27. Hypocycloid of n + 1 cusps harmonic function

ISSN: 1821-1291, URL: http://www.bmathaa.org Volume 3 Issue 3(2011), Pages 239-246.

HYPOCYCLOID OF n+ 1 CUSPS HARMONIC FUNCTION

(COMMUNICATED BY INDRAJIT LAHIRI)

TOSHIO HAYAMI AND SHIGEYOSHI OWA

Abstract. For a harmonic function h(z) = f(z) +g(z) in the open unit diskUwith holomorphic functionsf(z) andg(z) satisfyingg′_{(z) =z}n₋₁

f′_(z)

(n= 2,3,4,· · ·), a sufficient condition onf(z) forh(z) to be univalent inU

and the image ofUbyh(z) to be a hypocycloid ofn+ 1 cusps are discussed.

1. Introduction

For holomorphic functions f(z) and g(z) in a simply connected domain D_{, a}

complex-valued harmonic functionh(z) is given byh(z) =f(z) +g(z). The theory and applications of harmonic mappings are discussed by Duren [1]. Mocanu [3] has shown the following result for the univalence of harmonic functions.

Theorem 1.1. Let f(z)andg(z)be holomorphic functions in a domain D_{. If the}

function f(z)is convex and|g′_(z)|<|f′_{(z)|for z∈}D_{, then the harmonic function}

h(z) =f(z) +g(z)is univalent and sense preserving in D_.

In fact, considering the harmonic function

h(z) =f(z) +g(z) =z+ 1 nz

n

(n= 2,3,4,· · ·)

for allz∈U_={z∈C_:|z|<1}, it is clear thatf(z) =zis convex,|g′_(z)|<|f′_(z)| (z∈U_{) andh(z) is univalent and sense preserving in}U_{. For this harmonic function}

h(z) and

Dh(z) =z∂h(z) ∂z −z

∂h(z) ∂z , it follows that

Re

Dh(z) h(z)

= Re

reit_−rn_e−int

reit₊rn

ne−int

(z=reit)

≧ n(1−r n−1₎

n+rn₋1 > 0

2000Mathematics Subject Classification. 30C45.

Key words and phrases. Harmonic function; univalent function; hypocycloid ofn+ 1 cusps. c

2011 Universiteti i Prishtin¨es, Prishtin¨e, Kosov¨e. Submitted Jul 29, 2010. Published August 8, 2011.

which shows that h(z) is also starlike in U_{(see, [2]). Furthermore, h(z) maps} U

onto the region inside a hypocycloid ofn+ 1 cusps (for detail, [1, p. 115]).

This work is motivated by the following theorem due to Mocanu [4].

Theorem 1.2. Let f(z)be holomorphic in the closed unit disk U_{, withf}′_{(z)6= 0}

for z∈U_{and let}

F(t) = 3t+ 2 arg f′_(eit₎

(−π≦t≦π).

If for eachk∈K={0,±1,±2} the equation

F(t) = 2kπ

has at most a single root in [−π, π] and for allk ∈K there exist three such roots in [−π, π], then the harmonic functionh(z) =f(z) +g(z), with g′_{(z) =zf}′_(z)is

univalent in U_{, sense preserving and the image of}U _{by h(z) is a ”three-cornered}

hat” domain.

We obtain an extension result of the above theorem for the following generalized class of harmonic functionsh(z) inU_{of the form}

h(z) =f(z) +g(z)

wheref(z) andg(z) are holomorphic functions inU_{and satisfyg}′_{(z) =z}n₋1_f′_(z) (n= 2,3,4,· · ·). This shows that the harmonic function h(z) is well defined if a holomorphic functionf(z) inU_{is given.}

2. Main result

Our first result is contained in

Theorem 2.1. Let f(z)be holomorphic in the closed unit disk U_{, withf}′_{(z)6= 0}

for z∈U_{and let}

F(t) = (n+ 1)t+ 2 arg(f′_(eit_{)) (−π≦t < π), n= 2,3,4,· · ·. (2.1)} If for each k∈K=

0,±1,±2,· · ·,±n+3

2 where[ ]denotes the Gauss symbol,

the equation

F(t) = 2kπ (2.2)

has at most a single root in [−π, π) and for anyk∈K there exist exactly(n+ 1)

such roots in [−π, π), then the harmonic function

h(z) =f(z) +g(z)

with g′_{(z) = z}n₋1_f′_{(z)is univalent in} U_{, sense preserving and the image of} U _by

h(z)is a hypocycloid ofn+ 1cusps.

