The Riemann hypothesis and holomorphic index
in complex dynamics
Tomoki KawahiraTokyo Institute of Technology
July 2, 2016

Abstract
We present an interpretation of the Riemann hypothesis in terms of complex
and topological dynamics. For example, the Riemann hypothesis is true and all
zeros of the Riemann zeta function are simple if and only if a meromorphic function that is explicitly given in this note has no attracting fixed point. To obtainthis, we use the holomorphic index (residue fixed point index) that characterizes
the local properties of the fixed points in complex dynamics.

1

The Riemann zeta function

For s ∈ C, the series

11+ s + ···s2
3
converges if Re s > 1 and ζ(s) is analytic on the half-plane {s ∈ C : Re s > 1}. It
is also analytically continued to a holomorphic function on C − {1}, and s = 1 is a
simple pole. Hence we regard ζ(s) as a meromorphic function (a holomorphic map)
b = C ∪ {∞}. This is the Riemann zeta function ([T, p.13]).
ζ:C→C
It is known that ζ(s) = 0 when s = −2, −4, −6, . . .. These zeros are called trivial
zeros of the Riemann zeta function ([T, p.30]). We say the other zeros are non-trivial.
The Riemann hypothesis, the most important conjecture on the Riemann zeta function, concerns the alignment of the non-trivial zeros of ζ:ζ(s) = 1 +The Riemann hypothesis.
Re s = 1/2}.

All non-trivial zeros lie on the vertical line {s ∈ C :Research partially supported by JSPS KAKENHI Grant.

1

The line {s ∈ C : Re s = 1/2} is called the critical line. It has been numericallyverified that the first 1013 non-trivial zeros above the real axis lie on the critical line
[G].
It has also been conjectured that every zero of ζ is simple. We refer to this conjecture
as the simplicity hypothesis. (See [RS, p176] or [Mu, p177] for example. For somerelated results and observations, see [T, §10.29, §14.34, §14.36]. )The aim of this note is to reformulate the Riemann hypothesis in terms of complex
and topological dynamical systems. More precisely, we translate the locations of the
non-trivial zeros into dynamical properties of the fixed points of a meromorphic functionof the formg(z)νg (z) = z − ′ ,z g (z)where g is a meromorphic function on C that shares (non-trivial) zeros with ζ. For
example, we set g = ζ or the Riemann xi function(z )1
ξ(z) := z(1 − z)π −z/2 Γ
ζ(z).
2
2
The function νg (z) is carefully chosen such that
• If g(α) = 0 then νg (α) = α; and• the holomorphic index (or the residue fixed point index; see §2) of νg at α is αitself when α is a simple zero of g(z).
See §3 for more details.
b we say a fixed point α of g is
For a given meromorphic function g : C → C,
attracting if |g ′ (α)| < 1, indiﬀerent if |g ′ (α)| = 1, or repelling if |g ′ (α)| > 1. We show
the following translations of the Riemann hypothesis (plus the simplicity hypothesis):
Theorem 1 The following conditions are equivalent:(a) The Riemann hypothesis is true and every non-trivial zero of ζ is simple.
(b) Every non-trivial zero of ζ is an indiﬀerent fixed point of the meromorphic function
ζ(z)νζ (z) := z − ′.
z ζ (z)(c) The meromorphic function νζ above has no attracting fixed point.
(d) There is no topological disk D with νζ (D) ⊂ D.
b in contrast to theNote that (d) is a topological property of the map νζ : C → C
analytic (or geometric) nature of (a). We will present the proof and some variants ofthis theorem in §4.

2

§5 is devoted to some numerical observations and questions on the linearization
problem. §6 is an appendix wherein we apply Newton’s method to the Riemann zetafunction. We also give a “semi-topological” criterion for the Riemann hypothesis interms of the Newton map (Theorem 14).
Remarks.
• The following well-known facts are used in this note (see p.13 and p.45 of [T]):– The functional equationπsΓ(1 − s)ζ(1 − s)(∗)
2
implies that if α is a non-trivial zero of ζ(s), then so is 1 − α and they havethe same order.
ζ(s) = 2s π s−1 sin– Every non-trivial zero is located in the critical strip
S := {s ∈ C : 0 < Re s < 1}.
Hence the non-trivial zeros are symmetrically arrayed with respect to s = 1/2 in
S. In fact, there is another symmetry (which we do not use in this note) withrespect to the real axis, which comes from the relation ζ(s) = ζ(s).) By theseproperties, we will primarily consider the zeros that lie on the upper half of thecritical strip S.
• We can apply the method of this note to other zeta functions without extra eﬀort.
(See [I, §1.8] for examples of zeta functions.) However, we need a functional
equation, such as (∗), to obtain a result similar to Theorem 1. For example, it isvalid for the Dirichlet L-functions.
• We used Mathematica 10.0 for all the numerical calculations.
Acknowledgments.
The author would like to thank Masatoshi Suzuki and thereferee for helpful, interesting, and stimulating comments.

