# Analytical Solution for Finding the Second Zero of the Ahlfors Map for an Annulus Region

Authors: Nur H. A. A. Wahid; Ali H. M. Murid; Mukhiddin I. Muminov
Journal: Journal of Mathematics
 (2019)

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## Abstract

The Ahlfors map is a conformal mapping function that maps a multiply connected region onto a unit disk. It can be written in terms of the Szegö kernel and the Garabedian kernel. In general, a zero of the Ahlfors map can be freely prescribed in a multiply connected region. The remaining zeros are the zeros of the Szegö kernel. For an annulus region, it is known that the second zero of the Ahlfors map can be computed analytically based on the series representation of the Szegö kernel. This paper presents another analytical method for finding the second zero of the Ahlfors map for an annulus region without using the series approach but using a boundary integral equation and knowledge of intersection points.


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## Body

## 1. Introduction

A conformal mapping that maps a multiply connected regionΩ of connectivity n>1 onto a unit disk E=w:w<1 is known as the Ahlfors map. It generalizes the Riemann map for a simply connected region. The Ahlfors map with a base point a∈Ω is a n-to-one map. It maps each boundary Γ of Ω corresponding in a one-to-one manner onto the boundary of the unit disk and maps a to the origin [1–3]. The Ahlfors map of a multiply connected region has several applications in modern function theory. For examples, the Bergman kernel and Green’s function of a multiply connected region can be written in terms of finitely many Ahlfors map [4, 5]. The Ahlfors map has n zeros in Ω where one of the zeros is a∈Ω which can be freely prescribed, and the remaining n−1 zeros in Ω are unknown [1–3]. It is known that the Ahlfors map can be written in terms of the Szegö kernel and the Garabedian kernel. In [1–3], the zeros of the Ahlfors map with a base point a occur at the pole of the Garabedian kernel at a and the remaining n−1 zeros are those of the Szegö kernel.Kerzman and Stein [6] and Kerzman and Trummer [7] have shown that the Szegö kernel of a simply connected region satisfies an integral equation which is valid also for a multiply connected region [8]. The integral equations with the Kerzman–Stein kernel, Neumann kernel, and Szegö kernel related to the Ahlfors map for a doubly connected region have been derived in [9]. The integral equation with the generalized Neumann kernel for computing the Ahlfors map has been constructed in [10]. The method in [10] is useful provided the zeros are known. Several integral equations involving the Kerzman–Stein kernel, the Neumann kernel, the Szegö kernel, the Neumann-type kernel, and the Kerzman-Stein type kernel have also been derived in [11, 12].In [2], the formula of the second zero of the Ahlfors map for a doubly connected region has been derived in terms of an integral involving the Szegö kernel and the derivative of the Szegö kernel, but the second zero has not been computed numerically. In [3], the second zero of the Ahlfors map for an annulus region has been obtained analytically from the series representation of the Szegö kernel. The problem of finding the zeros of the Ahlfors map has not been studied in [9–12], but discussed in [13]. The second zero of the Ahlfors map for any doubly connected region has been computed numerically in [13] based on the integral equation and Newton iterative method.In this paper, we derive a boundary integral equation related to the Ahlfors map for a doubly connected region different from [12]. This integral equation together with knowledge of intersection points provides another analytical method for finding the second zero of the Ahlfors map for an annulus region. Unlike [3], the method does not depend on the series representation of the Szegö kernel.The organization of this paper is as follows: Section2 presents some auxiliary materials related to the Ahlfors map and the zeros of the Ahlfors map. In Section 3, we give a boundary integral equation related to the Ahlfors map for a doubly connected region. In Section 4, we present another analytical method for finding the second zero of the Ahlfors map for an annulus region. In Section 5, we give a conclusion.


