δ ( n , k ) a k ζ ( β + ( 1 γ μ ) ) (11)

This establishes our proof.

Corollary 2.1.

If $f\in {B}_{\lambda }^{n}\left(\mu ,\beta ,\zeta \right)$ then

${a}_{k}\le \frac{\zeta \left(\beta +\left(1-\gamma \mu \right)\right)}{k\left(1+\zeta \beta \right){\left[1+\left(k-1\right)\lambda \right]}^{n}\delta \left(n,k\right)},k=2,3,\cdots$ (12)

equality is attained for

$f\left(z\right)=z+\frac{\zeta \left(\beta +\left(1-\gamma \mu \right)\right)}{k\left(1+\zeta \beta \right){\left[1+\left(k-1\right)\lambda \right]}^{n}\delta \left(n,k\right)}{z}^{k},k=2,3,\cdots$ (13)

We shall state the growth and distortion theorems for the class ${B}_{\lambda }^{n}\left(\vartheta ,\varsigma ,\gamma ,\iota \right)$ The results of which follow easily on applying Theorem 2.1, therefore, we deem it necessary to omit the trivial proofs.

2.2. Growth and Distortion Theorems

Theorem 2.2.

Let the function $f\left(z\right)\in {B}_{\lambda }^{n}\left(\mu ,\beta ,\zeta ,\gamma \right)$ then for $|z|=r$

$r-\frac{\zeta \left(\beta +\left(1-\gamma \mu \right)\right)}{2\left(1+\zeta \beta \right){\left[1+\lambda \right]}^{n}\delta \left(n,2\right)}{r}^{2}\le |f\left(z\right)|\le r+\frac{\zeta \left(\beta +\left(1-\gamma \mu \right)\right)}{2\left(1+\zeta \beta \right){\left[1+\lambda \right]}^{n}\delta \left(n,2\right)}{r}^{2}$

Theorem 2.3.

Let the function $f\left(z\right)\in {B}_{\lambda }^{n}\left(\mu ,\beta ,\zeta \right)$ then for $|z|=r$

$1-\frac{\zeta \left(\beta +\left(1-\gamma \mu \right)\right)}{\left(1+\zeta \beta \right){\left[1+\lambda \right]}^{n}\delta \left(n,2\right)}r\le |{f}^{\prime }\left(z\right)|\le 1+\frac{\zeta \left(\beta +\left(1-\gamma \mu \right)\right)}{\left(1+\zeta \beta \right){\left[1+\lambda \right]}^{n}\delta \left(n,2\right)}r$

When $f\left(z\right)=z+\frac{\zeta \left(\beta +\left(1-\gamma \mu \right)\right)}{2\left(1+\zeta \beta \right){\left[1+\lambda \right]}^{n}\delta \left(n,2\right)}{z}^{2}$

we obtain a sharp result.

2.3. Radii of Close-to-Convexity, Starlikeness and Convexity

Theorem 2.4.

Let the function $f\left(z\right)\in {B}_{\lambda }^{n}\left(\mu ,\beta ,\zeta ,\gamma \right)$ , then $f$ is close-to-convex of order $\delta$ in $|z|<{R}_{\tau }1$

where

${R}_{\tau }1=\underset{k\ge 2}{\mathrm{inf}}{\left[\frac{k\left(1+\zeta \beta \right)\left(1-\delta \right){\left[1+\left(k-1\right)\lambda \right]}^{n}\delta \left(n,k\right)}{\zeta \left(\beta +\left(1-\gamma \mu \right)\right)}\right]}^{\frac{1}{k-1}}$

The result obtained is sharp.

Proof.

It is sufficient to show that $|{f}^{\prime }\left(z\right)-1|\le 1-\delta$ for $|z|<{R}_{\tau }$ Thus we can write

$|{f}^{\prime }\left(z\right)-1|=|-\underset{n=2}{\overset{\infty }{\sum }}\text{ }\text{ }k{a}_{k}{z}^{k-1}|\le \underset{n=2}{\overset{\infty }{\sum }}\text{ }\text{ }k{a}_{k}{|z|}^{k-1}$

Therefore $|{f}^{\prime }\left(z\right)-1|\le 1-\delta$ if

$\underset{n=2}{\overset{\infty }{\sum }}\left(\frac{k}{1-\delta }\right){a}_{k}{|z|}^{k-1}\le 1$ (14)

But we have from theorem 2.1. that

$\underset{k=2}{\overset{\infty }{\sum }}\frac{k\left(1+\zeta \beta \right){\left[1+\left(k-1\right)\lambda \right]}^{n}\delta \left(n,k\right){a}_{k}}{\zeta \left(\beta +\left(1-\gamma \mu \right)\right)}\le 1$ (15)

Relating (14) and (15) we have our desired result.

