The Ptolemy–Alhazen Problem and Spherical Mirror Reflection
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Abstract
An ancient optics problem of Ptolemy, studied later by Alhazen, is discussed. This problem deals with reflection of light in spherical mirrors. Mathematically, this reduces to the solution of a quartic equation, which we solve and analyze using a symbolic computation software. Similar problems have been recently studied in connection with raytracing, catadioptric optics, scattering of electromagnetic waves, and mathematical billiards, but we were led to this problem in our study of the socalled triangular ratio metric.
Keywords
Triangular ratio metric Ptolemy–Alhazen problem Reflection of lightMathematics Subject Classification
30C20 30C15 51M991 Introduction
The Greek mathematician Ptolemy (ca. 100–170) formulated a problem concerning reflection of light at a spherical mirror surface: Given a light source and a spherical mirror, find the point on the mirror where the light will be reflected to the eye of an observer.
Alhazen (ca. 965–1040) was a scientist who lived in Iraq, Spain, and Egypt and extensively studied several branches of science. For instance, he wrote seven books about optics and studied, e.g., Ptolemy’s problem as well as many other problems of optics and is considered to be one of the greatest researchers of optics before Kepler [2]. Often the above problem is known as Alhazen’s problem [10, p. 1010]. At the end of this introduction, we will point out various applications and earlier results connected with the Ptolemy–Alhazen problem.
We will consider the twodimensional version of the problem and present an algebraic solution for it. The solution reduces to a quartic equation which we solve with symbolic computation software.
We call this the interior problem—there is a natural counterpart of this problem for the case when both points are in the exterior of the closed unit disk, called the exterior problem. Indeed, this exterior problem corresponds to Ptolemy’s questions about light source, spherical mirror, and observer. As we will see below, the interior problem is equivalent to finding the maximal ellipse with foci at \(z_1, z_2\) contained in the unit disk, and the point of reflection \(u \in \partial {\mathbb {D}}\) is the tangent point of the ellipse with the circumference. Algebraically, this leads to the solution of a quartic equation as we will see below.
We study the Ptolemy–Alhazen interior problem and in our main result, Theorem 1.1, we give an equation of degree four that yields the reflection point on the unit circle. Standard symbolic computation software can then be used to find this point numerically. We also study the Ptolemy–Alhazen exterior problem.
Theorem 1.1
It should be noted that the Eq. (1.3) may have roots in the complex plane that are not on the unit circle, and of the roots on the unit circle, we must choose one root \(u\,,\) that minimizes the sum \( z_1u+z_2u\,.\) We call this root the minimizing root of (1.3).
Corollary 1.2
As we will see below, the minimizing root need not be unique.
We have used Risa/Asir symbolic computation software [20] in the proofs of our results. We give a short Mathematica code for the computation of \(s_{\mathbb {D}}(z_1, z_2)\,.\)
Theorem 1.1 is applicable not merely to light signals but whenever the angles of incidence and reflection of a wave or signal are equal, for instance, in the case of electromagnetic signals like radar signals or acoustic waves. H. Bach [4] has made numerical studies of Alhazen’s raytracing problem related to circles and ellipses. A.R. Miller and E. Vegh [18] have studied the exterior Ptolemy–Alhazen problem and computed the grazing angle of specular reflection (the complement of the equal angles of incidence and of reflection) using a quartic equation, which is not the same as (1.3). They did not consider the problem of finding the point of incidence in the case of specular reflection, which is solved through Eq. (1.3).
Mathematical theory of billiards also leads to similar studies: see for instance the paper by M. Drexler and M.J. Gander [9]. The Ptolemy–Alhazen problem also occurs in computer graphics and catadioptric optics [1]. The wellknown lithograph of M. C. Escher named “Hand with reflecting sphere” demonstrates nicely the idea of reflection from a spherical mirror.
2 Algebraic solution to the Ptolemy–Alhazen problem
In this section, we prove Theorem 1.1 and give an algorithm for computing \(s_{\mathbb {D}}(z_1,z_2)\) for \(z_1,z_2\in \mathbb {D}\).
Problem 2.1
For \(z_1,z_2\in \mathbb {D} \), find the point \(u \in \partial \mathbb {D} \) such that the sum \(z_1u+z_2u \) is minimal.
The point u is given as the point of tangency of an ellipse \( zz_1+zz_2=r \) with the unit circle.
Remark 2.2
2.1 Proof of Theorem 1.1
Remark 2.3
The solution of (1.3) includes all the tangent points of the ellipse \(zz_1+zz_2=2\overline{u}z_1u\overline{z_2} \) and the unit circle. (See Figs. 1, 2). Figure 2 displays a situation where all the roots of the quartic equation have unit modulus. However, this is not always the case for the Eq. (1.3). E.g., if \(z_1=0.5+(0.1 \cdot k ) i, k=1,..,5, \, z_2= 0.5,\) the Eq. (1.3) has two roots of modulus equal to 1 and two roots off the unit circle, see Fig. 3. Miller and Vegh [18] computed the grazing angle of specular reflection using a quartic selfinversive polynomial equation, which is not the same as (1.3). Note that all the roots of their equation have modulus equal to one. They have also studied the Ptolemy–Alhazen problem using a quartic equation, that is different from our equation and, moreover, all the roots of their equation have modulus equal to one Fig. 3.