## 2. Auxiliary Materials

LetΩ be a bounded doubly connected region with the boundary Γ consisting of two smoothly closed Jordan curves Γj, j=0,1, i.e., Γ=Γ0∪Γ1. The curve Γ1 lies in the interior of the boundary Γ0. The outer curve Γ0 has a counterclockwise orientation, and the inner curve Γ1 has a clockwise orientation. The curves Γj are parameterized by 2π−twice continuously differentiable complex-valued functions zjt with the first derivatives zj′t≠0, t∈Jj=0,2π. The total parameter domain J is defined as the disjoint union of two intervals of Jj. The parameterization zt of the whole boundary Γ on J is defined as zt=zjt. The unit tangent to the boundary Γj at zjt is given by(1)Tzjt=zj′tzj′t.It is known that the Szegö kernel satisfies the Kerzman–Stein integral equation [1, 6]:(2)Sz,a0+∫ΓAz,wSw,a0dw=gz,z∈Γ,a0∈Ω,where(3)Az,w=12πiTwz−w−Tz¯z¯−w¯,z≠w∈Γ,0,z=w∈Γ,gz=−12πiTz¯z¯−a0¯,z∈Γ.The functionAz,w is continuous on the smooth boundary of Ω, and it is known as a Kerzman–Stein kernel [6, 7]. If z and w are on a circle, then Az,w=0. Numerical implementations of computing the Szegö kernel based on (2) are discussed in [7]. With z=zt and w=zs, (2) becomes(4)Szt,a0+∫JAzt,zsSzs,a0z′sds=gzt,zt∈Γ,a0∈Ω.The derivative of the Szegö kernel has been derived in [13] as a solution of the integral equation:(5)S′zt,a0z′t=g′ztz′t−∫JddtAzt,zsz′tSzs,a0z′sds,where(6)g′ztz′t=−12πiT′ztz′t¯zt−a0¯−Tztz′t¯zt−a02¯,T′ztz′t=z″t2z′t−z′t2z″t¯2z′t3,ddtAzt,zs=12πi−Tzsz′tzt−zs2−T′ztz′t¯zt¯−zs¯+Tztz′t¯zt¯−zs¯2,zt≠zs∈Γ,14πz′t13Imz‴tz′t−Rez″tz′tImz″tz′t,zt=zs∈Γ.Letfz be the Ahlfors function which maps Ω conformally onto the unit disk E=w:w<1 which satisfies the conditions faj=0, j=0,1, and f′a0>0, where aj∈Ω are the zeros of the Ahlfors map and a0∈Ω can be freely chosen. The boundary values of fz is represented by(7)fzjt=eiθjt,Γj:z=zjt,0≤t≤2π∈Jj,where θjt, j=0,1, are the boundary correspondence functions of the Ahlfors map on Γj.Differentiating both sides of (7), we have(8)f′zjtzj′t=fzjtiθj′t.Taking modulus on both sides of (8) gives(9)f′zjtzj′t=fzjtiθj′t.Dividing (8) by (9) and using (1), we get(10)fzjt=Tzjtiθj′tf′zjtθj′tf′zjt.The image ofΓ0 remains in a counterclockwise orientation, so θ0′t>0, and the image of Γ1 is in reversed orientation, so θ1′t>0.Sinceθj′t/θj′t=1, thus from (10), we get(11)fzjt=Tzjtf′zjtif′zjt.In general,(12)fz=Tzf′zif′z,z∈Γ.The interior values of the Ahlfors mapfz which is analytic on Ω can be obtained by using the Cauchy integral formula:(13)fz=12πi∫Γfww−zdw,z∈Ω.For pointsz closed to Γ, the method of Helsing and Ojala [14] can be used for the efficient numerical computation of (13).The Ahlfors map can be represented in terms of the Szegö kernelSz,a0 and the Garabedian kernel Lz,a0. It is given as follows [1]:(14)fz=Sz,a0Lz,a0,z∈Ω∪Γ,a0∈Ω.SinceLz,a0=iSz,a0Tz¯ [1], (14) can be written as(15)fzt=Szt,a0TztiSzt,a0¯,zt∈Γ,a0∈Ω.It is shown that the boundary values of the Ahlfors map can be determined from the boundary values of the Szegö kernel since the boundary values of the Garabedian kernel can be determined from the boundary values of the Szegö kernel.Differentiating both sides of (15), we get(16)f′ztz′t=1iSzt,a0T′ztz′t+S′zt,a0z′tTztSzt,a0¯−S′zt,a0z′t¯Szt,a0TztSzt,a0¯2.In [13], an alternative formula for the derivative of the boundary correspondence function of the Ahlfors map has been derived by using (8), (15), and (16), i.e.,(17)θ′t=2ImS′zt,a0z′tSzt,a0+Imz″tz′t,zt∈Γ,a0∈Ω.In general, the zeros of the Ahlfors map are unknown except for an annulus region. In [2], the second zero of the Ahlfors map for a doubly connected region is represented by(18)g1≡a1=12πi∫ΓzS′z,a0Sz,a0dz,z∈Γ,a0∈Ω.However, numerical computation ofa1 based on (18) is not given in [2] where the boundary values of S′z,a0 can be computed by solving the integral equation (5).The second zero of the Ahlfors map for an annulus regionz:ρ≤z≤1 is shown to be a1=−ρ/a0¯, and it has been obtained from the series representation of the Szegö kernel [3]:(19)Sz,a0=12π∑n=−∞∞za¯0n1+ρ2n+1,z∈Γ,a0∈Ω.Another series representation of the Szegö kernel for an annulus region has been derived in [15] by solving the Kerzman–Stein integral equation (2) using the Adomian decomposition method, and it is given by(20)Sz,a0=12π∑n=−∞∞−1n+1ρn+1ρ2n+2−za¯0,z∈Γ,a0∈Ω.The two series (19) and (20) are shown to be equivalent. However, the series in (20) gives a faster convergence compared to the series in (19) [15].


