LOGARITHMIC COEFFICIENTS OF STARLIKE FUNCTIONS CONNECTED WITH k-FIBONACCI NUMBERS

Let A denote the class of analytic functions f in the open unit disc U normalized by f(0) = f ′(0) − 1 = 0, and let S be the class of all functions f ∈ A which are univalent in U. For a function f ∈ S, the logarithmic coefficients δn (n = 1, 2, 3, . . .) are defined by

Furthermore, if the function g is univalent in U, then we have the following equivalence f (z) ≺ g (z) (z ∈ U) ⇔ f (0) = g (0) and f (U) ⊂ g (U) . Let A denote the subclass of H consisting of functions f normalized by Each function f ∈ A can be expressed as We also denote by S the class of all functions in the normalized analytic function class A which are univalent in U.
A function f ∈ A is said to be starlike of order α (0 ≤ α < 1), if it satisfies the inequality We denote the class which consists of all functions f ∈ A that are starlike of order α by S * (α). It is well-known that S * (α) ⊂ S * (0) = S * ⊂ S. By means of the principle of subordination, YılmazÖzgür and Sokól [13] defined the following class SL k of functions f ∈ S, connected with a shell-like region described by a functionp k with coefficients depicted in terms of the k-Fibonacci numbers where k is a positive real number. The name attributed to the class SL k is motivated by the shape of the curve The curve Γ has a shell-like shape and it is symmetric with respect to the real axis. For more details about the class SL k , please refer to [11,13]. Definition 1. [13] Let k be any positive real number. The function f ∈ S belongs to the class SL k if it satisfies the condition that wherep For k = 1, the class SL k reduces to the class SL which consists of functions f ∈ A defined by (1) satisfying This class was introduced by Sokól [10].

Definition 2. [3]
For any positive real number k, the k-Fibonacci sequence {F k,n } n∈N0 is defined recurrently by with initial conditions F k,0 = 0, F k,1 = 1. Furthermore n th k-Fibonacci number is given by where τ k is given by (4) .
For k = 1, we obtain the classic Fibonacci sequence {F n } n∈N0 : For more details about the k-Fibonacci sequences please refer to [7,9,12,14]. YılmazÖzgür and Sokól [13] showed that the coefficients of the functionp k (z) defined by (3) are connected with k-Fibonacci numbers. This connection is pointed out in the following theorem. Theorem 1. [13] Let {F k,n } n∈N0 be the sequence of k-Fibonacci numbers defined in Definition 2. Ifp then we havẽ It can be found the more results related to Fibonacci numbers in [7,12,14].
Remark 1. [13] For each k > 0, that is, f ∈ SL k is a starlike function of order α k , and so is univalent.
For a function f ∈ S, the logarithmic coefficients δ n (n ∈ N) are defined by and play a central role in the theory of univalent functions. The idea of studying the logarithmic coefficients helped Kayumov [8] to solve Brennan's conjecture for conformal mappings. If f ∈ S, then it is known that (see [2]). The problem of the best upper bounds for |δ n | of univalent functions for n ≥ 3 is still open.
The main purpose of this paper is to determine the upper bound for |δ 1 | , |δ 2 | and |δ 3 | for functions f belong to the univalent function class SL k . To prove our main results we need the following lemmas. Lemma 1. [11] If p (z) = 1 + p 1 z + p 2 z 2 + · · · (z ∈ U) and then we have |p 1 | ≤ k |τ k | and |p 2 | ≤ k 2 + 2 τ 2 k . The above estimates are sharp.
The result is sharp.
S. BULUT then we have The above estimates are sharp. Then for some x, |x| ≤ 1, and for some z, |z| ≤ 1.
If the function f given by (1) is in the class SL k , then we have Each of these results is sharp.
If the function f given by (1) is in the class SL k , then If the function f given by (1) is in the class SL k , then The bound is sharp.

The coefficients of log (f (z)/z)
Theorem 2. Let f ∈ SL k be given by (1) and the coefficients of log (f (z)/z) be given by (10) . Then where τ k is defined by (4). Each of these results is sharp. The equalities are attained by the functionp k given by (3) .
Proof. Firstly, by differentiating (10) and equating coefficients, we have If f ∈ SL k , then by the principle of subordination, there exists a Schwarz function where the functionp k is given by (8). Therefore, the function is in the class P. Now, defining the function p(z) by it follows from (12) and (13) that Note that

