Walsh’s Conformal Map onto Lemniscatic Domains for Polynomial Pre-images I

We consider Walsh’s conformal map from the exterior of a compact set E⊆C\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$E \subseteq \mathbb {C}$$\end{document} onto a lemniscatic domain. If E is simply connected, the lemniscatic domain is the exterior of a circle, while if E has several components, the lemniscatic domain is the exterior of a generalized lemniscate and is determined by the logarithmic capacity of E and by the exponents and centers of the generalized lemniscate. For general E, we characterize the exponents in terms of the Green’s function of Ec\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$E^c$$\end{document}. Under additional symmetry conditions on E, we also locate the centers of the lemniscatic domain. For polynomial pre-images E=P-1(Ω)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$E = P^{-1}(\Omega )$$\end{document} of a simply-connected infinite compact set Ω\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Omega $$\end{document}, we explicitly determine the exponents in the lemniscatic domain and derive a set of equations to determine the centers of the lemniscatic domain. Finally, we present several examples where we explicitly obtain the exponents and centers of the lemniscatic domain, as well as the conformal map.


Introduction
The famous Riemann mapping theorem says that for any simply connected, compact and infinite set E there exists a conformal map R E : By imposing the normalization R E (z) = z cap(E) + O(1) as z → ∞, where cap(E) denotes the logarithmic capacity of E, this map is unique.In his 1956 article [13], J. L. Walsh found the following canonical generalization for multiply connected domains.
Theorem 1.1.Let E 1 , . . ., E ⊆ C be disjoint simply connected, infinite compact sets and let (1.1) In particular, E c = C \ E is an -connected domain.Then there exists a unique compact set of the form where a 1 , . . ., a ∈ C are distinct and m 1 , . . ., m > 0 are real numbers with j=1 m j = 1, and a unique conformal map Φ : If E is bounded by Jordan curves, then Φ extends to a homeomorphism from E c to L c .
(ii) The points a 1 , . . ., a (sometimes called 'centers' of L) and also m 1 , . . ., m in Theorem 1.1 are uniquely determined.The function U is analytic in C\{a 1 , . . ., a } and in general not single-valued, but its absolute value is single-valued.Note that the compact set L, defined in (1.2), consists of disjoint compact components L 1 , . . ., L , with a j ∈ L j for j = 1, . . ., .The components L 1 , . . ., L are labeled such that a Jordan curve surrounding E j is mapped by Φ onto a Jordan curve surrounding L j .This shows that Walsh's map onto lemniscatic domains is a canonical generalization of the Riemann map from simply to multiply connected domains.
(iv) The existence in Theorem 1.1 was first shown by Walsh; see [13,Thm. 3] and the discussion below.Other existence proofs were given by Grunsky [2], [3], and [4,Thm. 3.8.3],and also by Jenkins [5] and Landau [6].However, these articles do not contain any analytic or numerical examples.The first analytic examples were constructed by Sète and Liesen in [12], and, subsequently, a numerical method for computing the Walsh map was derived in [8] for sets bounded by smooth Jordan curves.
(v) The domain L c is usually called a lemniscatic domain.This term seems to originate in Grunsky [4, p. 106].
In this paper, we bring some light into the computation of the parameters m j and a j appearing in Theorem 1. 1.In Section 2, as a first main result, we derive a general formula (Theorem 2.3) for the exponents m j in terms of the Green's function of E c , denoted by g E .Of special interest is of course the case when E is real or when E or some component E j are symmetric with respect to the real line, i.e., E * = E or E * j = E j , where denotes the complex conjugate of a set K ⊆ C. We prove that E * = E and E * j = E j implies that a j ∈ R (Theorem 2.7).In the case that all components are symmetric, we give an interlacing property of the components E j and the critical points of g E (Theorem 2.8).
