**2. Main Results**

Throughout this paper, we assume that *ϕ* is an analytic univalent function in the unit disk U satisfying *ϕ*(0) = 1 such that it has series expansion of the form

$$\varphi(z) = 1 + B\_1 z + B\_2 z^2 + B\_3 z^3 + \cdots, \quad B\_1 \neq 0. \tag{4}$$

**Theorem 1.** *Let the function f* ∈ S∗(*ϕ*)*. Then the logarithmic coefficients of f satisfy the inequalities:*

*(i) If ϕ is convex, then*

$$|\gamma\_n| \le \frac{|B\_1|}{2n}, \ n \in \mathbb{N},\tag{5}$$

$$\sum\_{n=1}^{k} \left| \gamma\_n \right|^2 \le \frac{1}{4} \sum\_{n=1}^{k} \frac{|B\_n|^2}{n^2}, \ k \in \mathbb{N},\tag{6}$$

*and* <sup>∞</sup>

$$\sum\_{n=1}^{\infty} |\gamma\_n|^2 \le \frac{1}{4} \sum\_{n=1}^{\infty} \frac{|B\_n|^2}{n^2}. \tag{7}$$

*(ii) If ϕ is starlike with respect to* 1*, then*

$$|\gamma\_n| \le \frac{|B\_1|}{2}, \ n \in \mathbb{N}.\tag{8}$$

*All inequalities in* (5)*,* (7)*, and* (8) *are sharp such that for any <sup>n</sup>* <sup>∈</sup> <sup>N</sup>*, there is the function fn given by z f <sup>n</sup>*(*z*) *fn*(*z*) <sup>=</sup> *<sup>ϕ</sup>*(*zn*) *and the function f given by z f* (*z*) *<sup>f</sup>*(*z*) <sup>=</sup> *<sup>ϕ</sup>*(*z*)*, respectively.*

**Proof.** Suppose that *f* ∈ S∗(*ϕ*). Then considering the definition of S∗(*ϕ*), it follows that

$$z\frac{d}{dz}\left(\log\left(\frac{f(z)}{z}\right)\right) = \frac{zf'(z)}{f(z)} - 1 \prec q(z) - 1 =: \Phi(z), \; z \in \mathbb{U}\_2$$

which according to the logarithmic coefficients *γ<sup>n</sup>* of *f* given by (1), concludes

$$\sum\_{n=1}^{\infty} 2n\gamma\_n z^n \prec \phi(z), \; z \in \mathbb{U}.$$

Now, for the proof of inequality (5), we assume that *ϕ* is convex in U. This implies that *φ*(*z*) is convex with *φ* (0) = *B*1, and so by Lemma 4(i) we get

$$|2n|\gamma\_{\mathfrak{n}}| \le |\mathfrak{q}'(0)| = |B\_1|\_{\prime} \ n \in \mathbb{N}\_{\prime}$$

and concluding the result.

Next, for the proof of inequality (6), we define *<sup>h</sup>*(*z*) :<sup>=</sup> *<sup>f</sup>*(*z*) *<sup>z</sup>* , which is an analytic function, and it satisfies the relation (*z*) (*z*)

$$\frac{zh'(z)}{h(z)} = \frac{zf'(z)}{f(z)} - 1 \prec \phi(z), \; z \in \mathbb{U},\tag{9}$$

as *φ* is convex in U with *φ*(0) = 0.

On the other hand, it is well known that the function (see [17])

$$b\_0(z) = \log\left(\frac{1}{1-z}\right) = \sum\_{n=1}^{\infty} \frac{z^n}{n}$$

belongs to the class K, and for *f* ∈ H,

$$f(z) \* b\_0(z) = \int\_0^z \frac{f(t)}{t} dt. \tag{10}$$

Now, by Lemma 2 and from (9), we obtain

$$\frac{zh'(z)}{h(z)} \ast b\_0(z) \prec \phi(z) \ast b\_0(z).$$

Considering (10), the above relation becomes

$$\log\left(\frac{f(z)}{z}\right) \prec \int\_0^z \frac{\phi(t)}{t} dt.$$

In addition, it has been proved in [18] that the class of convex univalent functions is closed under convolution. Therefore, the function *<sup>z</sup>* 0 *φ*(*t*) *t dt* is convex univalent. In addition, the above relation considering the logarithmic coefficients *γ<sup>n</sup>* of *f* given by (1) is equivalent to

$$\sum\_{n=1}^{\infty} 2\gamma\_n z^n \prec \sum\_{n=1}^{\infty} \frac{B\_n z^n}{n}.$$