S. BULUT
Thus, by using (13) in (15), and considering the valuesp k,j (j = 1, 2, 3) given in (9), we obtain On the other hand, a simple calculation shows that which, in view of (14), yields Substituting for a 2 , a 3 and a 4 from (20), we obtain Using Lemma 1 and Lemma 2, we get the desired results. This completes the proof of theorem.
Conjecture. Let f ∈ SL k be given by (1) and the coefficients of log (f (z)/z) be given by (10) . Then where {F k,n } n∈N0 is the Fibonacci sequence given by (7) .
This conjecture has been verified for the values n = 1, 2, 3 by the Theorem 2.
Letting k = 1 in Theorem 2, we obtain the following consequence.
Corollary 1. Let f ∈ SL be given by (1) and the coefficients of log (f (z)/z) be given by (10) . Then where τ is defined by (6). Each of these results is sharp. The equalities are attained by the functionp given by (5) .
Theorem 3. Let f ∈ SL k be given by (1) and the coefficients of log (f (z)/z) be given by (10) . Then for any γ ∈ C, we have Proof. By using (21) , the desired result is obtained from the equality and Lemma 3.
Letting k = 1 in Theorem 3, we obtain the following consequence.
Corollary 2. Let f ∈ SL be given by (1) and the coefficients of log (f (z)/z) be given by (10) . Then for any γ ∈ C, we have If we take γ = 1 in Theorem 3, then we obtain the following result.
Corollary 3. Let f ∈ SL k be given by (1) and the coefficients of log (f (z)/z) be given by (10) . Then Letting k = 1 in Corollary 3, we obtain the following consequence.
Corollary 4. Let f ∈ SL be given by (1) and the coefficients of log (f (z)/z) be given by (10) . Then

The coefficients of the inverse function
Since univalent functions are one-to-one, they are invertible and the inverse functions need not be defined on the entire unit disk U. In fact, the Koebe one-quarter theorem [2] ensures that the image of U under every univalent function f ∈ S contains a disk of radius 1/4. Thus every function f ∈ A has an inverse f −1 , which is defined by

S. BULUT
In fact, for a function f ∈ A given by (1) the inverse function f −1 is given by A n w n .
(22) Since SL k ⊂ S, the functions f belonging to the class SL k are invertible.
Theorem 4. Let f ∈ SL k be given by (1) , and f −1 be the inverse function of f defined by (22) . Then we have Each of these results is sharp.
Proof. Let the function f ∈ A given by (1) be in the class SL k , and f −1 be the inverse function of f defined by (22) . Then using (20), we obtain and The upper bound for |A 2 | is clear from Lemma 1. Furthermore by considering Lemma 3 we obtain the upper bound of |A 3 | as Finally, for the sharpness, we have by (8) that p k (z) = 1 + kτ k z + k 2 + 2 τ 2 k z 2 + · · · andp k z 2 = 1 + kτ k z 2 + k 2 + 2 τ 2 k z 4 + · · · . From this equalities, we obtain p 1 = kτ k and p 2 = k 2 + 2 τ 2 k and p 1 = 0 and p 2 = kτ k , respectively. Thus, it is clear that the equality for |A 2 | is attained for the functioñ p k (z); and the equality for the first value of |A 3 | is attained for the functionp k (z 2 ), for the second value of |A 3 | is attained for the functionp k (z). This evidently completes the proof of theorem.

Remark 2.
It is worthy to note that the coefficient bound obtained for |A 3 | in Theorem 4 is the improvement of [11,Corollary 2.4].
Theorem 5. Let f ∈ SL be given by (1) , and f −1 be the inverse function of f defined by (22) . Then we have Each of these results is sharp.
Proof. Let f ∈ SL be given by (1) , and f −1 be the inverse function of f defined by (22) . Then the upper bounds for |A 2 | and |A 3 | are obtained as a consequence of Theorem 4 when k = 1. From (22) , we have −A 4 = 5a 3 2 − 5a 2 a 3 + a 4 . By using (20) in the above equality, we obtain

By (17)-(19), this equality gives
By means of Lemma 5, we get As per Lemma 4, it is clear that |c 1 | ≤ 2. Therefore letting c 1 = c, we may assume without loss of generality that c ∈ [0, 2] . Hence, by using the triangle inequality, it is obtained that Thus, for µ = |x| ≤ 1, we have Now, we need to find the maximum value of F (c, µ) over the rectangle Π, For this, first differentiating the function F with respect to c and µ, we get respectively. The condition ∂F (c,µ) ∂µ = 0 gives c = 2 or µ = 0, and such points (c, µ) are not interior point of Π. So the maximum cannot attain in the interior of Π. Now to see on the boundary, by elementary calculus one can verify the following: Comparing these results, we get max Π F (c, µ) = 2 |τ | 3 (see Figure 1). Also note that p (z) = 1 + τ z + 3τ 2 z 2 + 4τ 3 z 3 + · · · by (8) with k = 1. From this equality, we obtain On the other hand, the sharpness of the upper bounds of |A 2 | and |A 3 | is known from Theorem 4 and it is seen that the equality for |A 4 | is attained for the functioñ p(z). This evidently completes the proof of theorem.
Theorem 6. Let f ∈ SL k be given by (1) , and f −1 be the inverse function of f defined by (22) . Then for any γ ∈ C, we have Proof. By using (20) , the desired result is obtained from the equality and Lemma 3.
Letting k = 1 in Theorem 6, we obtain following consequence.
Corollary 5. Let f ∈ SL be given by (1) , and f −1 be the inverse function of f defined by (22) . Then for any γ ∈ C, we have If we take γ = 1 in Theorem 6, then we obtain the following result.
Corollary 6. Let f ∈ SL k be given by (1) , and f −1 be the inverse function of f defined by (22) . Then Corollary 7. Let f ∈ SL be given by (1) , and f −1 be the inverse function of f defined by (22) . Then A 3 − A 2 2 ≤ τ 2 . Theorem 7. Let f ∈ SL k be given by (1) , and f −1 be the inverse function of f defined by (22) . Then