2

Fixed points and holomorphic indicesMultiplier. Let g be a holomorphic function on a domain Ω ⊂ C. We say that α ∈ Ωis a fixed point of g with multiplier λ ∈ C if g(α) = α and g ′ (α) = λ. The multiplier
λ is the primary factor that determines the local dynamics near α. We say the fixed
point α is attracting, indiﬀerent, or repelling according to whether |λ| < 1, |λ| = 1, or|λ| > 1.
Topological characterization. Attracting and repelling fixed points of holomorphic
mappings have purely topological characterizations (cf. Milnor [Mi, §8]):3

Proposition 2 (Topological characterization of fixed points) Let g be a holomorphic function on a domain Ω ⊂ C. The function g has an attracting (respectively,
repelling) fixed point if and only if there exists a topological disk D ⊂ Ω such thatg(D) ⊂ D (respectively, g|D is injective and D ⊂ g(D) ⊂ Ω).
The condition that g is holomorphic is essential. For example, if we only assume thatg is C ∞ , any kind of fixed point may exist in a disk D with g(D) ⊂ D.

Figure 1: Topological disks that contain an attracting or a repelling fixed point.
Proof. If a topological disk D in Ω satisfies g(D) ⊂ D, we may observe the map
g : D → g(D) ⊂ D via the Riemann map, and it is suﬃcient to consider the case of
D = D, the unit disk. By the Schwarz-Pick theorem (see [Ah2, §1]) the map is strictly
contracting with respect to the distance d(z, w) := |z − w|/|1 − zw| (or the hyperbolic
distance) on D, and it must have an attracting fixed point.
The converse is easy, and the repelling case is analogous.
■
Holomorphic index. Let α be a fixed point of a holomorphic function g : Ω → C.
We define the holomorphic index (or residue fixed point index) of α by∫11ι(g, α) :=dz,
2πi C z − g(z)where C is a small circle around α in the counterclockwise direction. The holomorphic
index is mostly determined by the multiplier:
Proposition 3 If the multiplier λ := g ′ (α) is not 1, then we have ι(g, α) =

1
.
1−λ

See [Mi, Lem.12.2] for the proof.
Remark. Any complex number K can be the holomorphic index of a fixed point ofmultiplier 1. For example, the polynomial g(z) = z − z 2 + Kz 3 has a fixed point atzero with g ′ (0) = 1 and ι(g, 0) = K.
1
Since the Möbius transformation λ 7→
= ι sends the unit disk to the half1−λplane {ι ∈ C : Re ι > 1/2}, fixed points are classified as follows:4Proposition 4 (Classification by index) Suppose that the multiplier λ = g ′ (α) is
not 1. Then α is attracting, indiﬀerent, or repelling if and only if the holomorphic
index ι = ι(g, α) satisfies Re ι > 1/2, = 1/2, or < 1/2 respectively.

Figure 2: Relation between multipliers and holomorphic indices

Note that the “critical line” in the ι-plane corresponds to indiﬀerent fixed points whose
multipliers are not 1.

3

The nu function

b be a non-constant meromorphic function. (We regard such a meroLet g : C → C
morphic function as a holomorphic map onto the Riemann sphere.) We define the nu
b of g byfunction νg : C → Cg(z)νg (z) := z − ′ .
zg (z)
This( is also)ma non-constant meromorphic function on C unless g(z) = gm (z) :=1
C 1−
for some C ∈ C − {0} and m ∈ Z − {0}.
mzUsing the Taylor and Laurent series, it is not diﬃcult to check:
Proposition 5 (Fixed points of νg ) Suppose that α ̸= 0. Then, α is a fixed pointof νg if and only if α is a zero or a pole of g. Moreover,• if α is a zero of g of order m ≥ 1, then νg′ (α) = 1 −1and ι(νg , α) = mα;mα• if α is a pole of g of order m ≥ 1, then νg′ (α) = 1 +1and ι(νg , α) = −mα.
mα

