**3. Main Results and Their Demonstrations**

In this section, we will prove our main results.

**Theorem 1.** *A function f* ∈ A *is in the class k-*Q (*φ*, *λ*, *η*, *f* , *ψ*) *if:*

$$\sum\_{n=2}^{\infty} \hat{\mathcal{U}}\_n \left( p, \phi, \lambda, \eta, \xi^n \right) < p^2 (p - \eta) \lambda$$

*where:*

$$\begin{split} \bar{\mathcal{U}}\_{n}\left(p,\phi,\lambda,\eta,\xi\right) &= (k+1)\left[\left(e^{i\lambda} - \phi\cos\lambda\right)(p-\eta)p + p^{4}\phi\cos\lambda \\ &\quad + \left(e^{i\lambda} - \phi\cos\lambda\right)(p-\eta)(n+p)\left|a\_{n+p}\right| + (n+p)^{2}\left|a\_{n+p}\right| \\ &\quad + \left[\left(np\phi\cos\lambda + p^{3}(p-\eta)\right)(n+p)\left|b\_{n+p}\right| + np^{2}\phi\cos\lambda - p^{3}(p-\eta). \end{split} \right] \tag{5}$$

**Proof.** Let us assume that the relation (4) holds true. It now suffices to show that:

$$\mathbb{P}\left|\mathcal{M}\left(\phi,\lambda,\eta,f,\psi\right)-p\right|-\mathbb{P}\left|\mathcal{M}\left(\phi,\lambda,\eta,f,\psi\right)-p\right|<1.\tag{6}$$

We first consider:

$$\begin{split} & \left| \mathcal{M} \left( \boldsymbol{\phi}, \lambda, \boldsymbol{\eta}, \boldsymbol{f}, \boldsymbol{\Psi} \right) - p \right| \\ & \quad = \left| \left( \boldsymbol{e}^{i\lambda} - \boldsymbol{\phi} \cos \lambda \right) \frac{zf'(z)}{p\boldsymbol{\upupupleft}(z)} + \frac{\boldsymbol{\phi} \cos \lambda}{\left( p - \boldsymbol{\eta} \right)} \left( \frac{(zf'(z))'}{\boldsymbol{\upupupleft}'(z)} - \boldsymbol{\eta} \right) - p \right| \right| \\ & \quad = \left| \frac{\left( \boldsymbol{e}^{i\lambda} - \boldsymbol{\phi} \cos \lambda \right) \left( p - \boldsymbol{\eta} \right) f'(z)}{p \left( p - \boldsymbol{\upupupright} \right) \boldsymbol{\upupupleft}'(z)} + \frac{p \boldsymbol{\upupupright} \cos \lambda \left( zf'(z) \right)'}{p \left( p - \boldsymbol{\upupupright} \right) \boldsymbol{\upupupleft}'(z)} - \\ & \quad \left| - \frac{\boldsymbol{\upupupright} p \boldsymbol{\upupup} \cos \lambda \boldsymbol{\upupup}'(z)}{p \left( p - \boldsymbol{\upupup} \right) \boldsymbol{\upupup}'(z)} - \frac{p^{2} \left( p - \boldsymbol{\upupup} \right) \boldsymbol{\upupup}'(z)}{p \left( p - \boldsymbol{\upupup} \right) \boldsymbol{\upupup}'(z)} \right|. \end{split}$$

Now, by using the series form of the functions *f* and *ψ* given by:

$$f(z) = z^p + \sum\_{n=2}^{\infty} a\_{n+p} z^{n+p}$$

and:

$$\psi(z) = z^p + \sum\_{n=2}^{\infty} b\_{n+p} z^{n+p}$$

in the above relation, we have:

$$\begin{split} &|\mathcal{M}\left(\boldsymbol{\phi},\lambda,\eta,f,\psi\right)-p| \\ &=\left|\frac{\left(\boldsymbol{e}^{i\lambda}-\boldsymbol{\phi}\cos\lambda\right)\left(p-\eta\right)\left(p z^{p-1}\right)+p\phi\cos\lambda\left(p^{2}z^{p-1}\right)}{p\left(p-\eta\right)\left(p z^{p-1}+\sum\_{n=2}^{\infty}(n+p)b\_{n+p}z^{n+p-1}\right)}\right| \\ &+\frac{\sum\_{n=2}^{\infty}(n+p)a\_{n+p}z^{n+p-1}\left[\left(\boldsymbol{e}^{i\lambda}-\boldsymbol{\phi}\cos\lambda\right)\left(p-\eta\right)+\left(n+p\right)\right]}{p\left(p-\eta\right)\left(p z^{p-1}+\sum\_{n=2}^{\infty}(n+p)b\_{n+p}z^{n+p-1}\right)}-\frac{n\boldsymbol{\phi}\cos\lambda}{\left(p-\eta\right)}-p\bigg| \\ &\leq\frac{\left(\boldsymbol{e}^{i\lambda}-\boldsymbol{\phi}\cos\lambda\right)\left(p-\eta\right)\left(p+\eta\right)\left(p+\eta\right)\boldsymbol{\phi}\boldsymbol{\phi}\cos\lambda\left(p^{2}\right)}{p\left(p-\eta\right)\left(p+\sum\_{n=2}^{\infty}(n+p)\right)\left[b\_{n+p}\right]} \\ &+\frac{\sum\_{n=2}^{\infty}(n+p)\left|a\_{n+p}\right|\left\{\left(\boldsymbol{e}^{i\lambda}-\boldsymbol{\phi}\cos\lambda\right)\left(p-\eta\right)+\left(n+p\right)\right\}}{p\left(p-\eta\right)\left(p+\sum\_{n=2}^{\infty}(n+p)\right)\left[b\_{n+p}\right]}-\left\{\frac{n\boldsymbol{\phi}\cos\lambda}{\left(p-\eta\right)}+p\right\}.\end{split}$$

We now see that:

$$\begin{split} &k\left|\mathcal{M}\left(\phi,\lambda,\eta,f,\psi\right)-p\right|-\mathfrak{R}\left\{\mathcal{M}\left(\phi,\lambda,\eta,f,\psi\right)-p\right\} \\ & \stackrel{\scriptstyle \mathcal{M}}{\leq}(k+1)\left|\mathcal{M}\left(\phi,\lambda,\eta,f,\psi\right)-p\right| \\ & \stackrel{\scriptstyle \mathcal{M}}{\leq}(k+1)\left[\frac{\left(e^{i\lambda}-\phi\cos\lambda\right)\left(p-\eta\right)\left(p\right)+p\phi\cos\lambda\left(p^{2}\right)}{p\left(p-\eta\right)\left(p+\sum\_{n=2}^{\infty}(n+p)\left|b\_{n+p}\right|\right)} \\ & +\frac{\sum\_{n=2}^{\infty}(n+p)\left|a\_{n+p}\right|\left[\left(e^{i\lambda}-\phi\cos\lambda\right)\left(p-\eta\right)+(n+p)\right]}{p\left(p-\eta\right)\left(p+\sum\_{n=2}^{\infty}(n+p)\left|b\_{n+p}\right|\right)}-\left[\frac{n\phi\cos\lambda}{(p-\eta)}+p\right]}\right]. \end{split}$$

The above inequality is bounded above by one, if:

$$\begin{split} &(k+1)\left[\left(\epsilon^{j\lambda}-\phi\cos\lambda\right)(p-\eta)p\right]+\left(p\phi\cos\lambda\right)p^{2} \\ &\qquad\qquad\qquad +\left(\sum\_{n=2}^{\infty}(n+p)\left|a\_{n+p}\right|\right)\left\{\left(\epsilon^{j\lambda}-\phi\cos\lambda\right)(p-\eta)+(n+p)\right\}-\left[\frac{n\phi\cos\lambda}{(p-\eta)}-p\right] \\ &\qquad\qquad\qquad\qquad\qquad\qquad\left\{p\left(p-\eta\right)\left(p+\sum\_{n=2}^{\infty}(n+p)\left|b\_{n+p}\right|\right)\right\} \\ &\leq p(p-\eta)p+\sum\_{n=2}^{\infty}(n+p)\left|b\_{n+p}\right|. \end{split}$$

Hence: <sup>∞</sup>

$$\sum\_{n=2}^{\infty} \ddot{\mathcal{U}}\_n \left( p, \phi, \lambda, \eta, \xi^x \right) \stackrel{<}{=} p^2 (p - \eta)\_x$$

where *U*¨ *<sup>n</sup>* (*p*, *φ*, *λ*, *η*, *ξ*) is given by (5), which completes the proof of Theorem 1.