Theorem.2.5.

Let the function $f\left(z\right)\in {B}_{\lambda }^{n}\left(\mu ,\beta ,\zeta \right)$ , then $f$ is starlike of order $\delta$ , $0\le \delta <1$ in $|z|<{R}_{\tau }2$

where

${R}_{\tau }2=\underset{k\ge 2}{\mathrm{inf}}{\left[\frac{k\left(1+\zeta \beta \right)\left(1-\delta \right){\left[1+\left(k-1\right)\lambda \right]}^{n}\delta \left(n,k\right)}{\left(k-\delta \right)\zeta \left(\beta +\left(1-\gamma \mu \right)\right)}\right]}^{\frac{1}{k-1}}$

The result obtain here is sharp.

Proof.

We must show that $|\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-1|\le 1-\delta$ for $|z|<{R}_{\tau }2$ . Equivalently, we have

$\underset{k=2}{\overset{\infty }{\sum }}\frac{\left(k-\delta \right){a}_{k}|{z}^{k-1}|}{1-\delta }\le 1$ (16)

But we have from theorem 2.1. that

$\underset{k=2}{\overset{\infty }{\sum }}\frac{k\left(1+\zeta \beta \right){\left[1+\left(k-1\right)\lambda \right]}^{n}\delta \left(n,k\right){a}_{k}}{\zeta \left(\beta +\left(1-\gamma \mu \right)\right)}\le 1$ (17)

Relating (16) and (17) will have our desired result.

Theorem 2.6.

Let the function $f\left(z\right)\in {B}_{\lambda }^{n}\left(\vartheta ,\varsigma ,\gamma ,\iota \right)$ , then $f$ is convex of order $\delta$ , $0\le \delta <1$ in $|z|<{R}_{\tau }3$

where

${R}_{\tau }3=\underset{k\ge 2}{\mathrm{inf}}{\left[\frac{\left(1+\zeta \beta \right)\left(1-\delta \right){\left[1+\left(k-1\right)\lambda \right]}^{n}\delta \left(n,k\right)}{\left(k-\delta \right)\zeta \left(\beta +\left(1-\gamma \mu \right)\right)}\right]}^{\frac{1}{k-1}}$

The result obtain here is sharp.

Proof.

By using the technique of theorem 2.5 we easily show that $|\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}|\le 1-\delta$ this holds for $|z|<{R}_{\tau }3$ . The analogous details of theorem 2.5 are thus omitted, hence the proof.

3. Integral Operator

Theorem 3.1.

Let the function $f\left(z\right)$ defined by (2) be in the class ${T}_{n}\left(\mu ,\beta ,\gamma \zeta \right)$ and let $c$ be a real number such that $c>-1$ . Then the function defined by

$F\left(z\right)=\frac{c+1}{{z}^{c}}\int {t}^{c-1}f\left(t\right)\text{d}t$ (18)

also belong to the class ${T}_{n}\left(\mu ,\beta ,\gamma ,\zeta \right)$

Proof.

From the representation and definition of $F\left(z\right)$ we have that

$F\left(z\right)=z-\underset{k=2}{\overset{\infty }{\sum }}{b}_{k}{z}^{k}$ (19)

where

${b}_{k}=\left(\frac{c+1}{c+k}\right){a}_{k}$ (20)

Thus we have

$\underset{k=2}{\overset{\infty }{\sum }}\text{ }k\left(1+\zeta \beta \right){\left[1+\left(k-1\right)\lambda \right]}^{n}\delta \left(n,k\right){b}_{k}$ (21)

$=\underset{k=2}{\overset{\infty }{\sum }}\text{ }\text{ }k\left(1+\zeta \beta \right){\left[1+\left(k-1\right)\lambda \right]}^{n}\delta \left(n,k\right)\left(\frac{c+1}{c+k}\right){a}_{k}$

$\le \underset{k=2}{\overset{\infty }{\sum }}\text{ }\text{ }k\left(1+\zeta \beta \right){\left[1+\left(k-1\right)\lambda \right]}^{n}\delta \left(n,k\right){a}_{k}\le \zeta \left(\beta +\left(1-\gamma \mu \right)\right)$ (22)

since $f\left(z\right)\in {T}_{n}\left(\mu ,\beta ,\gamma ,\zeta \right)$ . By theorem 1.1 $F\left(z\right)\in {T}_{n}\left(\mu ,\beta ,\gamma ,\zeta \right)$ . This establishes our proof.

Conflicts of Interest

The authors declare no conflicts of interest.

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