If 0 is a zero or a pole of g of order m ≥ 1, then νg (0) = 1/m or −1/m respectively.
In particular, 0 is not a fixed point of νg .
An immediate corollary is the following:5

Corollary 6 The function νg has no fixed point with multiplier 1.
Hence we can always apply Proposition 4 to the fixed points of νg .
Remark. There is another way to calculate the holomorphic index. (This is actually
how the author found the function νg .) Suppose that α is a zero of g of order m.
A variant of the argument principle ([Ah1, p.153]) yields that for any holomorphic
function ϕ defined near α, we have∫1g ′ (z)ϕ(z)dz = mϕ(α),
2πi C
g(z)
where C is a small circle around α in the counterclockwise direction. Set ϕ(z) := z.
Then the equality above is equivalent to∫1
1
dz = mα.
2πi C z − νg (z)The same argument is also valid when α is a pole.
(z + 1)(z − 1/2)2. By Propositionsz 3 (z − 1)
4 and 5, the zeros −1 and 1/2 of g are repelling and attracting fixed points of νgrespectively. The pole 1 is a repelling fixed point, however, the other pole 0 is not a
fixed point of νg .
Example.

4

Consider a rational function g(z) =The Riemann hypothesis

Let us consider the nu function νζ of the Riemann zeta function, and prove Theorem
1.
Trivial zeros and the pole. It is known that the trivial zeros α = −2, −4, · · · of
ζ of ζ are all simple [T, p30]. By Propositions 4 and 5, their holomorphic indices are
α themselves and they are all repelling fixed points of νζ . Similarly, the unique polez = 1 of ζ is simple and it is a repelling fixed point of νζ . Hence we have the following:
Proposition 7 Every fixed point of νζ oﬀ the critical strip S is repelling.
Non-trivial zeros. Let α be a non-trivial zero of order m ≥ 1 in the critical strip
S. By Proposition 5, α is a fixed point of νζ with multiplier λ := 1 − 1/(mα) ̸= 1, andits holomorphic index is ι := mα.
If the Riemann hypothesis holds, then Re ι = mRe α = m/2. Thus Re ι = 1/2 if
m = 1 and Re ι ≥ 1 if m ≥ 2. By Proposition 4, we have the following:
Proposition 8 Under the Riemann hypothesis, any fixed point α of νζ in S is a zeroof ζ that lies on the critical line. Moreover,6• If α is a simple zero of ζ, then α is an indiﬀerent fixed point of νζ with multiplier
̸= 1.
• If α is a multiple zero of ζ, then α is an attracting fixed point of νζ .
In particular, if the simplicity hypothesis also holds, then all non-trivial zeros of ζ are
indiﬀerent fixed points of νζ .
Hence, (a) implies (b) in Theorem 1. Now we show the converse.
Proposition 9 If the fixed points of νζ in the critical strip S are all indiﬀerent, thenboth the Riemann hypothesis and the simplicity hypothesis are true.
Proof.
Let α be an indiﬀerent fixed point of νζ in the critical strip S. Since ζ
has no pole in S, α is a zero of ζ of some order m ≥ 1, and the holomorphic index isι(νζ , α) = mα by Proposition 5. By the functional equation (∗), the point 1−α is also azero of ζ of order m contained in S. and the holomorphic index is ι(νζ , 1−α) = m(1−α).
By assumption, both α and 1 − α are indiﬀerent fixed points of νζ . Hence by
Proposition 4, the real parts of ι(νζ , α) = mα and ι(νζ , 1 − α) = m(1 − α) are both
1/2. This happens only if m = 1 and Re α = 1/2.
■
Let us finish the proof of Theorem 1:
Proof of Theorem 1. The equivalence of (a) and (b) is shown by Proposition 8 and
Proposition 9 above. Condition (b) implies (c) by Proposition 7.
Suppose that (c) holds. Then any fixed point α of νζ in the critical strip S isrepelling or indiﬀerent, and it is also a zero of ζ of some order m by Proposition 5. Hencethe holomorphic index ι(νζ , α) = mα satisfies Re mα ∈ (0, 1/2]. By the functional
equation (∗), the same holds for 1 − α ∈ S and the index ι(νζ , 1 − α) = m(1 − α)satisfies Re m(1 − α) ∈ (0, 1/2]. This implies that m = 1 and Re α = 1/2, and we
conclude that (c) implies (a).
The equivalence of (c) and (d) comes from Proposition 2.
■
A more topological version. Note again that condition (d) of Theorem 1 is a purely
topological condition for the function νζ . This means that the condition is preserved
for any continuous function that are topologically conjugate to νζ . More precisely, we
have:
Theorem 10 Conditions (a) - (d) of Theorem 1 are equivalent to:
b →C
b with h(∞) = ∞, the continuous function(e) For any homeomorphism h : Cνζ,h := h ◦ νζ ◦ h−1 has no topological disk D with νζ,h (D) ⊂ D.
The proof is a routine. We may regard the dynamics of the map νζ,h as a topologicaldeformation of the original dynamics of νζ (Figure 3). Note that the critical line may
7Figure 3: Topological deformation of the dynamics of νζ . By Theorem 10, we can still
state the Riemann hypothesis and the simplicity hypothesis in the deformed dynamics.
not be a “line” any more when it is mapped by a homeomorphism. (That may evenhave a positive area!)
Variants. Next we consider νg for g = ξ, the Riemann xi function presented in thefirst section. It is known that ξ : C → C is an entire function whose zeros are exactly
the non-trivial zeros of the Riemann zeta function. Moreover, we have the functional
equation ξ(z) = ξ(1 − z). See [T, p.16] for example.
Following the same argument as in the proof of Theorem 1, we have its variant interms of ξ:
Theorem 11 The following conditions are equivalent to conditions (a) - (e) statedabove :(b’) All the fixed points of the meromorphic function νξ (z) := z −ξ(z)are indiﬀerz ξ ′ (z)ent.
(c’) The meromorphic function νξ above has no attracting fixed point.
(d’) There is no topological disk D with νξ (D) ⊂ D.
The proof is simpler because ξ has neither trivial zeros nor poles.
Remark. To have an entire function it is enough to consider the function
η(z) := (z − 1)ζ(z).
In this case, η(z) and ζ(z) share all the zeros and one can state a theorem similar to
Theorem 1. (We will apply Newton’s method to this η in §6.)8

