The Bohr inequality, first introduced in 1914 by Harald Bohr in his seminal work [3] and subsequently improved independently by M. Riesz, I. Shur and F. Wiener, essentially states that if f(z)=∑ m=0∞amzm is an analytic function in the open unit disc D={z∈ℂ:|z|<1} and |f(z)|<1 for all z∈D, then

for all z∈D with |z|=r≤r0=1/3 and 1/3 is the largest possible value of r₀, called the Bohr radius. Inequalities of type (1.1) have become famous by the name Bohr inequalities and the problems of finding the largest possible values of r₀ in different setups are now called Bohr radius problems. For a glimpse of the ongoing current research in this area we refer the reader to some recent articles, e.g. [1, 2, 5, 8, 9] and references therein.

In 2010, Abu Muhanna [8] investigated some Bohr radius problems using the concept of subordination. For two analytic functions f and g in D, g is said to be subordinate to f (written, g≺f) if there exists a function ψ analytic in D with ψ(0)=0 and |ψ(z)|<1 such that g=f∘ψ. In particular, when f is univalent, then g≺f is equivalent to g(0)=f(0) and g(D)⊂f(D). We shall denote by S(f), the class of all functions g subordinate to a fixed function f. A class of analytic (harmonic) functions in the unit disc D is said to possess classical Bohr's phenomenon if an inequalty of the type (1.1) is satisfied in |z|<r0, for some r0,0<r0≤1. It is known (see [8]) that not all classes of functions have classical Bohr's phenomenon. So, Abu Muhanna [8] reformulated classical Bohr's phenomenon and proved the following result:

Theorem 1.1. If f(z)=∑ m=0∞amzm is a univalent function and g(z)=∑ m=0∞bmzm∈S(f), then

for all |z|=r≤r0=3−8=0.17157.., where d(f(0),∂f(D)) is the Euclidean distance between f(0) and ∂f(D), the boundary of f(D). The value of r₀ is sharp for f(z)=z/(1−z)2, the Koebe function. Further, if f is convex univalent in D, then r0=1/3.

In the recent years, a number of research articles (for example see [2, 6, 7]) are published and many hidden facts of this subject are brought into broad daylight. In particular, Bhowmik and Das [2] successfully extended the Bohr inequalities of type (1.2) for certain harmonic functions. A complex valued function f(z)=u(x,y)+iv(x,y) of z=x+iy∈D is said to be harmonic if both u(x,y) and v(x,y) are real harmonic in D. It is known that such an f can be uniquely represented as f=h+g¯, where h and g are analytic functions in D with f(0)=h(0). It immediately follows from this representation that f is locally univalent and sense preserving whenever its Jacobian J_f, defined by Jf(z):=|h′(z)|2−|g′(z)|2, satisfies Jf(z)>0 for all z∈D; or equivalently if h′≠0 in D and the (second complex) dilatation w_f of f, defined by wf(z)=g′(z)/h′(z), satisfies the condition |wf(z)|<1 in D. A harmonic function f=h+g¯ defined in D is said to be K-quasiconformal if its dilatation w_f satisfies |wf|≤k,k=(K−1)/(K+1)∈[0,1). In view of the work of Schaubroeck in [10], aforesaid definitions and notations for subordination of analytic functions can be extended to harmonic functions without any change. This lead Bhowmik and Das [2] to extend Theorem 1.1. as under:

Theorem 1.2. Let f(z)=h(z)+g(z)¯=∑ m=0∞amzm+∑ m=1∞bmzm¯ be a sense preserving K-quasiconformal harmonic mapping defined in D such that h is univalent in D, and let f1(z)=h1(z)+g1(z)¯=∑ m=0∞cmzm+∑ m=1∞dm zm¯∈S(f). Then

for |z|=r≤r0=(5K+1−8K(3K+1)/(K+1). This result is sharp for the function p(z)=z/(1−z)2+kz/(1−z)2¯, where k=(K−1)/(K+1). Moreover, if we take h to be convex univalent then the inequality in (1.3) holds for |z|=r≤r0=(K+1)/(5K+1). This result is again sharp for the function q(z)=z/(1−z)+kz/(1−z)¯.

In this article, our aim is to establish the Bohr's phenomenon and compute Bohr radius for some subclasses of univalent harmonic functions. We also propose to improvise Theorem 1.1. and 1.2. stated above.

We close this section by setting certain notations for subsequent use in this paper. We denote by S_H, the class of univalent harmonic functions f normalized by the conditions f(0)=0 and fz(0)=1. In addition, if fz¯(0)=0 also, then the class is denoted by SH0. Further, KH0 is the usual subclass of SH0 consisting of convex functions. A domain Ω is said to be convex in the direction θ,0≤θ<π, if the intersection of the straight line through the origin and the point eiθ in the complex plane is connected or empty. A function f mapping the open unit disc D onto such a domain is called convex in direction 𝜃.

We begin this section by stating following lemma which immediately follows from the work of Bhowmik and Das [1].