## 3. A Boundary Integral Equation Related to the Ahlfors Map for an Annulus Region

In this section, a new boundary integral equation related to the Ahlfors map for a doubly connected region different from [12] is derived and applied in Section 4 to determine analytically the second zero of the Ahlfors map. It is given in the following theorem.Theorem 1.
For allz∈Γ, the function θ′t of the Ahlfors map for the doubly connected region satisfies(21)iθ′t+1πPV∫Jz′tzt−zsθ′sds=2z′t1zt−a0+1zt−a1,where a0 and a1 are the zeros of the Ahlfors map.Proof.
The Ahlfors map for a doubly connected region can be written as(22)fz=∏j=01z−ajgz,where gz is analytic in Ω and gz≠0. Applying log on both sides of (22) gives(23)logfz=∑j=01logz−aj+loggz,z∈Γ.
Differentiating on both sides of (23), we get(24)f′zfz=∑j=011z−aj+g′zgz.
Observe that(25)12πiPV∫Γ1w−zf′wfwdw=12πiPV∫Γ1w−z∑j=011w−aj+g′wgwdw=12πiPV∑j=01∫Γ1w−z1w−ajdw+12πiPV∫Γ1w−zg′wgwdw.
Applying the residue theory [16], Sokhotskyi formula [17], and (24) to (25), we get(26)12πiPV∫Γ1w−zf′wfwdw=12∑j=011z−aj+∑j=011aj−z+12g′zgz=12f′zfz−∑j=011z−aj.
Lettingz=zt and w=zs and multiplying (26) by z′t gives(27)12f′ztz′tfzt+12πiPV∫Jz′tzt−zsf′zsfzsz′sds=z′t∑j=011zt−aj.
Since from (8), iθ′t=f′ztz′t/fzt; thus, (27) yields(28)iθ′t+1πPV∫Jz′tzt−zsθ′sds=2z′t1zt−a0+1zt−a1.
This completes the proof of Theorem 1.If we take the imaginary part on both sides of (21), then it reduces the integral equation derived in [12], i.e.,(29)θ′t+∫JNt,sθ′sds=2Imz′tzt−a0+z′tzt−a1,where(30)Nt,s=1πImz′tzt−zs,t≠s,12πImz″tz′t,t=s,and θ′t is defined in (17). The kernel Nt,s is the classical Neumann kernel. If Ω is a simply connected region, then (29) reduces to the Warschawski’s integral equation for the Riemann map as given in [17] (p. 394-395) with R0=0, i.e.,(31)θ′t+∫JNt,sθ′sds=2Imz′tzt.Next we show how to find the second zeroa1 of the Ahlfors map for an annulus region using (21) and knowledge of intersection points.