In Section 3, we consider the case when E is a polynomial pre-image of a simply connected compact infinite set Ω, that is, E = P −1 n (Ω).In this case, we prove in Theorem 3.2 that the m j are always rational of the form m j = n j /n, where n is the degree of the polynomial P n and n j is the number of zeros of P n (z)−ω in E j , where ω ∈ Ω.Moreover, the unknowns a 1 , . . ., a are characterized by a system of equations which in particular can be solved explicitly in the case = 2.With the help of these findings, we obtain an analytic expression for the map Φ if P −1 n (Ω) is connected (Corrolary 3.7).Finally, Section 4 contains several illustrative examples when E = P −1 n (Ω) and when Ω = D, Ω = [−1, 1] or when Ω is a Chebyshev ellipse.In particular, we determine the exponents and centers of the corresponding lemniscatic domain and visualize the conformal map Φ.

Results for general compact sets
Let the notation be as in Theorem 1.1.The Green's function (with pole at ∞) of L c is Then Φ(Γ R ) = Λ R and Φ maps the exterior of Γ R onto the exterior of Λ R .Let R * > 1 be the largest number, such that g E has no critical point interior to Γ R * (if = 1, then R * = ∞; see Theorem 2.5 below).Then Φ is the conformal map of ext(Γ R ) onto the lemniscatic domain ext(Λ R ) for all 1 < R < R * ; see also [14, p. 31].
Here and in the following, we extensively use the Wirtinger derivatives where z = x + iy with x, y ∈ R. We relate the exponents and centers of the lemniscatic domain to the Wirtinger derivatives ∂ z g E and ∂ w g L of the Green's functions.Note that Lemma 2.1.The Green's functions g L and g E from (2.1) and (2.2) satisfy With the chain rule for the Wirtinger derivatives and (2.2), we find which is (2.3).Integrating this expression over γ yields In combination with (2.3), this yields (2.4).
We are now ready to express the exponents m j through the Wirtinger derivatives of the Green's function.For j = 1, . . ., , let γ j be a closed curve in C\E with wind(γ j ; z) = δ jk for z ∈ E k and k = 1, . . ., , where wind(γ; z 0 ) denotes the winding number of the curve γ about z 0 , and δ jk is the usual Kronecker delta.More informally, the curve γ j contains E j but no E k , k = j, in its interior.
Theorem 2.3.In the notation of Theorem 1.1, let g E and g L be the Green's functions of E c and L c , respectively.For each j ∈ {1, . . ., }, let γ j be a closed curve in C \ E with wind(γ j ; z) = δ jk for z ∈ E k and k = 1, . . ., , and let λ j = Φ • γ j .Then, Moreover, if the function f is analytic interior to λ j and continuous on trace(λ j ), then (2.9) Proof.Since 2∂ w log|w| = ∂ w log(ww) = 1/w, we obtain from (2.1) that which is a rational function.By construction, λ j is a closed curve in C \ L with wind(λ j ; a k ) = δ jk .Integrating over λ j yields the first equality in (2.8).The second equality follows by Lemma 2.1.Using (2.10) and the residue theorem, we obtain This shows the first equality in (2.9).Multiplying (2.3) by f (Φ(z)) and integrating yields the second equality in (2.9).
Remark 2.4.(i) By (2.8) in Theorem 2.3, the exponent m j of the lemniscatic domain is the residue of 2∂ w g L at a j .Moreover, m j is (up to the factor 1 2πi ) the module of periodicity (or period ) of the differential 2∂ z g E (z) dz; see [1, p. 147].The latter can be rewritten as where the middle integral is over the conjugate differential of dg E , and where ∂g E ∂n is the derivative with respect to the normal pointing to the right of γ j ; see [1, pp. 162-164] for a detailed discussion.
(ii) Since ∂ z g E is analytic in E c and ∂ w g L is analytic in C \ {a 1 , . . ., a } ⊇ L c , the integrals in (2.8) have the same value for all positively oriented closed curves that contain only E j or a j in their interior.
The following well-known result due to J. L. Walsh [15] establishes a relation between the critical points of the Green's function and the connectivity of E c .Theorem 2.5 ([15, pp. 67-68]).Let E ⊆ C be compact such that K = E c is connected and such that K possesses a Green's function g E with pole at infinity.If K is of finite connectivity , then g E has precisely − 1 critical points in C \ E, counted according to their multiplicity.If K is of infinite connectivity, g E has a countably infinite number of critical points.Moreover, all critical points of g E lie in the convex hull of E.