We say that a polynomial P(z) is selfinversive if \(P(1/\overline{u})= 0\) whenever \(u \ne 0\) and \(P(u)=0\,.\) It is easily seen that the quartic polynomial in (1.3) is selfinversive. Note that the points u and \(1/\overline{u}\) are obtained from each other by the inversion transformation \(w \mapsto 1/\overline{w} \,.\)
It is clear from the compactness of the unit circle, that the function \( z_1z+z_2z \) attains its maximum and minimum on the unit circle. However, as a property of the Eq. (1.3) itself, the following results can also be derived.
Lemma 2.4
The Eq. (1.3) always has at least two roots of modulus equal to 1.
Proof
Consider first the case, when \(z_1 z_2 =0 \,. \) In this case, the Eq. (1.3) has two roots \(u, u=1,\) with \(u^2={z_1}/{\overline{z_1}}\in \partial \mathbb {D} \) if \(z_2=0, z_1\ne 0\,. \) (The case \(z_1=z_2=0 \) is trivial.) Suppose that the equation has no root on the unit circle \(\partial \mathbb {D} \,.\)
By the invariance property pointed out above, if \(u_0\in \mathbb {C} \setminus ( \{0\} \cup \partial \mathbb {D}) \) is a root of (1.3), then \({1}/{\overline{u_0}} \) also is a root of (1.3). Hence, the number of roots off the unit circle is even and the number of roots on the unit circle must also be even. We will now show that this even number is either 2 or 4.
Remark 2.5
 Case 1.
 \(z_{1}\ne 0=z_{2}\) (cubic equation). The Eq. (1.3) is now \(\left( \overline{z_{1}}\right) u^{3}+z_{1}u=0\) and has the roots \(u_{1}=0\), \(u_{2,3}=\pm \frac{z_{1}}{\left z_{1}\right }\) and for \(z \in {\mathbb {D}}\)$$\begin{aligned} s_{\mathbb {D}}(0,z) = \frac{z}{2  z}\,. \end{aligned}$$
 Case 2.
 \(z_{1}+z_{2}=0\), \(z_{1}\ne 0\). The Eq. (1.3) reduces now to:The roots are: \(u_{1,2}=\pm \frac{z_{1}}{\left z_{1}\right }\), \(u_{3,4}=\pm i\frac{z_{1}}{\left z_{1}\right }\) (four distinct roots of modulus 1) and for \(z \in {\mathbb {D}}\)$$\begin{aligned} \left( \overline{z_{1}}^{2}\right) u^{4}+z_{1}^{2} =0\Leftrightarrow u^{4} =\left( \frac{z_{1}}{\overline{z_{1}}}\right) ^{2} \Leftrightarrow u^{4} =\left( \frac{z_{1}}{\left z_{1}\right }\right) ^{4}\,. \end{aligned}$$$$\begin{aligned} s_{\mathbb {D}}(z,z) = z\,. \end{aligned}$$
 Case 3.
 \(z_{1}=z_{2}\ne 0\,.\) Clearly \(s_{\mathbb {D}}(z,z)=0\,.\) Denote \(z:=z_{1}=z_{2}\). The Eq. (1.3) reduces now to:Then, we see that \(u_{1,2}=\pm \frac{z}{\left z\right }\) are roots. The other roots are:$$\begin{aligned} \overline{z}^{2}u^{4}2\overline{z}u^{3}+2zuz^{2} = (\overline{z} u^2 z)(\overline{z} u^22u +z)= 0\,. \end{aligned}$$
 1)

If \(\left z\right <1\), then \(u_{3,4}=\frac{1}{\overline{z}} \left( 1\pm \sqrt{1\left z\right ^{2}}\right) \) (with \(\left u_{3}\right >1\), \(\left u_{4}\right <1\))
 2)

If \(\left z\right >1\), then \(u_{3,4}=\frac{1}{\overline{z}} \left( 1\pm i\sqrt{\left z\right ^{2}1}\right) \) (with \(\left u_{3}\right =\left u_{4}\right =1\)).
 Case 4.
 \(\left z_{1}\right =\left z_{2}\right \ne 0\,.\) Denote \(\rho =\left z_{1}\right =\left z_{2}\right \). Using a rotation around the origin and a change of orientation, we may assume that \(\arg z_{2}=\arg z_{1}=:\alpha \), where \(0\le \alpha \le \frac{\pi }{2}\). The Eq. (1.3) reads now: \(\rho ^{2}u^{4}2\rho \left( \cos \alpha \right) u^{3}+2\rho \left( \cos \alpha \right) u\rho ^{2}=0\)The roots are: \(u_{1,2}=\pm 1\) and$$\begin{aligned} \rho ^{2}u^{4}2\rho \left( \cos \alpha \right) u^{3} +2\rho \left( \cos \alpha \right) u\rho ^{2} =\rho ^{2}\left( u^{2}1\right) \left( u^{2}\frac{ 2\cos \alpha }{\rho }u+1\right) \end{aligned}$$
 1)

If \(0<\rho <\cos \alpha \), then \(u_{3,4}=\frac{\cos \alpha }{\rho }\pm \sqrt{\left( \frac{\cos \alpha }{\rho }\right) ^{2}1}\) (here \(\left u_{3}\right >1\), \(\left u_{4}\right <1\))
 2)

If \(\rho \ge \cos \alpha \), then \(u_{3,4}=\frac{\cos \alpha }{\rho }\pm i \sqrt{1\left( \frac{\cos \alpha }{\rho }\right) ^{2}}\) (here \(\left u_{3}\right =\left u_{4}\right =1\)). Note that Case 4 includes Cases 2 and 3 (for \(\alpha =\frac{\pi }{2}\), respectively, \(\alpha =0\)).