Lemma 2.1. Let f(z)=∑ m=0∞amzm and g(z)=∑ m=0∞bmzm be two analytic functions in D and g≺f. Then

for |z|=r≤1/3.

Using this lemma, we now improvise Theorem 1.1 by taking univalent analytic function in D as f(z)=z+∑ m=2∞amzm. Making use of well known De Brange's theorem: |am|≤m,m=2,3,..., and after some simple calculations, we easily get:

Theorem 2.2. If f(z)=z+∑ m=2∞amzm is a univalent analytic function in D and g(z)=∑ m=1∞bmzm∈S(f), then

for all |z|=r≤1/3.

In a similar manner, we restate Theorem 1.2. as under;

Theorem 2.3. Let f(z)=h(z)+g(z)¯=z+∑ m=2∞amzm+∑ m=1∞bmzm¯ be a sense preserving K-quasiconformal harmonic mapping in D, such that h is analytic univalent in D. Then

for |z|=r≤r0=(2K+1−K(3K+2))/(K+1) and it is sharp for p(z)=z/(1−z)2+kz/(1−z)2¯. If we take h to be convex univalent then the inequality in (2.2) holds for |z|=r≤r0=(K+1)/(3K+1) and it is sharp for p(z)=z/(1−z)+kz/(1−z)¯, where k=(K−1)/(K+1). Further, let f1(z)=h1(z)+g1(z)¯=∑ m=1∞cmzm+∑ m=1∞dm zm¯∈S(f). Then

for |z|=r≤r0=min(1/3,(2K+1−K(3K+2))/(K+1)). If we take h to be convex univalent then the inequality in (2.3) holds for |z|=r≤r0=min(1/3,(K+1)/(3K+1)).

In next theorem, we establish Bohr's phenomenon for univalent harmonic functions f=h+g¯∈SH whose dilatation g'/h' is suitably chosen.

Theorem 2.4. Let f(z)=h(z)+g(z)¯=z+∑ m=2∞amzm+∑ m=1∞bmzm¯ be a univalent and K-quasiconformal harmonic mapping in D, where h is analytic univalent in D and g′(z)/h′(z)=keiθzn,k=(K−1)/(K+1)∈(0,1),n∈ℕ,θ∈ℝ. Then

for |z|=r≤r0, where r₀ is the only root of the equation

in (0,1) and this r₀ is best possible one.

Letting k→1 (equivalently, K→∞) we obtain the following result.

Corollary 2.5. Let f(z)=h(z)+g(z)¯=z+∑ m=2∞amzm+∑ m=1∞bmzm¯ be a univalent harmonic mapping in D, where h is analytic univalent in D and g′(z)/h′(z)=eiθzn,n∈ℕ,θ∈ℝ. Then

for |z|=r≤r0, where r₀ is the only root in (0,1) of the equation ϕ(r)=0, where

This r₀ is the best possible one.

By plotting the graph of ϕ(r) w.r.t r for different values of n, we observe that there is only one root of ϕ(r) in (0,1) which is the Bohr radius for that value of n in the dilatation function. Figure 1 illustrates the case when n=3 and in the following table we have listed values of r₀ computed for n=1,2,3 and 4.

Table 1

n	r0
1	0.3485...
2	0.3121...
3	0.1794...
4	0.0959...

Figure 1. ϕ(r) w.r.t r for n=3.

We observe that if n→∞, then r0→0.

Lemma 2.1 and Theorem 2.4 together lead us to the following result for the subordination class S(f).

Corollary 2.6. Let f1(z)=h1(z)+g1(z)¯=∑ m=1∞cmzm+∑ m=1∞dm zm¯∈S(f) where f is as defined in Theorem 2.4. Then

for |z|=r≤r1=min(1/3,r0), where r₀ is same as obtained in Theorem 2.4.

Next theorem shows the existence of Bohr's phenomenon for f∈SH with dilatation wf=(a+z)/(1+az),a∈(−1,1).

Theorem 2.7. Let f(z)=h(z)+g(z)¯=z+∑ m=2∞amzm+∑ m=1∞bmzm¯ be a univalent harmonic mapping in D, where h is univalent in D and g′(z)/h′(z)=a+z1+az,a∈(−1,1) Then

for |z|=r≤r0, where r₀=0.2291... is a unique root lying in (0,1) of r3−3r2+5r−1=0.

Remark 2.8. We observe that if we take g′/h′=a−z1−az,a∈(−1,1), in Theorem 2.7, then we obtain the same value of r₀.

In the following theorem we establish Bohr's phenomenon for univalent harmonic functions convex in one direction.

Theorem 2.9. Let f(z)=h(z)+g(z)¯=z+∑ m=2∞amzm+∑ m=1∞bmzm¯ be a harmonic mapping in D, where h is analytic univalent in D and h(z)+eiθg(z) is convex univalent in D for some θ∈ℝ. Then

for |z|=r≤r0=0.2192....

We can drop the condition of univalency of h in Theorem 2.9 if we take b₁=0.