As is typical for conformal maps with Φ(z) = z + O(1/z) at ∞, symmetry of E (e.g., rotational symmetry or symmetry with respect to the real line) leads to the same symmetry of L, and to "symmetry" in the map Φ.
In the last two results of this section, we consider the case where E and one or all of its components E j are symmetric with respect to the real line.This allows to locate the points a 1 , . . ., a and the critical points of the Green's function g E .
Theorem 2.7.In the notation of Theorem 1.1, suppose that Proof.Since E * = E, we have Φ(z) = Φ(z) by Lemma 2.6 and for some j ∈ {1, . . ., } then there exists a smooth Jordan curve γ j in C \ E symmetric with respect to the real line which surrounds E j in the positive sense, but no other component E k , k = j, i.e., wind(γ j ; z) = δ jk for z ∈ E k and k = 1, . . ., .By (2.9), In Theorem 2.7, if a component E j is not symmetric with respect to the real line, then the corresponding point a j is in general not real, as the example of the star in [12,Cor. 3.3] shows.
If E * j = E j for all components of E then we order the components "from left to right": By Lemma A.2, each E j ∩ R is a point or an interval, and we label E 1 , . . ., E such that x ∈ E j ∩ R and y ∈ E j+1 ∩ R implies x < y for all j = 1, . . ., − 1.
(i) The − 1 critical points of the Green's function g E are real.Moreover, each E j intersects R in a point or an interval, and the critical points of g E interlace the sets E j ∩ R, j = 1, . . ., .
Proof.(i) For each j = 1, . . ., , the set E j ∩ R is a point or an interval by Lemma A.2.For j = 1, . . ., − 1, denote the 'gap' on the real line between E j and E j+1 by The Green's function g E is positive on I j and can be continuously extended to I j with boundary values 0. Then g E has a maximum on I j at a point x j ∈ I j at which ∂ x g E (x j ) = 0.By (A.1), we have ∂ y g E (x j ) = −∂ y g E (x j ), i.e., ∂ y g E (x j ) = 0.This shows that x j is a critical point of g E for j = 1, . . ., − 1.These are the − 1 critical points of g E which are real and interlace the sets Then γ 1 intersects I 0 and I 1 (see (i)), hence the curve Φ(γ 1 ) intersects the images J 0 = Φ(I 0 ) and J 1 := Φ(I 1 ).This shows that L 1 is the leftmost component of L and a 1 is the minimum of a 1 , . . ., a .Proceeding in a similar way gives that the components L 1 , . . ., L are ordered from left to right, and therefore a 1 < a 2 < . . .< a .

Results for polynomial pre-images
Let Ω ⊆ C be a compact infinite set such that Ω c is a simply connected domain in C and let R Ω be the exterior Riemann map of Ω, i.e., the conformal map where R := max z∈Ω |z|, and Let P n be a polynomial of degree n ≥ 1, more precisely, and consider the pre-image of Ω under P n , that is The set E is compact and, by Theorem A.4, the complement E c is connected.Therefore, the Green's function of see [9, p. 134].Since 2∂ z log|f | = f /f for an analytic function f , we have The logarithmic capacity of E is if and only if P n has exactly − 1 critical points z 1 , . . ., z −1 (counting multiplicities) for which P n (z k ) / ∈ Ω for k = 1, . . ., − 1.Moreover, the number of zeros of P n (z) − ω in E j is the same for all ω ∈ Ω, and this number is denoted by n j .

Proof. By Theorem 2.5, E has components if and only if g
), the latter is equivalent to P n (z 0 ) = 0.
In the rest of this section, we assume that E has components E 1 , . . ., E , i.e., that P n has exactly − 1 critical points with critical values in C \ Ω. Theorem 3.2.Let E = P −1 n (Ω) and the numbers n 1 , . . ., n be defined as in Theorem 3.1.Then the exponents m j in the lemniscatic domain in Theorem 1.1 are given by m j = n j n , j = 1, . . ., .