 Case 5.
 \(z_{1}=tz_{2}\) (\(t\in \mathbb {R}\), \(z_{2}\ne 0\)). This case is generalization of cases \(z_{1}=0\ne z_{2}\), \(z_{1}+z_{2}=0\), \(z_{1}\ne 0\) and \(z_{1}=z_{2}\ne 0\). Denote \(P\left( u\right) =\overline{z_{1}z_{2}}u^{4}\left( \overline{z_{1}}+ \overline{z_{2}}\right) u^{3}+\left( z_{1}+z_{2}\right) uz_{1}z_{2}\). Denoting \(z_{2}=z\) we have:For \(t=0\) the roots of P are \(0,\pm \frac{z}{\left z\right }\). Let \(t\ne 0\). Besides \(\pm \frac{z}{\left z\right }\) there are two roots, which have modulus 1 if and only if \(\left z\right \ge \left \frac{1+t}{2t}\right \).$$\begin{aligned} P(u)&= t\overline{z}^{2}u^{4}\left( 1+t\right) \overline{z} u^{3}+\left( 1+t\right) zutz^{2} \\&= t\overline{z}^{2}\left( u^{4} \frac{z^{4}}{\left z\right ^{4}}\right)  \left( 1+t\right) \overline{z}u\left( u^{2} \frac{z^{2}}{\left z\right ^{2}}\right) .\\ P(u)&=\overline{z}\left( u\frac{z}{\left z\right }\right) \left( u+\frac{z}{\left z\right }\right) \left( t\overline{z}u^{2}\left( 1+t\right) u+tz\right) \end{aligned}$$
2.2 Exterior Problem
Given \(z_1,z_2\in \mathbb {C}\setminus \overline{\mathbb {D}} \), find the point \(u \in \partial \mathbb {D} \) such that the sum \(z_1u+z_2u \) is minimal.
Lemma 2.6
Remark 2.7
The above equation coincides with the Eq. (1.3) for the “interior problem”, since Theorem 1.1 could be proved without using the assumption \( z_1,z_2\in \mathbb {D} \).
Remark 2.8
Lemma 2.9
The boundary of \(B_s(z,t) =\{ w \in \mathbb {D}: s_{\mathbb {D}}(z,w)<t \} \) is included in an algebraic curve.
Proof
Remark 2.10
The algebraic curve \(\{ w: {\mathcal {B}}(w)=0 \}\) does not coincide with the boundary \(\partial B_s(c,t) \). There is an “extra” part of the curve since the Eq. (2.9) contains extraneous solutions.
One can also use numerical methods to compute \(s_{\mathbb {D}}\,, \) see [6].
3 Geometric Approach to the Ptolemy–Alhazen Problem
In this section, the unimodular roots of Eq. (1.3) are characterized as points of intersection of a conic section and the unit circle, then n such roots are studied, where \(n=4\) in the case of the exterior problem and \(n=2 \) in the case of the interior problem. We describe the construction of the conic section mentioned above. Except in the cases where \(0,z_{1},z_{2}\) are collinear or \( \left z_{1}\right =\left z_{2}\right , \) the construction cannot be carried out as rulerandcompass construction. Neumann [19] proved that Alhazen’s interior problem for points \(z_{1},z_{2}\) is solvable by ruler and compass only for \(({\mathrm {Re}}z_{1},{\mathrm {Im}} z_{1},{\mathrm {Re}}z_{2},{\mathrm {Im}}z_{2})\) belonging to a null subset of \( \mathbb {R}^{4}\), in the sense of Lebesgue measure.
We characterize algebraically condition (1.1) without assuming that \(z_{1},z_{2}\in \mathbb {D}\), or \(z_{1},z_{2}\in \mathbb {C} \setminus \overline{\mathbb {D}}\), or \(u\in \partial \mathbb {D}\).
Lemma 3.1
 (i)
\(\measuredangle (z_{1},u,0)=\measuredangle (0,u,z_{2})\).
 (ii)
\(\frac{u^{2}}{\left( uz_{1}\right) \left( uz_{2}\right) } =\frac{\overline{u}^{2}}{\left( \overline{u}\overline{z_{1}}\right) \left( \overline{u}\overline{z_{2}}\right) }\) and \(\frac{u^{2}}{\left( uz_{1}\right) \left( uz_{2}\right) } +\frac{\overline{u}^{2}}{\left( \overline{u}\overline{z_{1}}\right) \left( \overline{u}\overline{z_{2}}\right) }>0\);
 (iii)and$$\begin{aligned} \overline{z_{1}z_{2}}u^{2}\left( \overline{z_{1}}+\overline{z_{2}}\right) \overline{u}u^{2}+\left( z_{1}+z_{2}\right) \overline{u}^{2}uz_{1}z_{2} \overline{u}^{2}=0\ \end{aligned}$$(3.1)$$\begin{aligned} \overline{z_{1}z_{2}}u^{2}\left( \overline{z_{1}}+\overline{z_{2}}\right) \overline{u}u^{2}\left( z_{1}+z_{2}\right) \overline{u}^{2}u+z_{1}z_{2} \overline{u}^{2}+2u^{2}\overline{u}^{2}>0. \end{aligned}$$(3.2)
Proof
Let \(u\in \mathbb {C}^{*}{\setminus }\left\{ z_{k}:k=1,2\right\} \). Clearly, \( \measuredangle (z_{1},u,0) =\arg \frac{u}{uz_{1}}\) and \(\measuredangle (0,u,z_{2}) =\arg \frac{uz_{2}}{u}\). Denoting \(v:=\frac{u}{uz_{1}}:\frac{uz_{2}}{u}\), we see that \(\measuredangle (z_{1},u,0) =\measuredangle (0,u,z_{2})\) if and only if v satisfies both \(v=\overline{v}\) and \(v+\overline{v}>0\), i.e. if and only if (ii) holds.