5Linearization problem

Under the Riemann hypothesis and the simplicity hypothesis, each non-trivial zero ofζ is of the form α = 1/2 + γi (γ ∈ R), and α is an indiﬀerent fixed point of νζ withνζ′ (α) = 1 − 1/α (Theorem 1 and Proposition 5). In this section, we present somebrief observations on these fixed points. (This section is addressed to readers familiar
with complex dynamics. See [Mi, §10, §11] and references therein for more details on
this subject.)
Non-trivial zeros and rotation number.
fact:

We first point out the following easy

Proposition 12 Under the Riemann and simplicity hypotheses, non-trivial zero α =1/2 + γi (γ ∈ R) is an indiﬀerent fixed point of νζ with multiplier e2πiθ (θ ∈ R), where1
1
1, or equivalently, θ = arctan .
the values γ and θ are related by γ =2 tan πθπ2γSuch a θ is called the rotation number of the indiﬀerent fixed point α. We say α islinearizable if the local dynamics near α is conjugate to the rigid rotation w 7→ e2πiθ w.
A Siegel disk is the maximal domain where such a conjugacy exists. Now it is natural
to ask:Linearization problem. Can θ be a rational number? Is α linearizable?
That is, can νζ have a Siegel disk?
It is known that for Lebesgue almost every θ ∈ [0, 1) − Q, the irrationally indiﬀerent
fixed points with multiplier e2πiθ are all linearizable. However, there is a dense subsetof rotation numbers (consisting of some irrationals and all rationals in [0, 1)) thatalways gives non-linearizable indiﬀerent fixed points. The suﬃcient conditions for(non)-linearizability are often described in terms of the continued fraction expansionof the rotation number θ ∈ [0, 1), which we denote by1θ==: [a1 , a2 , a3 , . . .]

1

a1 +
a2 +

1
a3 + · · ·where a1 , a2 , a3 , . . . are all in N.
Numerical observation. Table 1 lists the continued fractions up to the 50th term
for some zeros calculated by built-in functions ZetaZero and ContinuedFraction of
Mathematica 10.0. (Here, 1/2 + γn i (γn > 0) is the nth non-trivial zero of ζ from below
and γn = 1/(2 tan πθn ). Note that θn → 0 as γn → ∞.)Figure 4 shows some orbits near the first four non-trivial zeros of ζ in the dynamicsof νζ and νξ .
9

Figure 4: First four Siegel disks (?) of νζ (left) and νξ (right).