## 4. Finding the Second Zero of the Ahlfors Map for an Annulus Region

We represent the left-hand side and the right-hand side of (21), respectively, as(32)ht=iθ′t+1πPV∫Jz′tzt−zsθ′sds,(33)gt=2z′t1zt−a0+1zt−a1.We consider an annulus regionΩ bounded by(34)Γ0:z0t=eit,Γ1:z1t=ρe−it,where 0<ρ<1 and 0≤t≤2π. Given a0 is known, and we assume a1 is unknown. In the next two examples, we show the graphs of ht for special values of ρ and a0. The graph of ht depends on θ′t which is computed based on (17). The function θ′t in (17) depends on Szt,a0 and S′zt,a0 which are computed by solving (2) and (5), respectively. Thus, the graph of ht does not depend on a1. If a1 is known, then the graph of gt is the same as the graph of ht.Example 1.
Givena0=−0.5−0.5i and ρ=0.2, and the annulus region is shown in Figure 1. The graphs of the functions h0t=hz0t and h1t=hz1t are shown in Figure 2.Figure 1
Annulus regionΩ for Example 1.Figure 2
Graphs ofh0t (top curve) and h1t (bottom curve) for Example 1.Example 2.
Givena0=0.2−0.4i and ρ=0.3, and the annulus region is shown in Figure 3. The graphs of the functions h0t=hz0t and h1t=hz1t are shown in Figure 4.
Figures2 and 4 show some common characteristics:(i)
C1: the graphs ofh0t and h1t have the same shape and size(ii)
C2: the graphs ofh0t and h1t are symmetrical with respect to the imaginary axis(iii)
C3: both graphs also have intersection points on the imaginary axis(iv)
C4: the graph ofh0t is 4i units above the graph of h1t
These characteristics are unique for the Ahlfors map for the annulus region. Note that the graphs in Figures2 and 4 can also be generated using (33) provided a1=−ρ/a0¯ is known. In Example 3, we consider plotting (33) using a1≠−ρ/a0¯.Figure 3
Annulus regionΩ for Example 2.Figure 4
Graphsh0t (top curve) and h1t (bottom curve) for Example 2.Example 3.
Givena0=0.2−0.4i and ρ=0.3, and the annulus region is shown in Figure 3. We choose a1=−0.2+0.6i≠−ρ/a0¯. The graphs of the functions g0t=gz0t and g1t=gz1t are shown in Figure 5.
In Example 4, we consider plotting (32) involving nonconcentric circles.Figure 5
Graphs ofg0t (top curve) and g1t (bottom curve) for Example 3.Example 4.