We have \(v=\overline{v}\) (respectively, \(v+\overline{v}>0\)) if and only if (3.1) (respectively, (3.2)) holds, therefore (ii) and (iii) are equivalent.
In the special case \(z_{1}=z_{2}=0\) (\(z_{1}=z_{2}\ne 0\)) (i), (ii) and (iii) are satisfied whenever \(u\in \mathbb {C}^{*}\) (respectively, if and only if \(u=\lambda z_{1}\) for some real number \(\lambda \ne 0,1\)). \(\square \)
Remark 3.2
Consider the interior problem, with \(z_{1},z_{2}\in \mathbb {D}\) and \(u\in \partial \mathbb {D}\). The unit circle is exterior to the circles of diameters \(\left[ 0,z_{1}\right] \), \(\left[ 0,z_{2}\right] \). An elementary geometric argument shows that \(\frac{\pi }{2}< \measuredangle (z_{1},u,0)<\frac{\pi }{2}\) and \(\frac{\pi }{2}<\measuredangle (0,u,z_{2}) <\frac{\pi }{2}\), therefore \(\left \measuredangle (z_{1},u,0) \measuredangle (0,u,z_{2})\right \ne \pi \). In this case (3.1) implies \(\measuredangle (z_{1},u,0) =\measuredangle (0,u,z_{2})\).
Remark 3.3
Lemma 3.4
Let \(z_{1},z_{2}\in \mathbb {C}^{*}\). The conic section \(\Gamma \) given by (3.3) has the center \(c=\frac{1}{2}\left( \frac{1}{\overline{z_{1}}} +\frac{1}{\overline{z_{2}}}\right) \) and it passes through \(0,\frac{1}{\overline{z_{1}}}\), \(\frac{1}{\overline{z_{2}}}\), \(\frac{1}{\overline{z_{1}}}+\frac{1}{\overline{z_{2}}}\). If \(\left z_{1}\right =\left z_{2}\right \) or \(\left \arg z_{1}\arg z_{2}\right \in \left\{ 0,\pi \right\} \), then \(\Gamma \) consists of the parallels \(d_{1}\), \(d_{2}\) through c to the bisectors (interior, respectively, exterior) of the angle \(\measuredangle (z_{1},0,z_{2})\). In the other cases, \(\Gamma \) is an equilateral hyperbola having the asymptotes \(d_{1}\) and \(d_{2}\).
Proof
 (1)
\(\frac{\overline{z_{1}}}{\overline{z_{2}}} +\frac{\overline{z_{2}}}{\overline{z_{1}}}\in \mathbb {R}\);
 (2)
\(\frac{z_{2}}{z_{1}}\in \mathbb {R}\) or \(\left \frac{z_{2}}{z_{1}}\right =1\);
 (3)
\(\left \arg z_{1}\arg z_{2}\right \in \left\{ 0,\pi \right\} \) or \(\left z_{1}\right =\left z_{2}\right \).
Lemma 3.5
(Sylvester’s theorem) In any triangle with vertices \(z_{1},z_{2},z_{3}\), the orthocenter \(z_{H}\) and the circumcenter \(z_{C}\) satisfy the identity \(z_{H}+2z_{C}=z_{1}+z_{2}+z_{3}\).
Proof
Let \(z_{G}\) be the centroid of the triangle. It is well known that \(z_{G}=\frac{z_{1}+z_{2}+z_{3}}{3}\). By Euler’s straightline theorem, \(z_{H}z_{G}=2(z_{G}z_{C})\). Then \(z_{H}+2z_{C}=3z_{G}=z_{1}+z_{2}+z_{3}\). \(\square \)
Lemma 3.6
Let \(z_{1},z_{2}\in \mathbb {C}^{*}\). The orthocenter of the triangle with vertices \(0,\frac{1}{\overline{z_{1}}}\), \(\frac{1}{\overline{z_{2}}}\) belongs to the conic section given by Eq. (3.3).
Proof
Consider a triangle with vertices \(z_{1},z_{2},z_{3}\) and denote by \(z_{H}\) and \(z_{C}\) the orthocenter and the circumcenter, respectively. By Sylvester’s theorem, Lemma 3.5, \(z_{H}=z_{1}+z_{2}+z_{3}2z_{C}\).