10

n
1

Im γn
θn
14.1347 0.0112552

2

21.0220 0.00756943

325.0109 0.00636259

430.4249 0.00523061

1049.7738 0.00319746

102
103
104

[a1 , a2 , · · · , a25 , a26 , · · · , a50 ][88,1,5,1,1,2,2,5,2,15,2,4,2,4,1,9,1,1,5,2,10,1,1,5,1,
2,7,100,9,2,2,3,2,5,1,8,179,23,1,1,35,1,3,1,2,8,7,34,4,1][132,9,14,1,1,1,2,1,52,1,9,3,4,1,1,1,1,2,2,3,2,1,10,1,1,
1,9,1,1,6,5,1,5,3,1,5,2,6,1,135,1,1,5,1,2,3,2,9,1,3][157,5,1,12,3,1,1,1,1,2,11,1,29,5,1,4,1,1,3,5,14,1,3,1,2,
3,6,14,4,1,41,4,1,1,7,4,1,3,21,1,4,3,1,2,2,1,16,1,2,3][191,5,2,15,3,2,2,7,2,1,2,46,2,1,1,6,1,4,2,2,4,1,6,1,1,
2,5,1,8,1,2,2,5,1,4,39,3,19,5,2,9,1,1876,2,12,1,4,4,1,6][312,1,2,1,48,1,1,4,1,3,1,2,5,1,21,1,4,1,3,2,1,1,8,1,6,
9,1,2,3,2,1,2,1,1,1,4,1,1,1,5,2,1,3,126,1,24,3,2,29,5]
236.524 0.00067289[1486, 7,1,4,1,2,1,53,2,8,1,4,6,3,1,3,1,13,3,1,7,2,18,3,1,
1,1,1,1,2,1,210,3,4,1,3,1,2,2,8,1,1,7,2,2,2,1,12,2,2]1419.42 0.000112127 [8918,2,48,220,1,15,1,1,6,1,1,6,2,4,149,1,15,3,6,1,4,1,1,64,11,
1,1,13,11,2,4,2,3,2,4,1,2,1,1,2,3,46,1,11,3,1,18,1,6,4]
9877.78 0.0000161124 [62063,1,15,2,1,2,1,8,1,2,6,2,1,2,2,1,5,1,1,186,3,5,4,14,9,
1,12,1,12,1,6,4,22,1,3,15,1,2,1,14,1,2,2,12,3,1,1,1,45,2]
Table 1: Continued fraction expansions for some zeros of ζ.

6

Appendix: Newton’s method

There are many root finding algorithms, but the most popular one is Newton’s method.
Let us apply it to the Riemann zeta function and its variants. The aim of this appendix is to describe it in terms of the holomorphic index. (See [S] for an intriguinginvestigation on how we can eﬃciently detect all the zeros of ζ by Newton’s method. )b be a non-constant meromorphic
Newton maps and fixed points. Let g : C → C
b byfunction. We define its Newton map Ng : C → C
Ng (z) = z −g(z),g ′ (z)which is again meromorphic. (See [B, §6].) Here is a version of Proposition 5 (andCorollary 6) for Ng :Proposition 13 (Fixed points of Ng ) A given point α ∈ C is a fixed point of Ng if
and only if α is a zero or a pole of g. Moreover,• if α is a zero of g of order m ≥ 1, then α is an attracting fixed point of multiplierNg′ (α) = 1 − 1/m and its index is ι(Ng , α) = m.
• if α is a pole of g of order m ≥ 1, then α is a repelling fixed point of multiplierNg′ (α) = 1 + 1/m and its index is ι(Ng , α) = −m.
11