Applying Lemma 3, from the above subordination this gives

$$4\sum\_{n=1}^{k} |\gamma\_n|^2 \le \sum\_{n=1}^{k} \frac{|B\_n|^2}{n^2},$$

which yields the inequality in (6). Supposing that *k* → ∞, we deduce that

$$4\sum\_{n=1}^{\infty} |\gamma\_n|^2 \le \sum\_{n=1}^{\infty} \frac{|B\_n|^2}{n^2}.$$

and it concludes the inequality (7).

Finally, we suppose that *ϕ* is starlike with respect to 1 in U, which implies *φ*(*z*) is starlike, and thus by Lemma 4(ii), we obtain

$$|2n|\gamma\_n| \le n|\phi'(0)| = n|B\_1|\_\prime, n \in \mathbb{N}\_\prime$$

This implies the inequality in (8).

For the sharp bounds, it suffices to use the equality

$$z\frac{d}{dz}\left(\log\left(\frac{f(z)}{z}\right)\right) = \frac{zf'(z)}{f(z)} - 1\_{\prime\prime}$$

and so these results are sharp in inequalities (5), (6), and (8) such that for any *<sup>n</sup>* <sup>∈</sup> <sup>N</sup>, there is the function *fn* given by *z f <sup>n</sup>*(*z*) *fn*(*z*) <sup>=</sup> *<sup>ϕ</sup>*(*zn*) and the function *<sup>f</sup>* given by *z f* (*z*) *<sup>f</sup>*(*z*) <sup>=</sup> *<sup>ϕ</sup>*(*z*), respectively. This completes the proof.

In the following corollaries, we obtain the logarithmic coefficients *γ<sup>n</sup>* for two subclasses S∗(*α* + (<sup>1</sup> <sup>−</sup> *<sup>α</sup>*)*ez*) and <sup>S</sup>∗(*<sup>α</sup>* + (<sup>1</sup> <sup>−</sup> *<sup>α</sup>*) <sup>√</sup><sup>1</sup> <sup>+</sup> *<sup>z</sup>*), which were defined by Khatter et al. in [19], and *<sup>α</sup>* + (<sup>1</sup> <sup>−</sup> *<sup>α</sup>*)*e<sup>z</sup>* and *α* + (1 − *α*) <sup>√</sup><sup>1</sup> <sup>+</sup> *<sup>z</sup>* are the convex univalent functions in <sup>U</sup>. For *<sup>α</sup>* <sup>=</sup> 0, these results reduce to the logarithmic coefficients *<sup>γ</sup><sup>n</sup>* for the subclasses <sup>S</sup>∗(*ez*) and <sup>S</sup>∗( <sup>√</sup><sup>1</sup> <sup>+</sup> *<sup>z</sup>*) (see [20,21]).

**Corollary 1.** *For* <sup>0</sup> <sup>≤</sup> *<sup>α</sup>* <sup>&</sup>lt; <sup>1</sup>*, let the function <sup>f</sup>* ∈ S∗(*<sup>α</sup>* + (<sup>1</sup> <sup>−</sup> *<sup>α</sup>*)*ez*)*. Then the logarithmic coefficients of <sup>f</sup> satisfy the inequalities*

$$|\gamma\_n| \le \frac{1-\mathfrak{a}}{2n}, \quad n \in \mathbb{N}$$

*and* <sup>∞</sup>

$$\sum\_{n=1}^{\infty} \left| \gamma\_n \right|^2 \le \frac{1}{4} \sum\_{n=1}^{\infty} \frac{(1-\alpha)^2/(n!)^2}{n^2}.$$