Let \(z_{1},z_{2}\in \mathbb {C}^{*}\) be such that \(\left z_{1}\right \ne \left z_{2}\right \) and \(\left \arg z_{1}\arg z_{2}\right \notin \left\{ 0,\pi \right\} \). Let h be given by (3.6). Note that \(h\left( \frac{1}{\overline{z_{1}}}+\frac{1}{\overline{z_{2}}}\right) =\frac{2\left( z_{2}z_{1}\right) }{z_{1}\overline{z_{2}}\overline{z_{1}}z_{2}} \ne 0\). If \(h\notin \left\{ 0,\frac{1}{\overline{z_{1}}}, \frac{1}{\overline{z_{2}}}\right\} \) then the hyperbola \(\Gamma \) passing through the five points \(0,\frac{1}{\overline{z_{1}}},\frac{1}{\overline{z_{2}}}, \frac{1}{\overline{z_{1}}}+\frac{1}{\overline{z_{2}}}\), h can be constructed using a mathematical software.
In the cases where \(h\in \left\{ 0,\frac{1}{\overline{z_{1}}}, \frac{1}{\overline{z_{2}}}\right\} \), we choose a vertex of the hyperbola \(\Gamma \) as the fifth point needed to construct \(\Gamma \) . The vertices of the equilateral hyperbola \(\Gamma \) are the intersections of \(\Gamma \) with the line passing through the center of the hyperbola, with the slope \(m=1\) if \(\left z_{1}\right >\left z_{2}\right \), respectively, \(m=1\) if \(\left z_{1}\right <\left z_{2}\right \). Let \(\alpha :=\frac{\arg z_{2}\arg z_{1}}{2}\). Using (3.5) it follows that the distance d between a vertex and the center of \(\Gamma \) is \(d=\frac{\sqrt{\left \left z_{1}\right ^{2}\left z_{2}\right ^{2}\right }}{2\left z_{1}z_{2}\right } \sqrt{\sin 2\alpha }\).
If \(h=0\) we have \(\alpha =\frac{\pi }{4}\) and \(d=\frac{1}{2}\sqrt{\Big  \big  \frac{1}{\overline{z_{2}}}\big ^{2}\big  \frac{1}{\overline{z_{1}}} \big ^{2}\Big  }\). Assume that \(h=\frac{1}{\overline{z_{1}}}\), the case \(h=\frac{1}{\overline{z_{2}}}\) being similar. Then \(\left z_{2}\right =\left z_{1}\right \cos 2\alpha <\left z_{1}\right \) and \(\left \left z_{1}\right ^{2} \left z_{2}\right ^{2}\right =\left z_{1}z_{2}\right ^{2}\), therefore \(d=\frac{1}{2}\left \frac{1}{\overline{z_{2}}}\frac{1}{\overline{z_{1}}}\right \sqrt{\sin 2\alpha }\). Let \(z_{3}\) be the orthogonal projection of \(\frac{1}{\overline{z_{1}}}\) on the line joining \(\frac{1}{\overline{z_{2}}}\) to the origin. Then \(d=\frac{1}{2}\sqrt{\left \frac{1}{\overline{z_{2}}} \frac{1}{\overline{z_{1}}}\right \cdot \left \frac{1}{\overline{z_{2}}} z_{3}\right }\). We see that a vertex of \(\Gamma \) can be constructed with ruler and compass if \(h\in \left\{ 0,\frac{1}{\overline{z_{1}}}, \frac{1}{\overline{z_{2}}}\right\} \).
Remark 3.7
Being symmetric with respect to the center of \(\Gamma \), \(\frac{1}{\overline{z_{1}}}\) and \(\frac{1}{\overline{z_{2}}}\) belong to distinct branches of \(\Gamma \), each branch being divided by \(\frac{1}{\overline{z_{1}}}\) or \(\frac{1}{\overline{z_{2}}}\) into two arcs. If \(z_{k}\in \mathbb {C}\setminus \overline{\mathbb {D}}\), \(k\in \left\{ 1,2\right\} \), then each of these arcs joins \(\frac{1}{\overline{z_{k}}}\), that is in the unit disk, with some point exterior to the unit disk; therefore, it intersects the unit circle. It follows that, in the case of the exterior problem, \(\Gamma \) intersects the unit circle at four distinct points.
In the following, we identify the points of intersection of the conic section \(\Gamma \) given by (3.3) with the unit circle. After finding the points \(u\in \partial \mathbb {D}\cap \Gamma \), it is easy to select among these the points u for which (1.1) holds, respectively, for which \(\left uz_{1}\right +\left uz_{2}\right \) attains its minimum or its maximum on \(\partial \mathbb {D}\).
First assume that \(\Gamma \) is a pair of lines \(d_{1},d_{2}\), parallel to the interior bisector and to the exterior bisector of the angle \(\measuredangle (z_{1},0,z_{2})\), respectively. Let \(\alpha =\frac{1}{2}\left \arg z_{2}\arg z_{1}\right \). Then \(\alpha \in \left\{ 0,\frac{\pi }{2}\right\} \) or \(\left z_{1}\right =\left z_{2}\right \). The distances from the origin to \(d_{1}\) and \(d_{2}\) are \(\delta _{1}=\frac{\left \left z_{2}\right \left z_{1}\right \right }{2\left z_{1}z_{2}\right }\sin \alpha \) and \(\delta _{2}=\frac{\left z_{1}\right +\left z_{2}\right }{2\left z_{1}z_{2}\right }\cos \alpha \). Then, \(\Gamma \) intersects the unit circle at four distinct points in the following cases: (i) \(z_{1},z_{2}\in \mathbb {C}\setminus \mathbb {D}\); (ii) \(z_{1},z_{2}\in \mathbb {D}\) with \(\frac{1}{2}\left \frac{1}{\left z_{1}\right } \frac{1}{\left z_{2}\right }\right <1\) or with \(\left z_{1}\right =\left z_{2}\right >\cos \alpha \). In the other cases for \(z_{1},z_{2}\in \mathbb {D}\) the intersection of \(\Gamma \) with the unit circle consists of two distinct points.