Proof. Let us define the functionw(t) by

w(t) =h(eit_{) =}

f(eit_{) +g(e}it_{) (z=e}it_).

Supposing that

w′_{(t) =i(zf}′_(z)−zg′_{(z)) =i(zf}′_(z)−zn

then we need the equation

zn+1f ′_(z)

f′_(z) = 1.

Therefore, it follows that

zn+1f ′_(z)

f′_(z)

= 1 and arg zn+1f ′_(z)

f′_(z)

!

= 2kπ

that is, that

(n+ 1)t+ 2 arg(f′_(eit_{)) = 2kπ (−π≦t < π)}

forz =eit _{which gives us the equation (2.2). By the assumption of the theorem,}

there exist (n+ 1) distinct roots on the unit circle|z|= 1 and they divide the unit circle onto (n+ 1) arcs.

Sinceg′′_{(z) = (n−1)z}n₋2_f′_{(z) +z}n₋1_f′′_{(z), we have}

w′′_{(t) = −zf}′_{(z) +z}2_f′′_{(z) +zg}′_{(z) +z}2_g′′_(z)

= −zf′_{(z) +z}2

f′′_{(z) +nz}n

f′_{(z) +z}n+1_f′′_(z)

and therefore, we obtain that

w′′_(t)w′_{(t) =−zf}′_{(z) +z}2

f′′_{(z) +nz}n

f′_{(z) +z}n+1_f′′_(z)(−i)zf′_(z)−zn_f′_(z)

=i−(n−1)|f′_(z)|2_+zf′(z)f′′_(z)−zf′_(z)f′′_(z)−zn+1_f′_(z)2_+zn+1_f′_(z)2

+(n−1)zn+1_f′(z)2_−zn+2_f′_(z)f′′_{(z) +z}n+2_f′_(z)f′′_(z)

=i(n−1)zn+1f′_(z)2_{− |f}′_(z)|2

+2iImzf′_(z)f′′_(z)−zn+1_f′_(z)2_−zn+2_f′_(z)f′′_(z).

This implies that

Imw′′_(t)w′_{(t)= (n−1)|f}′_(z)|2

"

Re zn+1

f′_(z) |f′_(z)|

2

−1

!#

≦0.

Thus, we derive

Im

w′′_(t) w′_(t)

= 1

|w′_(t)|2Im

w′′_(t)w′_(t)≦0.

This shows that the image byw(t) is concave. By the help of a simple geometrical observation, we know that the image of the unit circle, as a union of the (n+ 1) concave arcs, is a simple curve. Namely, h(z) is univalent and the domainh(U_{) is}

3. Some illustrative examples and image domains

In this section, we enumerate several illustrative examples and the image domains for the harmonic functions h(z) =f(z) +g(z) satisfying the condition of Theorem 2.1.

Example 3.1. Let f(z) =z. Then, we immediately obtain

h(z) =f(z) +g(z) =z+ 1 nz

n

and the equation(2.2)becomes

(n+ 1)t= 2kπ

k= 0,±1,±2,· · ·,±

n+ 3 2

which has at most a single root in [−π, π) and for any k, just (n+ 1) roots in

[−π, π). Hence, h(z) is univalent in U _{and h(}U_{) is a hypocycloid of n+ 1 cusps.}

For example, settingn= 4, the following hypocycloid of five cusps as the image of

U_{by the harmonic function}

h(z) =z+1 4z

4

is obtained.

The image ofU_{byh(z) =}z+1 4z

4_.

Remark. The inequalityF′_{(t)≧0, withF(t) defined by (2.1), is equivalent to}

Re

1 + zf′′(z) f′_(z)

≧−n−1

2 (z=e

Noting the above, we derive

Example 3.2. Let f(z) =z+ p mz

m _{(m = 2,3,4,· · ·). Then, the equation (3.1)} becomes

Re

1 + zf′′(z) f′_(z)

=m−(m−1)Re

₁

1 +pzm−1

≧−n−1 2

and the function F(t) given by(2.1)satisfies

F(−π) =−(n+ 1)π and F(π) = (n+ 1)π.