Consider the region bounded by nonconcentric circlesΓ0:z0t=2eit and Γ1:z1t=−0.4+0.4i+ρe−it with a0=1+i and ρ=0.5. The nonconcentric circles region is shown in Figure 6. The graphs of the functions h0t=hz0t and h1t=hz1t are shown in Figure 7.
The graphs in Figures5 and 7 no longer satisfy the characteristics C1, C2, C3, and C4. Based on these observations, we prove the following theorem to determine the second zero a1 of the Ahlfors map for an annulus region.Figure 6
Nonconcentric circles region for Example 4.Figure 7
Graphsh0t (top curve) and h1t (bottom curve) for Example 4.Theorem 2.
Letgt be defined in (33), where g0t=gz0t,g1t=gz1t,z0t=eit,t∈J0,z1t=ρe−it, t∈J1, and a0=reiϕ be given. Then, there exists t1,t2,t1∗,t2∗∈0,2π, t1≠t2, and t1∗≠t2∗ such that(35)g0t1=g0t2,g1t1∗=g1t2∗.
Furthermore,a1=−ρ/a0¯.Proof.
From (33),(36)gt=2z′t1zt−a0+1zt−a1.
We consider for two cases which arez0t=eit, t∈J0, and z1t=ρe−it, t∈J1.Case 1.
z
0
t
=
e
i
t
, t∈J0.
Letz0t=eit, t∈J0, and a0=reiϕ be given, and a1=seiθ is unknown. Since g0t is periodic, there exists t1,t2∈0,2π with t1≠t2 such that(37)g0t1=g0t2.
This gives(38)Reg0t1=Reg0t2,(39)Img0t1=Img0t2.
From (38) and (39), respectively, we have(40)−2ssinθ−t11−2scosθ−t1+s2+rsinϕ−t11−2rcosϕ−t1+r2=−2ssinθ−t21−2scosθ−t2+s2+rsinϕ−t21−2rcosϕ−t2+r2,(41)21−scosθ−t11−2scosθ−t1+s2+1−rcosϕ−t11−2rcosϕ−t1+r2=21−scosθ−t21−2scosθ−t2+s2+1−rcosϕ−t21−2rcosϕ−t2+r2.
From (40) and (41), we seek t1 and t2 such that(42)cosϕ−t1=cosϕ−t2,that is,(43)cosϕ−t1−cosϕ−t2=2sinϕ−t1+ϕ−t22sinϕ−t2−ϕ−t12.
Sincet1≠t2, thus this implies(44)ϕ−t1+ϕ−t22=πk,t2=2ϕ−t1−2πk,k=0,±1,….
Choosingk=0 gives(45)t2=2ϕ−t1.
Then,(46)sinϕ−t2=sinϕ−2ϕ−t1=−sinϕ−t1.
Using (42) and (46) into (40) and (41) gives(47)ssinθ−t11−2scosθ−t1+s2−ssinθ−t21−2scosθ−t2+s2=−2rsinϕ−t11−2rcosϕ−t1+r2,(48)1−scosθ−t11−2scosθ−t1+s2=1−scosθ−t21−2scosθ−t2+s2.
Simplifying (48) to solve θ, we get(49)1−scosθ−t11−2scosθ−t2+s2=1−scosθ−t21−2scosθ−t1+s2,ss2−1cosθ−t1−cosθ−t2=0.
Sinces≠0 and s2−1≠0, we have(50)cosθ−t1−cosθ−t2=0.
Solving (50) and applying (45) gives(51)θ−t1+θ−t22=πk,2θ−t1−2ϕ−t1=2πk,θ=ϕ+πk,k=0,±1,….
Choosingk=±1 gives(52)θ=ϕ±π.