**Theorem 2.** *A function f* ∈ A(*p*) *satisfies the condition:*

$$\left| \left| \frac{1}{e^{ij}F(z)} - \frac{1}{2\rho} \right| < \frac{1}{2\rho} \right| \qquad (0 \le \rho < 1; \ j \in \mathbb{R}) \tag{7}$$

*if and only if f* ∈ 0*-*K(*p*, *λ*), *where*

 

$$F(z) = \frac{zf'(z)}{p\psi(z)}.$$

**Proof.** Suppose that *f* satisfies (7). We then can write:

$$\begin{split} \left| \frac{2\rho - \epsilon^{ij}F(z)}{\epsilon^{ij}F(z)2\rho} \right| &< \frac{1}{2\rho} \\ \Longleftrightarrow \left( \left| \frac{2\rho - \epsilon^{ij}F(z)}{\epsilon^{ij}F(z)2\rho} \right| \right)^{2} &< \left(\frac{1}{2\rho} \right)^{2} \\ \Longleftrightarrow \left( 2\rho - \epsilon^{ij}F(z) \right) \left( \overline{2\rho - \epsilon^{ij}F(z)} \right) &< e^{-i\overline{jF(z)}\epsilon^{ij}F(z)} \\ \Longleftrightarrow 4\rho^{2} - 2\rho \left[ e^{-i\overline{jF(z)}} + \epsilon^{ij}F(z) \right] + F(z)\overline{F(z)} &< F(z)\overline{F(z)} \\ \Longleftrightarrow 4\rho^{2} - 2\rho \left[ e^{-i\overline{jF(z)}} + \epsilon^{ij}F(z) \right] &< 0 \\ \Longleftrightarrow 2\rho - 2\mathbb{R} \left[ \epsilon^{ij}F(z) \right] &< 0 \\ \Longleftrightarrow \& \left| \mathcal{R} \left[ \epsilon^{ij}F(z) \right] > \rho \\ \Longleftrightarrow \& \left( \epsilon^{ij}\frac{zf'(z)}{p\Psi(z)} \right) > \rho. \end{split}$$

This completes the proof of Theorem 2.

**Theorem 3.** *For* 0 *ϕ*<sup>1</sup> < *ϕ*2, *it is asserted that:*

$$k \cdot \mathcal{Q}\left(p, \not\!p\_2, \lambda, \eta\right) \subset 0 \text{-} \mathcal{Q}\left(p, \not\!p\_1, \lambda, \eta\right) \dots$$

**Proof.** Let *f*(*z*) ∈ *k*-Q (*p*, *ϕ*2, *λ*, *η*). Then:

$$\begin{split} &\frac{1}{p-\eta} \left[ \left( \epsilon^{j\lambda} - \phi\_1 \cos \lambda \right) \left( p-\eta \right) \frac{zf'(z)}{p\psi(z)} + \varphi\_1 \cos \lambda \left( \frac{(zf'(z))'}{\psi(z)} - \eta \right) \right] \\ &= \frac{\varrho\_1}{\varrho\_2} \left[ \left( \epsilon^{j\lambda} - \eta\_2 \cos \lambda \right) \frac{zf'(z)}{p\psi(z)} + \frac{\varrho\_2 \cos \lambda}{(p-\eta)} \left( \frac{(zf'(z))'}{p\psi(z)} - \eta \right) \right] \\ &\quad - \left( \frac{\varrho\_1 - \varrho\_2}{\varrho\_2} \right) \epsilon^{j\lambda} \frac{sf'(z)}{p\psi(z)} \\ &= \frac{\varrho\_1}{\varrho\_2} H\_1(z) + \left( 1 - \frac{\varrho\_1}{\varrho\_2} \right) H\_2(z) = H(z) \,, \end{split}$$

where:

$$H\_1\left(z\right) = \left(\epsilon^{i\lambda} - \varrho\_2 \cos \lambda\right) \frac{zf'\left(z\right)}{p\psi\left(z\right)} + \frac{\varrho\_2 \cos \lambda}{\left(p - \eta\right)} \left(\frac{\left(zf'z\right)'}{\Psi'\left(z\right)} - \eta\right) \in \mathcal{P}\left(\mathfrak{h}\_{k,\rho}\right) \subset \mathcal{P}\left(\rho\right)$$

and:

$$H\_2\left(z\right) = e^{i\lambda} \frac{zf'\left(z\right)}{p\psi\left(z\right)} \in \mathcal{P}\left(\rho\right).$$