Proposition 3.8
 (i)
four distinct points if \(z_{1},z_{2}\in \mathbb {C}\setminus \mathbb {D}\), one in the interior of each angle determined by the lines that pass through the origin and \(z_{1}\), respectively, \(z_{2}\);
 (ii)
at least two distinct points if \(z_{1},z_{2}\in \mathbb {D}\), one in the interior of the angle determined by the rays passing starting at the origin and passing through \(z_{1}\), respectively, \(z_{2}\) and the other in the interior of the opposite angle.
Proof
 (i)
Assume that \(z_{1},z_{2}\in \mathbb {C}\setminus \mathbb {D}\). Then \(g\left( \alpha \right) <0\) and \(g\left( \alpha \right) >0\). If \(\left z_{1}\right <\left z_{2}\right \), then \(g\left( \pi \right) <0\) \(<g\left( \alpha \pi \right) >0\), \(g\left( \alpha \right)<0<g\left( 0\right) \) and \(g(\alpha )>0>g\left( \pi \alpha \right) \). Since g is continuous on \(\mathbb {R}\), Eq. (3.7) has at least one root in each of the open intervals \(\left( \pi ,\alpha \pi \right) \), \(\left( \alpha \pi ,\alpha \right) \), \(\left( \alpha ,0\right) \) and \(\left( \alpha ,\pi \alpha \right) \). If \(\left z_{2}\right <\left z_{1}\right \), then \(g\left( \alpha \pi \right)>0>g\left( \alpha \right) \), \(g\left( 0\right) <0\) \(<g(\alpha )\) and \(g\left( \pi \alpha \right)<0<g\left( \pi \right) \). The Eq. (3.7) has at least one root in each of the open intervals \(\left( \alpha \pi ,\alpha \right) \), \(\left( 0,\alpha \right) \), \(\left( \alpha ,\pi \alpha \right) \) and \(\left( \pi \alpha ,\pi \right) \).
 (ii)
Now assume that \(z_{1},z_{2}\in \mathbb {D}\). Then \(g\left( \alpha \right) >0\) and \(g\left( \alpha \right) <0\). If \(\left z_{1}\right <\left z_{2}\right \), then \(g\left( \pi \right) <0\) \(<g\left( \alpha \pi \right) \) and \(g\left( 0\right) >0\) \(>g(\alpha )\). Since g is continuous on \(\mathbb {R}\), Eq. (3.7) has at least one root in each of the open intervals \(\left( \pi ,\alpha \pi \right) \) and \(\left( 0,\alpha \right) \). If \(\left z_{1}\right >\left z_{2}\right \), then \(g\left( 0\right)>0>g(\alpha )\) and \(g\left( \pi \alpha \right)<0<g\left( \pi \right) \). The Eq. (3.7) has at least one root in each of the open intervals \(\left( 0,\alpha \right) \) and \(\left( \pi \alpha ,\pi \right) \).
Corollary 3.9
The Eq. (1.3) has four distinct unimodular roots in the case of the exterior problem and has at least two distinct unimodular roots in the case of the interior problem.
4 Remarks on the Roots of the Equation (1.3)
In this section, we study the number of the unimodular roots of the Eq. (1.3) (i.e., the roots lying on the unit circle) and their multiplicities. Denote \(P(u)=\overline{z_{1}}\overline{z_{2}}u^{4} (\overline{z_{1}}+\overline{z_{2}})u^{3}+(z_{1}+z_{2})uz_{1}z_{2}\) . If either \(z_{1}=0\) or \(z_{2}=0\) then the cubic Eq. (1.3) \(P\left( u\right) =0\) has a root \(u=0\) and two simple roots on the unit circle.
We will assume in the following that \(z_{1}\ne 0\) and \(z_{2}\ne 0\). As we observed in Sect. 2, the quartic polynomial P is selfinversive. Then, P has an even number of zeros on the unit circle, each zero being counted as many times as its multiplicity. According to Lemma 2.4, P has at least two unimodular zeros, distinct or not, that is P has four or two unimodular zeros. There is a rich literature dealing with the location of zeros of a complex selfinversive polynomial with respect to the unit circle [5, 7, 8, 15, 16, 17].
Lemma 4.1
\(P(u)=\overline{z_{1}}\overline{z_{2}}u^{4} (\overline{z_{1}}+\overline{z_{2}})u^{3}+(z_{1}+z_{2})uz_{1}z_{2}\) cannot have two double zeros on the unit circle.
Proof
Similarly, we rule out another case.
Lemma 4.2
For \(P(u)=\overline{z_{1}}\overline{z_{2}}u^{4} (\overline{z_{1}}+\overline{z_{2}})u^{3}+(z_{1}+z_{2})uz_{1}z_{2}\) it is not possible to have a double zero on the unit circle and two zeros not on the unit circle.
Proof
Lemma 4.3
If \(P(u)=\overline{z_{1}}\overline{z_{2}}u^{4} (\overline{z_{1}}+\overline{z_{2}})u^{3}+(z_{1}+z_{2})uz_{1}z_{2}\) has a triple zero a and a simple zero b, then \(b=a\), with a and b lying on the unit circle and \(\left z_{1}+z_{2}\right =2\left z_{1}z_{2}\right \).