*These results are sharp such that for any <sup>n</sup>* <sup>∈</sup> <sup>N</sup>*, there is the function fn given by z f <sup>n</sup>*(*z*) *fn*(*z*) <sup>=</sup> *<sup>α</sup>* + (<sup>1</sup> <sup>−</sup> *<sup>α</sup>*)*ez<sup>n</sup> and the function f given by z f* (*z*) *<sup>f</sup>*(*z*) <sup>=</sup> *<sup>α</sup>* + (<sup>1</sup> <sup>−</sup> *<sup>α</sup>*)*ez.*

**Corollary 2.** *For* 0 ≤ *α* < 1*, let the function f* ∈ S∗(*α* + (1 − *α*) <sup>√</sup><sup>1</sup> <sup>+</sup> *<sup>z</sup>*)*. Then the logarithmic coefficients of f satisfy the inequalities*

$$|\gamma\_{\mathfrak{n}}| \le \frac{1-\mathfrak{a}}{4n}, \quad \mathfrak{n} \in \mathbb{N}$$

*and*

$$\sum\_{n=1}^{\infty} |\gamma\_n|^2 \le \frac{1}{4} \sum\_{n=1}^{\infty} \frac{\left((1-\alpha)\binom{\frac{1}{n}}{n}\right)^2}{n^2}.$$

*These results are sharp such that for any <sup>n</sup>* <sup>∈</sup> <sup>N</sup>*, there is the function fn given by z f <sup>n</sup>*(*z*) *fn*(*z*) <sup>=</sup> *<sup>α</sup>* + (<sup>1</sup> <sup>−</sup> *α*) <sup>√</sup><sup>1</sup> <sup>+</sup> *<sup>z</sup><sup>n</sup> and the function f given by z f* (*z*) *<sup>f</sup>*(*z*) <sup>=</sup> *<sup>α</sup>* + (<sup>1</sup> <sup>−</sup> *<sup>α</sup>*) <sup>√</sup><sup>1</sup> <sup>+</sup> *z.*

The following corollary concludes the logarithmic coefficients *γ<sup>n</sup>* for a subclass S∗(1 + sin *z*) defined by Cho et al. in [22], in which considering the proof of Theorem 1 and Corollary 1, the convexity radius for *q*0(*z*) = 1 + sin *z* is given by *r*<sup>0</sup> ≈ 0.345.

**Corollary 3.** *Let the function f* ∈ S∗(1 + sin *z*) *where q*0(*z*) *is a convex univalent function for r*<sup>0</sup> ≈ 0.345 *in* U*. Then the logarithmic coefficients of f satisfy the inequalities*

$$|\gamma\_n| \le \frac{1}{2n'} \quad n \in \mathbb{N}$$

*and* <sup>∞</sup>

$$\sum\_{n=1}^{\infty} |\gamma\_n|^2 \le \frac{1}{4} \sum\_{n=1}^{\infty} \frac{1}{((2n+1)!n)^2}.$$

*These results are sharp such that for any <sup>n</sup>* <sup>∈</sup> <sup>N</sup>*, there is the function fn given by z f <sup>n</sup>*(*z*) *fn*(*z*) <sup>=</sup> *<sup>q</sup>*0(*zn*) *and the function f given by z f* (*z*) *<sup>f</sup>*(*z*) <sup>=</sup> *<sup>q</sup>*0(*z*)*.*

In the following result, we get the logarithmic coefficients *γ<sup>n</sup>* for a subclass S∗(*pk*(*z*)) defined by Kanas and Wisniowska in [23] (see also [24,25]), in which

$$p\_k(z) = 1 + P\_1(k)z + P\_2(k)z^2 + \cdots \cdot \cdot \cdot \cdot$$

where *pk*(*z*) is a convex univalent function in U and

$$P\_1(k) = \begin{cases} \frac{2A^2}{1-k^2} & \text{if } \quad 0 \le k < 1, \\\frac{8}{\pi^2} & \text{if } \quad k = 1, \\\frac{\pi^2}{4\kappa^2(t)(k^2-1)(1+t)\sqrt{t}} & \text{if } \quad k > 1. \end{cases}$$

*A* = <sup>2</sup> *<sup>π</sup>* arccos *k* and *κ*(*t*) is the complete elliptic integral of the first kind.