Since P(*ρ*) is a convex set (see [27]), we therefore have *H*(*z*) ∈ P(*ρ*). This implies that *f* ∈ 0-Q (*p*, *ϕ*1, *λ*, *η*). Thus:

$$k\text{-}\mathcal{Q}\left(p,\,\varphi\_2,\lambda,\,\eta\right) \subset 0\text{-}\mathcal{Q}\left(p,\,\varphi\_1,\lambda,\,\eta\right)\dots$$

The proof of Theorem 3 is now completed.

**Theorem 4.** *Let φ* > 0 *and λ* < *<sup>π</sup>* <sup>2</sup> *. Then:*

$$k\text{-}\mathbb{Q}(p,\phi,\lambda,\eta,\xi^{\circ}) \subset k\text{-}\mathbb{K}(p,0,\xi).$$

**Proof.** Let *f* ∈ *k*-Q(*p*, *φ*, *λ*, *η*, *ξ*), and suppose that:

$$\frac{f'(z)}{\psi'(z)} = p\left(z\right),\tag{8}$$

where *p* (*z*) is analytic and *p* (0) = 1. Now, by differentiating both sides of (8) with respect to *z*, we have:

$$\frac{(zf'(z))'}{\Psi'(z)} = zp'(z) + p(z)\varepsilon(z),\tag{9}$$

where:

$$\varepsilon(z) = \frac{\left(z\psi'\left(z\right)\right)'}{\psi'(z)}.$$

By using (8) and (9) in (4), we arrive at:

$$\begin{split} \mathcal{M}\left(\phi,\lambda,\eta,f,\psi\right) &= \left(e^{i\lambda} - \phi\cos\lambda\right) \frac{p(z)}{p} + \frac{\phi\cos\lambda}{p-\eta} \left(zp'(z) + p(z)\varepsilon(z) - \eta\right) \\ &= \frac{\phi\cos\lambda}{p-\eta} z p'(z) + \left(\frac{e^{i\lambda}}{p} - \frac{\phi\cos\lambda}{p-\varepsilon(z)}\right) \left(\frac{\phi\cos\lambda}{p-\eta}\right) p(z) - \frac{\eta\phi\cos\lambda}{p-\eta} \\ &= B\left(z\right) z p'\left(z\right) + \mathcal{C}\left(z\right) p\left(z\right) + D\left(z\right) .\end{split} \tag{10}$$

where:

$$B\left(z\right) = \frac{\phi\cos\lambda}{p-\eta}\prime$$

$$C(z) = \frac{e^{i\lambda} \left(p - \eta\right) - \phi \cos \lambda \left(p - \eta\right) + \phi \cos \lambda \varepsilon(z) p}{p(p - \eta)}$$

and:

$$D\left(z\right) = \frac{\eta\phi\cos\lambda}{p-\eta}.$$

Now, since *f* ∈ *k*-Q(*p*, *φ*, *λ*, *η*, *ξ*), we have:

$$B\left(z\right)z p'\left(z\right) + \mathcal{C}\left(z\right)p\left(z\right) + D\left(z\right) \prec p\_k\left(z\right),\tag{11}$$

which, upon replacing *p* (*z*) by:

$$p\_\*\left(z\right) = p\left(z\right) - 1\_{\prime}$$

and *pk* (*z*) by:

$$p\_k^\*\left(z\right) = p\_k\left(z\right) - 1\_{\prime}$$

shows that the above subordination in (11) becomes as follows:

$$B\left(z\right)z p\_{\ge}^{\prime}\left(z\right) + \mathbb{C}\left(z\right)p\_{\ge}\left(z\right) + D\_{\ast}\left(z\right) \prec p\_{k}^{\ast}\left(z\right),\tag{12}$$

where:

$$D\_\*\left(z\right) = \mathcal{C}\left(z\right) + D\left(z\right) - 1.$$

We now apply Lemma 2 with:

*A* = 0

and

$$p\_\*\left(z\right) \prec p\_k^\*\left(z\right).$$

We thus find that:

$$\frac{f'(z)}{\psi'(z)} = p\left(z\right) \prec p\_k^\*\left(z\right). \tag{13}$$

This complete the proof of Theorem 4.