Proof
Assume that P has a triple zero a and a simple zero b, \(P(u)=\overline{z_{1}z_{2}}(za)^{3}(zb)\), where \(a,b\in \mathbb {C},\ a\ne b\). Since P is selfinversive, \(a=b=1\) and \(b=\frac{1}{\overline{a}}=a\). Also, the fact that the coefficient of \(u^{2}\) in P(u) vanishes already implies \(a(a+b)=0\). But \(\overline{z_{1}z_{2}}a^{2}b=z_{1}z_{2}\ne 0\), therefore \(b=a\). Considering the coefficient of \(u^{3}\) in \(P(u)=\overline{z_{1}z_{2}}(ua)^{3}(u+a)\), it follows that \(2a\overline{z_{1}}\overline{z_{2}}=\overline{z_{1}}+\overline{z_{2}}\), hence \(\left z_{1}+z_{2}\right =2\left z_{1}z_{2}\right \). \(\square \)
Example 4.4
Find the relation between \(z_{1},z_{2}\) such that \(P(u)=\overline{z_{1}}\overline{z_{2}}u^{4}(\overline{z_{1}} +\overline{z_{2}})u^{3}+(z_{1}+z_{2})uz_{1}z_{2}\) has the triple zero 1 and the simple zero \((1)\).
 Case 1.

P has four simple unimodular zeros.
 Case 2.

P has two simple unimodular zeros and two zeros that are not unimodular.
 Case 3.

P has a double unimodular zero and two simple unimodular zeros.
Proposition 4.5
 a)
P has four simple unimodular zeros if \(\left z_{1}+z_{2}\right <\left z_{1}z_{2}\right \) and
 b)
P has exactly two unimodular zeros, that are simple, if \(\left z_{1}+z_{2}\right >2\left z_{1}z_{2}\right \).
 c)
If P has four simple unimodular zeros, then \(\left z_{1}+z_{2}\right <2\left z_{1}z_{2}\right \).
 d)
If P has exactly two unimodular zeros, that are simple, then \(\left z_{1}+z_{2}\right >\left z_{1}z_{2}\right \).
Proof
Let f be a complex polynomial. The location of the zeros of the derivative \(f^{\prime }\) of f is connected with the location of the zeros of f. Gauss–Lucas theorem [17, Thm. 6.1] shows that the zeros of the derivative \(f^{\prime }\) lie within the convex hull of the set of zeros of f. In particular, if all the zeros of f lie on the unit circle, then all the zeros of \(f^{\prime }\) lie in the closed unit disk (and f is selfinversive). The converse holds by a theorem of Cohn [8] stating that a complex polynomial has all its zeros on the unit circle if and only if the polynomial is selfinversive and its derivative has all its zeros in the closed unit disk.
 a)Assume that \(\left z_{1}+z_{2}\right <\left z_{1}z_{2}\right \). Then for \(u\in \partial \mathbb {D}\) we haveIt follows by Rouché’s theorem [22, 3.10] that the derivative \(P^{\prime }\) has all its zeros in the unit disk. By Cohn’s theorem cited above, P has all its four zeros on the unit circle \(\partial \mathbb {D}\). Assume that the polynomial f is selfinversive. By [7, Thm.1], the following are equivalent:$$\begin{aligned} \left 4\overline{z_{1}z_{2}}u^{3}\right =4\left z_{1}z_{2}\right >4\left z_{1}+z_{2}\right \ge \left 3\left( \overline{z_{1}}+\overline{z_{2}}\right) u^{2}+\left( z_{1}+z_{2}\right) \right \, \end{aligned}$$In our case, \(P\left( u\right) =u^{m}Q\left( z\right) +e^{i\theta }Q^{*}\left( u\right) \) for \(m=3\), \(\theta =\pi \) and \(Q\left( u\right) = \overline{z_{1}z_{2}}u^{3}+\left( z_{1}+z_{2}\right) \). The roots of Q have modulus \(\root 3 \of {\frac{\left z_{1}+z_{2}\right }{\left z_{1}z_{2}\right }}<1\). The implication \((ii)\Rightarrow (i)\) from [7, Thm. 1] shows that P has four simple zeros on the unit circle.
 (i)
all the zeros of f are simple and unimodular;
 (ii)
there exist a polynomial g having all its zeros in the unit disk \(\left z\right <1\), a nonnegative integer m and a real number \(\theta \) such that \(f\left( z\right) =z^{m}g\left( z\right) +e^{i\theta }g^{*}\left( z\right) \) for all \(z\in \mathbb {C}\). Here \(g^{*}\left( z\right) :=z^{n}\overline{g}\left( \frac{1}{z}\right) \), where \( n=\deg Q\).