Then, forp=1,2, we have(53)sinθ−tp=sinϕ±π−tp=sinϕ±πcostp−cosϕ±πsintp=−sinϕ−tp,(54)cosθ−tp=cosϕ±π−tp=cosϕ±πcostp+sinϕ±πsintp=−cosϕ−tp.
Applying (46), (50), (53), and (54) into (47), we get(55)−ssinϕ−t11+2scosϕ−t1+s2+ssinϕ−t21+2scosϕ−t2+s2=−2rsinϕ−t11−2rcosϕ−t1+r2,−ssinϕ−t11+2scosϕ−t1+s2−ssinϕ−t11+2scosϕ−t1+s2=−2rsinϕ−t11−2rcosϕ−t1+r2,s1+2scosϕ−t1+s2=r1−2rcosϕ−t1+r2.
It shows thats≠r. This implies(56)cosϕ−t1=s−r1−rs4rs≡A,or(57)s−r4rs=cosϕ−t11−rs.Case 2.
z
1
t
=
ρ
e
−
i
t
, t∈J1.
Letz1t=ρe−it, t∈J1. If g1t is periodic, then there exists t1∗,t2∗∈0,2π, t1∗≠t2∗, such that(58)g1t1∗=g1t2∗.
This gives(59)Reg1t1∗=Reg1t2∗,(60)Img1t1∗=Img1t2∗.
From (59) and (60), respectively, we have(61)2ρssinθ+t1∗ρ2−2ρscosθ+t1∗+s2+rsinϕ+t1∗ρ2−2ρrcosϕ+t1∗+r2=2ρssinθ+t2∗ρ2−2ρscosθ+t2∗+s2+rsinϕ+t2∗ρ2−2ρrcosϕ+t2∗+r2,(62)−2ρρ−scosθ+t1∗ρ2−2ρscosθ+t1∗+s2+ρ−rcosϕ+t1∗ρ2−2ρrcosϕ+t1∗+r2=−2ρρ−scosθ+t2∗ρ2−2ρscosθ+t2∗+s2+ρ−rcosϕ+t2∗ρ2−2ρrcosϕ+t2∗+r2.
From (61) and (62), we seek t1∗ and t2∗ such that(63)cosϕ+t1∗=cosϕ+t2∗,that is,(64)cosϕ+t1∗−cosϕ+t2∗=2sinϕ+t1∗+ϕ+t2∗2sinϕ+t2∗−ϕ+t1∗2.
This implies(65)ϕ+t1∗+ϕ+t2∗2=πk,t2∗=−2ϕ−t1∗+2πk,k=0,±1,….
Choosingk=0 gives(66)t2∗=−2ϕ−t1∗.
Then,(67)sinϕ+t2∗=sinϕ+−2ϕ−t1∗=−sinϕ+t1∗.
Sinceθ=ϕ±π, then for p=1,2, we have(68)sinθ+tp∗=sinϕ±π+tp∗=sinϕ±πcostp∗+cosϕ±πsintp∗=−sinϕ+tp∗,(69)cosθ+tp∗=cosϕ±π+tp∗=cosϕ±πcostp∗−sinϕ±πsintp∗=−cosϕ+tp∗.
Applying (63), (67), (68), and (69) into (61), we get(70)ssinθ+t1∗ρ2−2ρscosθ+t1∗+s2+rsinϕ+t1∗ρ2−2ρrcosϕ+t1∗+r2=ssinθ+t2∗ρ2−2ρscosθ+t2∗+s2+rsinϕ+t2∗ρ2−2ρrcosϕ+t2∗+r2,−ssinϕ+t1∗ρ2+2ρscosϕ+t1∗+s2+rsinϕ+t1∗ρ2−2ρrcosϕ+t1∗+r2=−ssinϕ+t2∗ρ2+2ρscosϕ+t2∗+s2+rsinϕ+t2∗ρ2−2ρrcosϕ+t2∗+r2,−ssinϕ+t1∗ρ2+2ρscosϕ+t1∗+s2+rsinϕ+t1∗ρ2−2ρrcosϕ+t1∗+r2=ssinϕ+t1∗ρ2+2ρscosϕ+t1∗+s2−rsinϕ+t1∗ρ2−2ρrcosϕ+t1∗+r2,sρ2+2ρscosϕ+t1∗+s2=rρ2−2ρrcosϕ+t1∗+r2.
It shows thats≠r. This implies(71)cosϕ+t1∗=s−rρ2−rs4ρrs≡B,or(72)s−r4rs=ρcosϕ+t1∗ρ2−rs.
Equating (56) and (71) and solving for s, we get(73)s=ρρA−BrA−ρB.
Adding both sides of (56) and (71) gives(74)A+B=s−r4rs1−rs+ρ2−rsρ.
Using (73) into (74), we obtain(75)A+B=ρρA−B−r2A−BA+B1−ρ24ρrρA−BA−B,A+B1−1−ρ241rA−ρB−rρρA−B=0,A+B1−1−ρ24ρrAρ2−r2+ρBr2−1ρA2+B2−AB1+ρ2=0.
Since1−1−ρ2/4ρrAρ2−r2+ρBr2−1/ρA2+B2−AB1+ρ2≠0, this implies(76)A+B=0.