Proof.For j = 1, . . ., , let γ j be a positively oriented Jordan curve in C \ E with wind(γ j ; z) = δ jk for z ∈ E k and k = 1, . . ., .Using (2.8) and (3.7), we obtain dz. (3.12) Substituting u = P n (z) yields Since wind(P n • γ j ; u 0 ) = n j for u 0 ∈ Ω, the integral in (3.13) can be replaced by n j times an integral over a positively oriented Jordan curve Γ in C \ Ω, i.e., The integral is with U from (1.2).This follows by considering the identity (2.2) between the corresponding Green's functions.In Theorem 3.3, we establish a stronger result.By Theorem 3.2, the exponents of U satisfy m j = n j /n.Together with (3.8), we see that Next, we show that equality is also valid without the absolute value.Moreover, we derive a relation between the points a j and the coefficients p n−1 , p n of P n for n ≥ 2. The case Theorem 3.3.Let E = P −1 n (Ω) be as in (3.5).We then have with that branch of and, by (3.17), (3.23) The function (Q•Φ)/(R Ω •P n ) is analytic in C\E with constant modulus one, see (3.18), therefore constant (maximum modulus principle) and where c ∈ C with |c| = 1.By comparing the coefficients of z n of the Laurent series at ∞, we see that c = 1, which shows (3.19).Comparing the coefficients of z n−1 then yields (3.21).
Remark 3.4.In the case n = 1, i.e., P 1 (z) = p 1 z + p 0 is a linear transformation, the conformal map and lemniscatic domain are given explicitly as follows.In this case, E = P −1 1 (Ω) consists of a single component, i.e., = 1 and m 1 = 1, and Q(w) = d 1 p 1 (w −a 1 ).Comparing the constant terms at infinity of R Ω (P n (z) and Formula (3.19) does not lead to separate expressions for Q and Φ, even if R Ω and P n are known.However, if the polynomial Q(w) = d 1 p n j=1 (w − a j ) n j is known, equation (3.20) yields an expression for Φ.Since the numbers n j are already known (Theorem 3.1), our next aim is to determine a 1 , . . ., a .Lemma 3.5.Let E = P −1 n (Ω) be as in (3.5) and with components.
hence a 1 , . . ., a are critical points of Q with multiplicity j=1 (n j − 1) = n − .The remaining − 1 critical points of Q are the zeros of the polynomial in (3.29).
In principle, the right hand side in (3.28) can be computed when P n and R Ω are given.If also Q(w * ) can be computed, (3.28) yields − 1 equations for a 1 , . . ., a .
has n components, m j = 1/n for j = 1, . . ., n, the points a 1 , . . ., a n are given by with the n distinct values of the n-th root and and the Walsh map is with that branch of the n-th root such that Φ(z) = z + O(1/z) at infinity.
and the conformal map of E c onto a lemniscatic domain is Next, let us determine α 1 , β 1 , γ 1 in terms of α, β, γ.We have with distinct a 1 , . . ., a n ∈ C. In particular, n j = 1 for j = 1, . . ., n. Equating the coefficients of w n−1 in (3.35) and using Theorem 3.3, we obtain with the n distinct values of the n-th root.By (3.17In [12,Thm. 3.1], the lemniscatic domain and conformal map Φ have been explicitly constructed under the additional assumptions that Ω is symmetric with respect to R (i.e., Ω * = Ω), γ ∈ R is left of Ω, α > 0 and β = 0.A shift β = 0 can be incorporated with [12,Lem. 2.3].In Proposition 3.6 we can relax the assumptions on Ω and the coefficients α, β, γ.
The proof of Proposition 3.6 (ii) generalizes to arbitrary polynomials P n of degree n ≥ 2, which yields the following result for a connected polynomial pre-image.Corollary 3.7.Let Ω ⊆ C be a simply connected infinite compact set.Let P n be a polynomial of degree n ≥ 2 as in (3.4) such that E = P −1 n (Ω) is connected, i.e., = 1.
npn , and with that branch of the n-th root such that Φ(z) = z + O(1/z) at infinity.

Proof.