 (i)
 b)Now assume that \(\left z_{1}+z_{2}\right >2\left z_{1}z_{2}\right \). For \(u\in \partial \mathbb {D}\) we haveand it follows using Rouché’s theorem that \(P^{\prime }\) has exactly two zeros in the closed unit disk. Cohn’s theorem shows that P cannot have all its zeros on \(\partial \mathbb {D}\). By Lemma 2.4, P has at least two unimodular zeros; therefore, P has exactly two unimodular zeros. By Lemma 4.2, these unimodular zeros are simple. An alternative way to prove that P has exactly two unimodular zeros is indicated below. Assume by contrary that P has four unimodular zeros. Using the Gauss–Lucas theorem two times, it follows that each of the derivatives \(P^{\prime }\)and \(P^{\prime \prime }\) has all its zeros in the closed unit disc \(z\le 1\). The zeros of \(P^{\prime \prime }\) are 0 and \(\frac{\overline{z_{1}}+\overline{z_{2}}}{2\overline{z_{1}z_{2}}}\). Then, under the assumption \(\left z_{1}+z_{2}\right >2\left z_{1}z_{2}\right \), the second derivative \(P^{\prime \prime }\) has a zero in \(z>1\), which is a contradiction.$$\begin{aligned} \left 3\left( \overline{z_{1}}+\overline{z_{2}}\right) u^{2}\right =3\left z_{1}+z_{2}\right >4\left z_{1}z_{2}\right +\left z_{1}+z_{2}\right \ge \left 4 \overline{z_{1}z_{2}}u^{3}+\left( z_{1}+z_{2}\right) \right \end{aligned}$$
 c)
Assume that P has four simple unimodular zeros. Then, \(P^{\prime }\) has all its zeros in the closed unit disk. Given a selfinversive polynomial f, it is proved in [5, Lem.] that each unimodular zero of the derivative \(f^{\prime }\) is also a zero of f. If \(P^{\prime }\) has a unimodular zero a, then \(P\left( a\right) =0\); therefore, a is a zero of P of multiplicity at least 2, a contradiction. It follows that \(P^{\prime }\) has all its zeros in the unit disk. By Gauss–Lucas theorem, the second derivative \(P^{\prime \prime }\) also has all its zeros in the unit disk; therefore, \(\left z_{1}+z_{2}\right <2\left z_{1}z_{2}\right \).
 d)Now suppose that P has exactly two simple unimodular zeros, a and b. Let c and \(\frac{1}{\overline{c}}\) the other zeros of P, with \(\left c\right <1\). Then \(P\left( u\right) =\overline{z_{1}} \overline{z_{2}}\left( ua\right) \left( ub\right) \left( uc\right) \left( u\frac{1}{\overline{c}}\right) \). The coefficient of \(u^{2}\) in \(P\left( u\right) \) vanishes; therefore,and \(a+b=\frac{ab}{c+\frac{1}{\overline{c}}}\frac{c}{\left c\right ^{2}+1}\). Because \(\left \frac{ab}{c+\frac{1}{\overline{c}} }\right =\frac{1}{\left c+\frac{1}{\overline{c}}\right }< \frac{1}{2}\) and \(\frac{\left c\right }{\left c\right ^{2}+1}<\frac{1}{2}\), we get \(\left a+b\right <1\). Considering the coefficient of \(u^{3}\) in \(P\left( u\right) \) we obtain \(\frac{\overline{ z_{1}}+\overline{z_{2}}}{\overline{z_{1}z_{2}}}=a+b+c+\frac{1}{\overline{c}}\) . Then \(\frac{\left z_{1}+z_{2}\right }{\left z_{1}z_{2}\right }\ge \left \left c+\frac{1}{\overline{c}} \right \left a+b\right \right >1\).$$\begin{aligned} ab+\frac{c}{\overline{c}}+\left( a+b\right) \left( c+\frac{1}{\overline{c}} \right) =0, \end{aligned}$$
Example 4.6
Let \(z_{1}=\left( 1+t\right) e^{i\alpha }\) and \(z_{2}=(1+t)e^{i\left( \alpha +t\right) }\), where \(t>0\) and \(\alpha \in (\pi ,\pi ]\). By Corollary 3.9, the Eq. (1.3) has four simple unimodular roots in this case. On the other hand, \(\frac{\left z_{1}+z_{2}\right }{\left z_{1}z_{2}\right } =\left( 1+t\right) \left( 1+e^{it}\right) \rightarrow 2\) as \(t\rightarrow 0\), therefore the constant 2 in Proposition 4.5 c) cannot be replaced by a smaller constant.
We give a direct proof for the following consequence of Proposition 4.5.
Corollary 4.7
If \(P(u)=\overline{z_{1}}\overline{z_{2}}u^{4}(\overline{z_{1}} +\overline{z_{2}})u^{3}+(z_{1}+z_{2})uz_{1}z_{2}\) has one double zero and two simple zeros on the unit circle, then \(\left z_{1}z_{2}\right \le \left z_{1}+z_{2}\right \le 2\left z_{1}z_{2}\right \).
Proof
Assume that P has one double unimodular zero a and two simple unimodular zeros b, c. Then \(P\left( u\right) =\overline{z_{1}}\overline{z_{2}}\left( za\right) ^{2}\left( zb\right) \left( zc\right) \).
Notes
Acknowledgements
Open access funding provided by University of Turku (UTU) including Turku University Central Hospital. This research was begun during the RomanianFinnish Seminar in Bucharest, Romania, June 20–24, 2016, where the authors P.H., M.M., and M.V. met. During a workshop at the Tohoku University, Sendai, Japan, in August 2016 organized by Prof. T. Sugawa, M.F., P.H., and M.V. met and had several discussions about the topic of this paper. P.H. and M.V. are indebted to Prof. Sugawa for his kind and hospitable arrangements during their visit. This work was partially supported by JSPS KAKENHI Grant Number 15K04943. The second author was supported by University of Turku Foundation and CIMO. The authors are indebted to Prof. G.D. Anderson for a number of remarks on this paper and to the referee for several useful remarks and comments.
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