Applying (56) and (71) into (76) and solving for s, we get(77)s1+r2−r1+s24rs+sρ2+r2−rρ2+s24ρrs=0,rs2−sρ+r2+ρr=0,s=ρr,r.
Sinces≠r, we must have(78)s=ρr.
Thusa1=seiθ=ρ/reiϕ±π=−ρ/reiϕ=−ρ/a0¯. This completes the proof of Theorem 2.Remark.
Usings=ρ/r, (53), (54), and (56), it can be shown that Reg0t1=Reg0t2=0, and similarly by using s=ρ/r, (68), (69), and (71), it can be shown that Reg1t1∗=Reg1t2∗=0.
Now we are in the position to prove the characteristic C4. The following theorem shows that there existsq such that h0t−h1t+q=4i.Theorem 3.
Letgt be defined in (33), where g0t=gz0t,g1s=gz1s,z0t=eit,t∈J0,z1s=ρe−is, s∈J1, and a0=reiϕ be given. Then, there exists a constant q such that(79)g0t−g1t+q=4i,which is equivalent to h0t−h1t+q=4i.Proof.
Sincez0t=eit, t∈J0, (33) becomes(80)g0t=2ieit1eit−a0+1eit−a1=2i2e2it−a0+a1eite2it−a0+a1eit+a0a1.
Sincez1s=ρe−is, s∈J1, (33) becomes(81)g1s=−2iρe−is1ρe−is−a0+1ρe−is−a1.
Lettings=t+q, we get(82)g1t+q=−2iρe−it+q1ρe−it+q−a0+1ρe−it+q−a1=−2iρ2ρ−a0+a1eit+qρ2−ρa0+a1eit+q+a0a1e2it+q=−2iρa0a1e2iq2ρ−a0+a1eit+qρ2/a0a1e2iq−ρa0+a1eit−q/a0a1+e2it.
Comparing the denominators (80) and (82), we seek q such that(83)1=ρe−iqa0a1,(84)a0a1=ρ2a0a1e2iq.
Observe that (84) is just the square of (83). Note that Theorem 2 gives a1=−ρ/a0¯.
Hence, (83) becomes(85)eiq=ρa0−ρ/a0¯=−a0¯a0=−re−iϕreiϕ=−e−2iϕ=e−2iϕe±iπ.
This yields(86)q=−2ϕ±π.
Thus, (82) simplifies to(87)g1t+q=−2i2a0a1−a0+a1eite2it−a0+a1eit+a0a1.
Hence,(88)g0t−g1t+q=2i2e2it−a0+a1eite2it−a0+a1eit+a0a1+2i2a0a1−a0+a1eite2it−a0+a1eit+a0a1=4ie2it−a0+a1eit+a0a1e2it−a0+a1eit+a0a1=4i.
Sinceht=gt, h0t−h1t+q=4i. This completes the proof of Theorem 3.


## 5. Conclusion

In this paper, we have presented a boundary integral equation related to the Ahlfors map for a doubly connected region and an analytical method for finding the second zero of the Ahlfors map for an annulus region. The method does not depend on the series representation of the Szegö kernel. The method is based on some geometrical properties and knowledge of intersection points. Probably the method can be extended to some other doubly connected regions.



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*Source: 6961476-2019-08-28.xml*