The assumption = 1 implies m 1 = 1 and n 1 = n.By Theorem 3.3, we have a 1 = − p n−1 npn , which yields the expressions for L, Q(w) = d 1 p n (w − a 1 ) n and Φ.
Let us consider the case = 2 in more detail.In this case, P n has exactly one critical point outside E. Theorem 3.8.Let E = P −1 n (Ω) in (3.9) consist of two components, and let z * be the critical point of P n in C \ E. Then a 1 , a 2 satisfy Proof.By Theorem 3.3, the centers a 1 , a 2 of L satisfy or, equivalently, By Lemma 3.5 (ii), the only critical point The corresponding critical value is where we used (3.39) in the last step.Since (R Ω •P n )(z * ) = Q(w * ) by Lemma 3.5 (i), formula (3.36) is established.
In order to specify the branch of the n-th root in (3.36), some additional information is needed.We show this for a set Ω which is symmetric with respect to the real axis and contains the origin, which covers the important examples Ω = D and Ω = [−1, 1].Lemma 3.9.Suppose that Ω * = Ω and 0 ∈ Ω.Let P n be a polynomial of degree n as in (3.4) with real coefficients such that , where we label E 1 , . . ., E from left to right along the real line.
(ii) For each j ∈ {1, . . ., − 1} and each which holds in particular for z = z j .(Chebyshev ellipse), R > 1.We have the exterior Riemann maps where the branch of the square root is chosen such that In particular, the coefficients of z at infinity are We take the branch of the square root with |z n + √ z 2n − 1| > 1.In the second representation of Φ we take the principal branch of the n-th root; see [12,Thm. 3.1].In particular, the logarithmic capacity of E is 2 −1/n .Figure 2 illustrates the case n = 5.The left panel shows a phase plot of Φ.In a phase plot, the domain is colored according to the phase f (z)/|f (z)| of the function f ; see [16] for an introduction to phase plots.The middle and right panels show E and ∂L (in black) as well as a grid and its image under Φ. ∈ Ω, the set E consists of n = 5 components, while for γ = 0.75 ∈ Ω, the set E has only one component; see Proposition 3.6.Figure 3 shows phase plots of Φ (left), the sets ∂E and ∂L in black and a grid and its image.The phase plots show Φ and an analytic continuation to the interior of E. The discontinuities in the phase (in the interior of E) are branch cuts of this analytic continuation.
In the case α = 2 and β = 1/2, we have P 3 (z − ) ≈ 0.5076 ∈ D and P 3 (z + ) ≈ 1.9375 ∈ C\D, hence E = P −1 3 (D) has = 2 components by Theorem 3.1; see Figure 6 (left).Note that E c is not a lemniscatic domain (in contrast to the case considered in Example 4.5).Write E = E 1 ∪ E 2 , where E 1 is the component on the left (with ±β ∈ E 1 ).Then m 1 = 2/3 and m 2 = 1/3 by Theorem 3.2.Moreover, E * 1 = E 1 and E * 2 = E 2 , since P n is real and D is symmetric with respect to the real line, which implies that a 1 , a 2 ∈ R by Theorem 2.7.Then, by Theorem 3.8, Since a 2 − α 3 is real, taking the real third root yields Here, with a branch of Q −1 such that Φ(z) = z + O(1/z) at infinity.Here, we can obtain the boundary values of Φ for z ∈ ∂E by solving Q(w) = P 3 (z) and identifying the boundary points in the correct way.Then, since Φ(z) − z is analytic in E c and zero at infinity, we have where ∂E is negatively oriented, such that E c lies to the left of ∂E. Figure 6 also shows a cartesian grid (left) and its image under Φ (right).For the computation, we numerically approximate the integral in (4.1) with the trapezoidal rule.

A Appendix
Lemma A.1.Let E ⊆ C be compact, such that E c has a Green's function g E .If E * = E, then g E (z) = g E (z) and ∂ z g E (z) = ∂ z g E (z).Moreover, the critical points of g E are real or appear in complex conjugate pairs.