In particular, Ng has no fixed point of multiplier 1.
The proof is similar to that of Proposition 5 and is left to the reader.
Newton’s method. The idea of Newton’s method is to use the attracting fixed pointsof Ng to detect the zeros of g. More precisely, by taking an initial value z0 suﬃciently{}
close to a zero α, the sequence Ngn (z0 ) n≥0 converges rapidly to the attracting fixed
point α.
Holomorphic index and the argument principle. For the fixed point α of Ng
its holomorphic index is
∫
∫ ′
1
1
1
g (z)
ι(Ng , α) =dz =dz
2πi C z − Ng (z)2πi C g(z)where C is a small circle around α in the counterclockwise direction. This is exactly
the argument principle applied to g.
The Riemann hypothesis. When we apply Newton’s method to the Riemann zetafunction, all zeros of ζ are converted to attracting fixed points of Nζ (z) = z−ζ(z)/ζ ′ (z).
By Proposition 2, we have:
Theorem 14 The Riemann hypothesis is true if and only if there is no topological disk
D contained in the strip S ′ = {z ∈ C : 1/2 < Re z < 1} satisfying Nζ (D) ⊂ D.
Some pictures. It is easier to draw pictures of the chaotic locus (the Julia set) of
Newton’s map Nζ than of νζ . After the list of references, we present some picturesof the chaotic loci for the Newton maps of ζ(z), η(z) := (z − 1)ζ(z) (this is an entire
function), ξ(z) and cosh(z) with comments.

References[Ah1] L.R. Ahlfors. Complex analysis (3rd Edition). McGraw-Hill, 1979.
[Ah2] L.R. Ahlfors. Conformal invariants. Chelsea, 1991.
[B] W. Bergweiler. Iteration of meromorphic functions. Bull. Amer. Math. Soc.,
26(1999), 151–188.
[G] X. Gourdon. The 1013 first zeros of the Riemann zeta function, and zeros computation at very large height.
http://numbers.computation.free.fr/
Constants/Miscellaneous/zetazeros1e13-1e24.pdf[I]

A.Ivić. The Riemann zeta-function. Dover, 1985.
12

[Mi] J. Milnor. Dynamics in one complex variable. (3rd edition). Ann. of Math. Studies
160, Princeton University Press, 2006.
[Mu] J. Mueller. Arithmetic equivalent of essential simplicity of zeta zeros. Trans.
Amer. Math. Soc. 275 (1983), no. 1, 175-183.
[MS] S. Mayer and D. Schleicher. Immediate and virtual basins of Newton’s method
for entire functions. Ann. Inst. Fourier 56 (2006), no. 2, 325-336.
[RS] M. Rubinstein and P. Sarnak, Chebyshev’s bias. Experimental Math. 3 (1994),
173-197.
[S] D. Schleicher. Newton’s method as a dynamical system: Eﬃcient root findingof polynomials and the Riemann ζ function. Fields Inst. Commun. 53 (2008),
213-224.
[T] E.C. Titchmarsh. The theory of the Riemann zeta function. (2nd edition). Oxford
University Press, 1986.

13

Figure 5: Left: The Julia set of Nζ in {Re z ∈ [−20, 10], Im z ∈ [−1, 39]}. (It issymmetric with respect to the real axis.) The orange dots are disks that are close to
the zeros of ζ. More bluish colors indicate the points that require more iteration to getclose to the zeros. Right: The same region in diﬀerent colors. Colors distinguish thezeros to converge.

14Figure 6: The Julia set of Nη in {Re z ∈ [−20, 20], Im z ∈ [−1, 39]}, drawn in the samecolors as in Figure 5. Probably because η is entire, the dynamics of Nη is simpler thanthat of Nζ . It is known that for any transcendental entire function g and its zeros,
their immediate basins (the connected components of the Fatou set of Ng that containthe zeros) are simply connected and unbounded ([MS]).

15

Figure 7: Details of the Julia set of Nη in {Re z ∈ [−2, 10], Im z ∈ [10, 60]} (“Heads ofChickens”).

16Figure 8: Details of the Julia set of Nξ in {Re z ∈ [0, 40], Im z ∈ [−1, 39]}. (It issymmetric with respect to the real axis and the critical line.) The dynamics seems
surprisingly simple. Compare with the case of the hyperbolic cosine in Figure 10.

Figure 9: Details of the Julia set of Nξ in {Re z ∈ [0.5, 6.5], Im z ∈ [0, 2.5]}.
17

Figure 10: The Julia set of the Newton map Ncosh of the hyperbolic cosine cosh z in
{Re z ∈ [0, 40], Im z ∈ [−1, 39]}. Does the Julia set of Ncosh have the same topology asthat of Nξ ? Do they belong to the same deformation space?

Figure 11: Details of the Julia set of the Newton map Ncosh in {Re z ∈ [0, 1.5], Im z ∈[0, 3.5]}.

18