Let f(z)=h(z)+g(z)¯=z+∑ m=2∞amzm+∑ m=2∞bmzm¯∈SH0 be a harmonic mapping in D, where h(z)+eiθg(z) is convex univalent in D for some θ∈ℝ. Then

for |z|=r≤r0=0.3134..., where r₀ is a unique root in (0,1) of 4r3−9r2+12r−3=0. This result is sharp for Koebe function K(z)=z−1/2z2+1/6z3(1−z)3+1/2z2+1/6z3(1−z)3¯.

Our last theorem gives Bohr radius for convex univalent harmonic functions in SH0.

Theorem 2.11. Let f(z)=h(z)+g(z)¯=z+∑ m=2∞amzm+∑ m=2∞bmzm¯∈KH0,z∈D. Then

for |z|=r≤r0=(3−5)/2=0.3819.... This value of r₀ is sharp for L(z)=12z1−z+z(1−z)2+z1−z−z(1−z)2¯.

We begin this section by stating a lemma which is easy to prove.

Lemma 3.1. Let h(z)=∑ m=0∞amzm and g(z)=∑ m=0∞bmzm be two holomorphic functions in D such that h(z)=g(z). Then

for all |z|=r<1.

Proof of Theorem 2.4. From g′(z)=keiθznh′(z), we get

where a₁=1 and on integrating we obtain

Now, applying Lemma 3.1, we get

for all |z|=r<1. Since h is analytic univalent in D and according to De Brange's theorem, |am|≤m,m=2,3,..., therefore, from (3.2), we have

Thus ∑ m=1∞|am|rm+∑ m=1∞|bm|rm≤1 if

Now, we need to verify that inequality (3.3) holds for r≤r0, where r₀ is the unique root of the equation (2.5) lying in (0,1). For this let

Then ϕ(r) is continuous in (0,1),ϕ(0)=−1<0 and limr→1−ϕ(r)>0 implies that there is atleast one root of ϕ(r)=0 in (0,1). But ϕ′(r)>0 for all r∈(0,1),k∈(0,1) and for all n∈ℕ shows that ϕ is strictly increasing in (0,1). Hence ϕ(r)=0 has a unique root r₀ in (0,1).

To see that this r₀ is best possible one, we consider f0(z)=z/(1−z)2+z/(1−z)2−2z/(1−z)−log(1−z)¯. f₀ maps |z|<0.3485... onto the region given in the Figure 2 from which it is evident that r₀ is sharp and can not be improved further.

Figure 2. Image of |z|<0.3485 under f₀(z).

Proof of Theorem 2.7. From g′(z)=a+z1+azh′(z) we obtain

where a₁=1 and this gives

Thus we have

On integrating from 0 to r, we get

and this implies

Now, we have

From (3.5), we get

Therefore, we get

Multiplying both sides with r and then differentiating w.r.t r, we get

As h is univalent, so |am|≤m by De Branges's Theorem. From (3.6), we have

for r3−3r2+5r−1≤0 and this happens for r≤0.2291....

Proof of Theorem 2.9. Let h(z)+eiθg(z)=ψ(z), where ψ(z)=∑ m=1∞Cmzm is a convex univalent function. So, we have |am+eiθbm|=|Cm|≤1 for all m∈ℕ. This implies |bm|≤1+|am|,m∈ℕ. We have with |a1|=1,

Since h is univalent in D, so by De Brange's Theorem, we have |am|≤m and hence we get

for 2r(1−r)2+r1−r≤1 i.e. for 2r2−5r+1≥0. This is true for r≤r0=5−174=0.2192....

Now to prove Theorem 2.10., we first state the following result of Sheil-Small [11].

Lemma 3.2. If f(z)=h(z)+g(z)¯=z+∑ m=2∞amzm+∑ m=2∞bmzm¯∈SH0 is convex in one direction, then

Proof of Theorem 2.10. h+eiθg is convex univalent implies that f is convex in the direction −θ/2, by the well known result of Clunie and Sheil-Small [4]. Therefore, from Lemma 3.2, we have

and so,

if 4r3−9r2+12r−3≤0. This inequality holds for r≤r0=0.3134..., where r₀ is unique root of 4r3−9r2+12r−3=0 in (0,1). This result is sharp for K(z)=z−1/2z2+1/6z3(1−z)3+1/2z2+1/6z3(1−z)3¯,where K is harmonic mapping in D, which maps |z|<0.3134 onto region given in Figure 3. It is clear from Figure 3 that |K(z)|<1 for |z|<0.3134... and 0.3134... can not be improved. Hence this r₀ is sharp for inequality (2.11) also.

Figure 3. Image of |z|<0.3134 under K(z).

To prove Theorem 2.11., we need following result of Clunie and Sheil-Small [4].

Lemma 3.3. If a harmonic function

then

Proof of Theorem 2.11. In view of Lemma 3.3., f(z)∈KH0 implies that

This gives for a₁=1,

for r≤r0=0.3819... This value of r₀ is best possible, as the result is sharp for L(z)=12z1−z+z(1−z)2+z1−z−z(1−z)2¯. For L(z) we have

for r≤0.3819... Thus r₀ is sharp.