Imz = 0} , (36)
and then
Ga \{z € C : Rez = 0} C f (Q,a).
This leads to
Ga \{z € C : Rez = 0} C Kt(2) (Qa).
Our next goal is to prove that the line segment Ga n{z € C : Rez = 0} is also included in Kt(2) (^a).
Let us suppose that there exists a point iyo such that iyo € Kt(2) (^a). It means there exists a function f.* € T(2) such that f*(Qa) ^ iyo.
Let ao be taken in such a way that 0 < ao < a and iyo € Ga0 (existence of such ao follows from the definition of Ga and from the fact that iyo is an interior point of this segment). We have Hao C £la. Moreover, these sets have only two common points z = —1 and z = 1.
Since f* is a typically real function, we can see that f*( — 1) = iyo and f*(1) = iyo. Hence there exists a neighborhood U of the point iyo such that U n f*(Qao) = 0.
This gives U n Ga0 = 0, a contradiction, because
Ga0 \ {z € C : Rez = 0} C f (&a0 ) .
The above given argument leads to Ga C Kt(2) (^a).
From Corollary 1 and the symmetry of f (Qa) with respect to both axes of the complex plane we deduce nee[o 1] fe(^a) = Ga and then
Kt(2) (^a) C Ga. Hence Kt(2) (^a) = Ga.
II. Let a < 0. We will prove that
Ga n {f (z) : f € T(2), z € dtoa, z = ±1} = 0 . (37)
According to Theorem 4,
Ea n {f (z) : f € T(2), z € dAa, RezImz = 0} = 0 .
All functions belonging to T(2) are univalent in the lens H [1,3], and then in toa (now toa C H). From this we obtain
f (Aa) C {z € C : Rez > 0Imz < 0},
and, as a consequence,
Ga n {f (z) : f € T(2), z € dtoa, RezImz = 0} = 0 . (38)
Observe that
dtoa n {z € C : Rez = 0} = < ± — (V8 — 2a — V4 — 2a) i 1
The inclusion (36) also holds for a < 0. Combining (38), (39) and (36) we can write that for each f € T(2)
e£ [o, 1]
From the univalence of fe in toa it follows that p|e£[o 1] fe(toa) = Ga. □
The specific significance of the set to 3 is presented in Theorem 6.
We know that the equation g'e (w) = 0, where ge is defined by (6) and e €
and
Since 1 f (1 (a/8 — 2a — \J4 — 2a) i) > ^—2a, we have
Ga C f(toa) .
Hence
Ga C Kt(2) (toa) C P| fe (toa) .
(0, 9), has four different solutions. From the univalence of z ^ z +1, while z € A, we conclude that the equation f (z) = 0, where fe is defined by (7) and e € (0, 9), has also four different solutions in A: ze ,ze, —ze , —ze (we choose ze to satisfy Reze > 0 , Imze > 0). Moreover, zo = 1, z8/9 = ^i are the only solutions of f (z) = 0 and f8/9(z) = 0 respectively, in the set {z € A : Reze > 0 , Imze > 0}.
Theorem 6. dKT(2) (to3) n {z € C : Rez > 0 , Imz > 0} = {fe(ze) :
e € [0, 9]}. 3
Proof. By definition of fe and ge, f'e(z) =0 if and only if g'F(z + 1) = 0. Let w = z + 1. For e € [0, 9] we have g'F (w) =0 iff
w = ± ^\J(V1 — e + 1)(3V1 — e — 1) ± i\J(1 — V1 — e)(1 + 3V1 — e)) .
Since Rez > 0 , Imz > 0, z satisfies the equation
1
z +— — y (%/1 — e + 1)(3\/1 — e — 1) — i^J(1 — %/ 1 — e)(1 + 3\/1 — e). From this f (z ) =
ge ^\J(V1 —~ + 1)(3V1 — e — 1) — i^/(1 — V1 — e)(1 + 3V1 — e)^ =
(^4 — 3e + i^/e(8 — 9e^^2 — 3e + 2VT—e + V—2 + 3e + 2V1—e)
16VT—e .
Substituting p = 2\f\—e (then e € [0, 9 ] iff p € [ §, 2]) in the above we obtain
f ( ) = 3%/3
fe(ze) —
(2+ p)(p — 3)3 + ij (2 — p)(p +2)3
32
which completes the proof. □
One can define toa also for a € (3, 2]. It is easily seen that if a1 < a2 < 2 then toai C toa2. Let Ga, a € (4, 2] be defined analogously as for a € (0, 4]. Certainly, for a < 4 we have
Ga = Kt(2) (toa) C Kt(2) (to 4) = G3
From Theorem 6 we know for а G (3,2] that
Ga С G 4 ,Ga = G 4,
3 ' 3 '
which means
Ga С (2) (toa),Ga = (2) (toa).
The above presented argument shows that the set KT(2) (to 4) is the largest subset of KT(2) (A) (the set KT(2) (A) is still unknown) which one can compute applying the method of the envelope.
Список литературы
[1] Asnevic I. On the regions of values which have Stieltjes integral representation / I. Asnevic, G. V. Ulina // Vestn. Leningr. Univ. N 11. (1955). P. 31-45. (Russian).
[2] Golusin G. On Typically-Real Functions / G. Golusin // Mat. Sb. 27(69). (1950) P. 201-218. (Russian).
[3] Koczan L. On typically real functions with n-fold symmetry / L. Koczan, P. Zaprawa // Ann. Univ. Mariae Curie Sklodowska Sect.A (1998) V. LII. 2. P. 103-112.
[4] Koczan L. Domains of univalence for typically-real odd functions / L. Koczan, P. Zaprawa // Complex Variables. (2003). V. 48. N. 1 P. 1-17.
[5] Koczan L. Covering problems in the class of typically real functions / L. Koczan, P. Zaprawa // Ann. Univ. Mariae Curie Sklodowska Sect.A. (to appear).
Department of Applied Mathematics,
Lublin University of Technology, ul. Nadbystrzycka 38D,
20-618 Lublin, Poland E-mail: [email protected] E-mail: [email protected]