Proof.Since E * = E, the function z → g E (z) is also a Green's function with pole at infinity of E c , hence g E (z) = g E (z) for all z ∈ E c by the uniqueness of the Green's function.Write g(x, y) = g E (z), then g(x, y) = g(x, −y) and The critical points of g E are the zeros of the analytic function ∂ z g E .Since ∂ z g E (z) = ∂ z g E (z), if z * is a zero of ∂ z g E then also z * is a zero.
Lemma A.2. Let K ⊆ C be a non-empty compact, simply connected set with K * = K, then K ∩ R is either an interval or a single point.
Proof.Since K * = K and K is connected, K ∩ R is not empty.Since K c is connected, K ∩R must be connected (otherwise the symmetry and simply-connectedness of K would imply that K c is not connected).Thus, K ∩ R is a point or an interval.
Lemma A.3.Let γ be a smooth Jordan curve symmetric with respect to the real line and let f be integrable with f (z) = f (z) on γ.Then Proof.Since γ is symmetric with respect to the real line, we can write γ = γ 1 + γ 2 with γ 2 := −γ 1 .Then Though the following theorem must be known, we did not find it in the literature.For completeness, we include a proof.Next, we show that ∂P (G 1 ) ⊆ ∂Ω.Let w ∈ ∂P (G 1 ).Then there exists w k ∈ P (G 1 ) with w k → w.For each k, there exists z k ∈ G 1 with P (z k ) = w k .Since G 1 is bounded, the sequence (z k ) k has a convergent subsequence (z k j ) j with z k j → z ∈ G 1 .This implies that P (z) = w.Since w ∈ ∂P (G 1 ), we have z ∈ ∂G 1 (otherwise, z ∈ G 1 would imply P (z) ∈ P (G 1 ) and, since P (G 1 ) is open, w = P (z) / ∈ ∂P (G 1 )).Since G is open, this implies that z / ∈ G and hence that z ∈ P −1 (Ω) and w = P (z) ∈ Ω.Since w k ∈ P (G 1 ) ⊆ C \ Ω and w k → w, we obtain that w ∈ ∂Ω.
where C = C ∪ {∞} denotes the extended complex plane, D the open and D the closed unit disk.

Figure 1 :
Figure 1: Commutative diagram of the maps in Theorem 3.3.
.21) Proof.Consider the Laurent series at infinity of R Ω •P n and Q • Φ.By (3.1) and (3.4), .28) (ii) The polynomial Q has − 1 critical points in C \ L and these are the zeros of k=1 n k j=1,j =k (w − a j ).(3.29) Proof.(i) By Theorem 3.1, P n has − 1 critical points in C \ E. The functions P n and R Ω •P n have the same critical points in .34) with that branch of the n-th root such that Φ(z) = z + O(1/z) at infinity.Proof.(i) Since P n (β) = γ, the assumption γ / ∈ Ω is equivalent to β / ∈ E. The only critical point of P n is z * = β with multiplicity n − 1, hence E has = n components by Theorem 3.1.The point ), we have L = {w ∈ C : |Q(w)| ≤ 1}, which is equivalent to (3.31).Then (3.32) follows from (3.20).(ii) The assumption γ ∈ Ω is equivalent to β ∈ E, thus P n has no critical point in C \ E and E is connected, i.e., = 1.Then m 1 = 1 and n 1 = n.By Theorem 3.3, na 1 = − p n−1 pn = nβ, hence a 1 = β.Together with (3.8), we obtain the expression (3.33) for L. In contrast to case (i), we have Q(w) = d 1 α(w − β) n , which yields (3.34) by (3.20).

Theorem A. 4 .
Let P be a non-constant polynomial and Ω ⊆ C be a simply connected compact set.Then C \ P −1 (Ω) is open and connected, i.e., a region.Proof.Clearly, G := C \ P −1 (Ω) = P −1 ( C \ Ω) is open and contains ∞.Let G ∞ ⊆ G be that component of G that contains ∞.Suppose that G is not connected, i.e., G = G ∞ .Then there exists another component G 1 ⊆ G, and G 1 is a bounded region.Then P (G 1 ) is a bounded region with P (G 1 ) ⊆ C \ Ω.