**Corollary 4.** *For* 0 ≤ *k* < ∞*, let the function f* ∈ S∗(*pk*(*z*))*. Then the logarithmic coefficients of f satisfy the inequalities*

$$|\gamma\_n| \le \frac{P\_1(k)}{2n}, \quad n \in \mathbb{N}.$$

*This result is sharp such that for any n* <sup>∈</sup> <sup>N</sup>*, there is the function fn given by z f <sup>n</sup>*(*z*) *fn*(*z*) <sup>=</sup> *pk*(*zn*)*.*

The following result concludes the logarithmic coefficients *γ<sup>n</sup>* for a subclass S<sup>∗</sup> √<sup>2</sup> <sup>−</sup> ( <sup>√</sup><sup>2</sup> <sup>−</sup> 1) <sup>1</sup>−*<sup>z</sup>* 1+2( <sup>√</sup>2−1)*<sup>z</sup>* defined by Mendiratta et al. in [26], in which

$$\varphi\_0(z) = \sqrt{2} - (\sqrt{2} - 1)\sqrt{\frac{1 - z}{1 + 2(\sqrt{2} - 1)z}} = 1 + \frac{5 - 3\sqrt{2}}{2}z + \frac{71 - 51\sqrt{2}}{8}z^2 + \dotsb,$$

where *ϕ*<sup>0</sup> is a convex univalent function in U.

**Corollary 5.** *Let the function f* ∈ S<sup>∗</sup> √<sup>2</sup> <sup>−</sup> ( <sup>√</sup><sup>2</sup> <sup>−</sup> <sup>1</sup>) <sup>1</sup>−*<sup>z</sup>* 1+2( <sup>√</sup>2−1)*<sup>z</sup> . Then the logarithmic coefficients of f satisfy the inequalities* √2

$$|\gamma\_n| \le \frac{5 - 3\sqrt{2}}{4n}, \quad n \in \mathbb{N}.$$

*This result is sharp such that for any n* <sup>∈</sup> <sup>N</sup>*, there is the function fn given by z f <sup>n</sup>*(*z*) *fn*(*z*) <sup>=</sup> *<sup>ϕ</sup>*0(*zn*)*.*

The following results conclude the logarithmic coefficients *<sup>γ</sup><sup>n</sup>* for two subclasses <sup>S</sup>∗(*<sup>z</sup>* <sup>+</sup> <sup>√</sup> 1 + *z*2) and <sup>S</sup>∗ 1 + *<sup>z</sup>* (1−*αz*2) defined by Krishna Raina and Sokół in [27] and Kargar et al. in [28], where

$$z + \sqrt{1 + z^2} = 1 + z + \sum\_{n=1}^{\infty} \binom{\frac{1}{2}}{n}^2 z^{2n},$$

and

$$1 + \frac{z}{(1 - az^2)} = 1 + z + \sum\_{n=1}^{\infty} a^n z^{2n+1}, \quad (0 \le n < 1),$$

respectively. These functions are univalent and starlike with respect to 1 in U.

**Corollary 6.** *Let the function <sup>f</sup>* ∈ S∗(*<sup>z</sup>* <sup>+</sup> <sup>√</sup> 1 + *z*2)*. Then the logarithmic coefficients of f satisfy the inequalities*

$$|\gamma\_{\mathfrak{n}}| \le \frac{1}{2}, \quad \mathfrak{n} \in \mathbb{N}.$$

*This result is sharp such that for any n* <sup>∈</sup> <sup>N</sup>*, there is the function fn given by z f <sup>n</sup>*(*z*) *fn*(*z*) <sup>=</sup> *<sup>z</sup><sup>n</sup>* <sup>+</sup> <sup>√</sup> 1 + *z*<sup>2</sup>*n.*

**Corollary 7.** *Let the function <sup>f</sup>* ∈ S∗ 1 + *<sup>z</sup>* (1−*αz*2) *, where* 0 ≤ *α* < 1*. Then the logarithmic coefficients of f satisfy the inequalities*

$$|\gamma\_n| \le \frac{1}{2}, \quad n \in \mathbb{N}.$$

*This result is sharp such that for any <sup>n</sup>* <sup>∈</sup> <sup>N</sup>*, there is the function fn given by z f <sup>n</sup>*(*z*) *fn*(*z*) <sup>=</sup> <sup>1</sup> <sup>+</sup> *<sup>z</sup>* (1 − *αz*2*n*) *.*