Therefore, if we take − n+ 1

n+ 2m−1 ≦ p ≦

n+ 1

n+ 2m−1, then the conditions of

Theorem 2.1 are satisfied, so that the harmonic function

h(z) =z+ p mz

m₊ 1

nz

n₊ p

n+m−1z

n+m₋1

is univalent inU_andh(U_{)is a hypocycloid ofn+1cusps. In particular, considering}

h(z)withn= 7,m= 2andp= 8

15, the following hypocycloid of eight cusps as the

image ofU_{by the harmonic function}

h(z) =z+ 4 15z

2₊1

7z

7₊ 1

15z

8

is obtained.

The image ofU_{byh(z) =z+} 4

15z

2₊1

7z

7₊ 1

15z

4. A problem for harmonic functions

A problem related to the elementary transform of harmonic functions is the following.

Problem 4.1. For each holomorphic functionf(z) in certain domain withf(0) = f′_{(0)−1 = 0, can we find the largest domain}D_{c, such that the harmonic function}

hc(z) =fc(z) +gc(z), wherefc(z) =

1

cf(cz), with g ′

c(z) =zn−1fc′(z), is univalent

for allc∈D_c?

Letc=reiθ _and

F(t, r, θ) = (n+ 1)t+ 2 argf′_(rei(t+θ)₎

forz=eit_{(−π≦t < π). Then, the boundary of the domainD}

c is obtained by the

elimination oftfrom the system

        

F(t, r, θ) = 2kπ

k= 0,±1,±2,· · ·,±

n+ 3 2

∂F(t, r, θ) ∂t = 0. where [ ] is the Gauss symbol.

For example, we consider this problem for the case f(z) = ez_{−1. Then, we}

know that

fc(z) =

ecz₋₁

c (c=re

iθ₎

and the equation (2.2) implies that

(n+ 1)t+ 2rsin(t+θ) = 2kπ.

Differentiating the both sides with respect tot, we have that

(n+ 1) + 2rcos(t+θ) = 0

or

t+θ= cos−1

_{−(n+ 1)}

2r

=π−cos−1

n+ 1 2r

,

which means that

t=− 2r n+ 1sin

π−cos−1

n+ 1 2

+ 2kπ n+ 1. This gives us that

θ= 2r n+ 1sin

π−cos−1

n+ 1 2r

−cos−1

n+ 1 2r

+π− 2kπ

n+ 1. (4.1) Further, we have that

max

k,t r

2₌1

4

(

4

n+ 3 2

π+ (n+ 1)π

2

+ (n+ 1)2

)

and

min

k,t r

2 ₌(n+ 1)2

Letting Γ be the boundary of the domainD_{c, we have that the polar equations}

of Γ are given by

Γ =

 



x=rcosθ

y=rsinθ whererandθ satisfy the condition (4.1) with

 



k= 0,±1,±2,· · ·,±l (n= 2l)

k= 0,±1,±2,· · ·,±l,−(l+ 1) (n= 2l+ 1). Remark. Γ has a form of (n+ 1)-valently clover.

Example 4.1. For the case n= 3, the harmonic function

hc(z) =

ecz₋₁

c

−1 c

2 c2(1−e

cz_{) +}2

cze

cz_−z2_ecz

is univalent inU_{where cis in the domain} D_{c as follows:}

Acknowledgments. We express our sincere thanks to the referees for their valu-able suggestions and comments for improving this paper.

References

[1] P. L. Duren, Harmonic Mappings in the Plane, Cambridge University Press, Cambridge, 2004.

[2] P. T. Mocanu,Starlikeness and convexity for non-analytic functions in the unit disc, Math-ematica (Cluj)22_{(1980), 77–83.}

[4] P. T. Mocanu,Three-cornered hat harmonic functions, Complex Var. Elliptic Equ.54_(2009),

1079–1084.

Toshio Hayami

School of Science and Technology, Kwansei Gakuin University, Sanda, Hyogo 669-1337, Japan

E-mail address: ha ya [email protected]

Shigeyoshi Owa

Department of Mathematics, Kinki University, Higashi-Osaka, Osaka 577-8502, Japan