**Remark 1.** *1. Letting*

$$\begin{aligned} \varphi(z) &= \frac{1+Az}{1+Bz} \\ &= 1 + (A-B)z - B(A-B)z^2 + B^2(A-B)z^3 + \cdots \\ &= 1 + \begin{cases} \frac{A-B}{B} \sum\_{n=1}^{\infty} (-1)^{n-1} B^n z^n, & \text{if } \quad B \neq 0 \\ Az, & \text{if } \quad B = 0, \end{cases} \quad (-1 \le B < A \le 1), \end{aligned}$$

*which is convex univalent in* U *in Theorem 1, then we get the results obtained by Ponnusamy et al. [7] (Theorem 2.1 and Corollary 2.3).*

*2. For <sup>A</sup>* <sup>=</sup> *<sup>e</sup>iα*(*ei<sup>α</sup>* <sup>−</sup> <sup>2</sup>*<sup>β</sup>* cos *<sup>α</sup>*)*, where <sup>β</sup>* <sup>∈</sup> [0, 1) *and <sup>α</sup>* <sup>∈</sup> (−*π*/2, *<sup>π</sup>*/2) *in the above expression, then we get the results obtained by Ponnusamy et al. [7] (Theorem 2.5).*

*3. Taking*

$$\varphi(z) = \left(\frac{1+z}{1-z}\right)^a = 1 + 2az + 2a^2z^2 + \frac{8a^3 + 4a}{6}z^3 + \dots$$

$$= 1 + \sum\_{n=1}^{\infty} A\_n(a)z^n, \quad (0 < a \le 1),$$

*which is convex univalent in* <sup>U</sup>*, and An*(*α*) = *<sup>n</sup>* ∑ *k*=1 ( *n*−1 *<sup>k</sup>*−1)(*<sup>α</sup> <sup>k</sup>*)2*<sup>k</sup> in Theorem 1, then we get the results obtained by Ponnusamy et al. [7] (Theorem 2.6).*

*4. Setting*

$$\varphi(z) = 1 + \frac{\beta - \alpha}{\pi} i \log \left( \frac{1 - e^{2\pi i \frac{1 - \alpha}{\beta - \alpha} z}}{1 - z} \right) = 1 + \sum\_{n=1}^{\infty} \mathbb{C}\_n z^n, \quad (\alpha > 1, \ \beta < 1),$$

*which is convex univalent in* <sup>U</sup>*, and Cn* <sup>=</sup> *<sup>β</sup>* <sup>−</sup> *<sup>α</sup> <sup>n</sup><sup>π</sup> <sup>i</sup>* 1 − *e* <sup>2</sup>*nπ<sup>i</sup>* <sup>1</sup>−*<sup>α</sup> <sup>β</sup>*−*<sup>α</sup> in Theorem 1, then we get the results obtained by Kargar [5] (Theorems 2.2 and 2.3).*

*5. Letting*

$$\varphi(z) = 1 + \frac{1}{2i\sin\delta} \log\left(\frac{1 + z\varepsilon^{i\delta}}{1 + z\varepsilon^{-i\delta}}\right) = 1 + \sum\_{n=1}^{\infty} D\_n z^n, \quad (\pi/2 \le \delta < \pi),$$

*which is convex univalent in* <sup>U</sup>*, and Dn* <sup>=</sup> (−1)*n*−<sup>1</sup> sin *<sup>n</sup><sup>δ</sup> <sup>n</sup>* sin *<sup>δ</sup> in Theorem 1, then we get the results obtained by Kargar [5] (Theorems 2.5 and 2.6).*

*6. Letting*

$$\varphi(z) = \left(\frac{1+cz}{1-z}\right)^{(a\_1+a\_2)/2} = 1 + \sum\_{n=1}^{\infty} \lambda\_n z^n, \quad \left(0 < a\_1, a\_2 \le 1, \ c = e^{\pi i \theta}, \ \theta = \frac{a\_2 - a\_1}{a\_2 + a\_1}\right),$$

*which is convex univalent in* U*, and*

$$\lambda\_{\mathfrak{U}} = \sum\_{k=1}^{n} \binom{n-1}{k-1} \binom{(\mathfrak{a}\_1 + \mathfrak{a}\_2)/2}{k} (1+c)^k$$

*in Theorem 1, then we get the results obtained for* |*γn*| *by Kargar et al. [29] (Theorem 3.1). Moreover, for α*<sup>1</sup> = *α*<sup>2</sup> = *β, we get the result presented by Thomas in [30] (Theorem 1).*

*7. Let the function <sup>f</sup>* ∈ K <sup>1</sup> <sup>−</sup> *cz* 1 − *z* <sup>=</sup> <sup>K</sup>(<sup>1</sup> <sup>−</sup> *cz* <sup>−</sup> *cz*<sup>2</sup> <sup>−</sup> *cz*<sup>3</sup> <sup>+</sup> ...)*, where <sup>c</sup>* <sup>∈</sup> (0, 1]*. It is equivalent to*

$$\operatorname{Re}\left(1+\frac{zf''(z)}{f'(z)}\right) < 1+\frac{c}{2}.$$

*Then we have (see e.g., [31] (Theorem 1))*

$$\frac{zf'(z)}{f(z)} \prec \frac{(1+c)(1-z)}{1+c-z},$$

*where* (<sup>1</sup> <sup>+</sup> *<sup>c</sup>*)(<sup>1</sup> <sup>−</sup> *<sup>z</sup>*) <sup>1</sup> <sup>+</sup> *<sup>c</sup>* <sup>−</sup> *<sup>z</sup> is a convex univalent function in* <sup>U</sup>*, and*

$$\frac{(1+c)(1-z)}{1+c-z} = 1 - \frac{c}{c+1}z - \frac{c}{(c+1)^2}z^2 + \dots = 1 - c \sum\_{n=1}^{\infty} \frac{z^n}{(1+c)^n}.$$

*Thus, applying Theorem 1, we get the results obtained by Obradovi´c et al. [4] (Theorem 2 and Corollary 2).*

**Theorem 2.** *Let the function f* ∈ K(*ϕ*)*. Then the logarithmic coefficients of f satisfy the inequalities*

$$|\gamma\_1| \le \frac{|B\_1|}{4} \,\, \, \, \tag{11}$$

$$|\gamma\_2| \le \begin{cases} \frac{|B\_1|}{12} & \text{if } & |4B\_2 + B\_1^2| \le 4|B\_1| \\\\ \frac{|4B\_2 + B\_1^2|}{48} & \text{if } & |4B\_2 + B\_1^2| > 4|B\_1| \end{cases} \tag{12}$$

*and if B*1, *B*2*, and B*<sup>3</sup> *are real values,*

$$|\gamma\_3| \le \frac{|B\_1|}{24} H(q\_1; q\_2),\tag{13}$$

*where H*(*q*1; *<sup>q</sup>*2) *is given by Lemma 5, q*<sup>1</sup> <sup>=</sup> *<sup>B</sup>*1<sup>+</sup> <sup>4</sup>*B*<sup>2</sup> *B*1 <sup>2</sup> *, and q*<sup>2</sup> <sup>=</sup> *<sup>B</sup>*2<sup>+</sup> <sup>2</sup>*B*<sup>3</sup> *B*1 <sup>2</sup> *. The bounds* (11) *and* (12) *are sharp.*

**Proof.** Let *f* ∈ K(*ϕ*). Then by the definition of the subordination, there is a *ω* ∈ Ω with *ω*(*z*) = ∑<sup>∞</sup> *<sup>n</sup>*=<sup>1</sup> *cnz<sup>n</sup>* so that

$$\begin{aligned} 1 + \frac{z f''(z)}{f'(z)} &= \varrho(\omega(z)) \\ &= 1 + B\_1 c\_1 z + (B\_1 c\_2 + B\_2 c\_1^2) z^2 + (B\_1 c\_3 + 2c\_1 c\_2 B\_2 + B\_3 c\_1^3) z^3 + \dotsb \dotsb \end{aligned}$$

From the above equation, we get that

$$\begin{cases} 2a\_2 = B\_1 c\_1 \\ 6a\_3 - 4a\_2^2 = B\_1 c\_2 + B\_2 c\_1^2 \\ 12a\_4 - 18a\_2 a\_3 + 8a\_2^3 = B\_1 c\_3 + 2c\_1 c\_2 B\_2 + B\_3 c\_1^3 \end{cases} \tag{14}$$

By substituting values *an* (n = 1, 2, 3) from (14) in (3), we have

$$\begin{cases} \begin{aligned} 2\gamma\_1 &= \frac{B\_1 c\_1}{2} \\ 2\gamma\_2 &= \frac{8B\_1 c\_2 + (8B\_2 + 2B\_1^2)c\_1^2}{48} \\ 2\gamma\_3 &= \frac{B\_1}{12} \left[ c\_3 + \frac{4B\_2}{2} c\_1 c\_2 + \frac{B\_2 + \frac{2B\_3}{B\_1}}{2} c\_1^3 \right] \end{aligned} \end{cases}$$

Firstly, for *<sup>γ</sup>*1, by applying Lemma <sup>1</sup> we get <sup>|</sup>*γ*1| ≤ <sup>|</sup>*B*1<sup>|</sup> <sup>4</sup> , and this bound is sharp for <sup>|</sup>*c*1<sup>|</sup> <sup>=</sup> 1. Next, applying Lemma 1 for *γ*2, we have

$$\begin{aligned} |\gamma\_2| &\le \frac{4|B\_1|(1-|c\_1|^2)+|4B\_2+B\_1^2||c\_1|^2}{48} \\ &= \frac{4|B\_1|+\left[|4B\_2+B\_1^2|-4|B\_1|\right]|c\_1|^2}{48} \\ &\le \begin{cases} \frac{4|B\_1|}{48} & \text{if} \quad |4B\_2+B\_1^2| \le 4|B\_1| \\\\ \frac{|4B\_2+B\_1^2|}{48} & \text{if} \quad |4B\_2+B\_1^2| > 4|B\_1|. \end{cases} \end{aligned}$$

These bounds are sharp for *c*<sup>1</sup> = 0 and |*c*1| = 1, respectively.

Finally, using Lemma 5 for *γ*3, we obtain

$$|2|\gamma\_3| \le \frac{|B\_1|}{12} \left| c\_3 + \frac{B\_1 + \frac{4B\_2}{B\_1}}{2} c\_1 c\_2 + \frac{B\_2 + \frac{2B\_3}{B\_1}}{2} c\_1^3 \right| \le H(q\_1; q\_2),$$

where *<sup>q</sup>*<sup>1</sup> <sup>=</sup> *<sup>B</sup>*1<sup>+</sup> <sup>4</sup>*B*<sup>2</sup> *B*1 <sup>2</sup> and *<sup>q</sup>*<sup>2</sup> <sup>=</sup> *<sup>B</sup>*2<sup>+</sup> <sup>2</sup>*B*<sup>3</sup> *B*1 <sup>2</sup> . Therefore, this completes the proof.

**Remark 2.** *1. Letting*

$$\begin{aligned} \varphi(z) &= 1 + \frac{cz}{1 - z} \\ &= 1 + cz + cz^2 + cz^3 + \dots \quad (c \in (0, 3]) \end{aligned}$$

*in Theorem 2, (for* |*γ*3| *with respect to D*6*) then we get the results obtained by Ponnusamy et al. [7] (Theorem 2.7 and Corollary 2.8).*

*2. Taking*

$$\begin{aligned} q(z) &= 1 - \frac{cz}{1 - z} \\ &= 1 - cz - cz^2 - cz^3 + \dots \quad (c \in (0, 1]) \end{aligned}$$

*in Theorem 2, (for* |*γ*3| *respect to D*2*) then we get the results obtained by Ponnusamy et al. [7] (Theorem 2.10).*

**Author Contributions:** All authors contributed equally.

**Funding:** The authors would like to express their gratitude to the referees for many valuable suggestions regarding the previous version of this paper. This research was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science, and Technology (No. 2016R1D1A1A09916450).

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**


c 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).