For *f* ∈ A, we next consider the integral operator defined by:

$$F\left(z\right) = I\_{\text{ll}}\left[f\right] = \frac{m+1}{z^m} \int\_0^z t^{m-1} f\left(t\right) dt. \tag{14}$$

This operator was given by Bernardi [28] in the year 1969. In particular, the operator *I*<sup>1</sup> was considered by Libera [29]. We prove the following result.

**Theorem 5.** *Let f*(*z*) ∈ *k-*Q (*p*, *φ*, *λ*, *η*, *ξ*). *Then, Im* [ *f* ] ∈ K (*p*, 0, *ξ*).

**Proof.** Let the function *ψ* (*z*) be such that:

$$\mathcal{M}\left(\phi,\lambda,\eta,f,\psi\right) = \left(e^{i\lambda} - \phi\cos\lambda\right)\frac{zf'(z)}{p\psi(z)} + \frac{\phi\cos\lambda}{\left(p-\eta\right)}\left(\frac{\left(zf'(z)\right)'}{\psi'(z)} - \eta\right)^2$$

Then, according to [14], the function *G* = *Im* [ *f* ] ∈ CD (*k*, *δ*). Furthermore, from (14), we deduce that:

$$(1+m)\,f\,\left(z\right) = \left(1+m\right)F\left(z\right) + z\left(F\left(z\right)\right)'\tag{15}$$

.

and:

$$(1+m)\,\mathrm{g}\,(z) = (1+m)\,\mathrm{G}\,(z) + z\,(\mathrm{G}\,(z))'.\tag{16}$$

If we now put:

$$p\left(z\right) = \frac{F'\left(z\right)}{G'\left(z\right)}$$

and:

$$q\left(z\right) = \frac{1}{\left(m+1\right) + \left(\frac{zG''(z)}{G'(z)}\right)'}$$

then, by simple computations, we find that:

$$\frac{f'(z)}{\psi\_{\,\,}(z)} = \frac{(1+m)\,F'(z) + zF'''(z)}{(1+m)\,G'(z) + zG'''(z)}$$

or, equivalently, that:

$$\frac{f'(z)}{\psi'(z)} = p\left(z\right) + zp'\left(z\right)q\left(z\right). \tag{17}$$

We now let:

$$\frac{f'(z)}{\psi'(z)} = p\left(z\right) + zp'\left(z\right)q\left(z\right) = h\left(z\right),\tag{18}$$

where the function *h* (*z*) is analytic in E with *h* (0) = 1. Then, by using (18), we have:

$$\frac{\left(zf\left(z\right)\right)'}{\left(\psi'\left(z\right)\right)'} = zh'\left(z\right) + \varepsilon\left(z\right)h\left(z\right),\tag{19}$$

where:

$$
\varepsilon(z) = \frac{(z\psi'(z))'}{\psi'(z)}.
$$

Furthermore, by using (18) and (19) in (4), we obtain:

$$\begin{split} \mathcal{M}\left(a,\emptyset,\gamma,\lambda,\delta,\delta'\right) &= \left(\epsilon^{i\lambda} - \theta\cos\lambda\right) \frac{zf'(z)}{\psi'(z)} + \frac{\phi\cos\lambda}{p-\eta} \left(\frac{\left(zf'\left(z\right)\right)'}{\psi'\left(z\right)} - \eta\right) \\ &= \left(\epsilon^{i\lambda} - \theta\cos\lambda\right) + \frac{\phi\cos\lambda}{p-\eta} zh'(z) + \left[zh'\left(z\right) + \epsilon\left(z\right)h\left(z\right) - \eta\right] \\ &= \frac{\phi\cos\lambda}{p-\eta} zh'(z) + \left(\epsilon^{i\lambda} - \phi\cos\lambda + \frac{\phi\cos\lambda}{p-\eta}\right) h\left(z\right) - \frac{\eta\left(\phi\cos\lambda\right)}{p-\eta} zh'(z) \\ &= B\left(z\right) zh'\left(z\right) + C\left(z\right) h\left(z\right) + D\left(z\right), \end{split}$$

where:

$$B\left(z\right) = \frac{\oint \cos \lambda}{p - \eta},$$

$$C\left(z\right) = \frac{\left(\left(p - \eta\right)e^{i\lambda} - \left(p - \eta\right)\phi\cos\lambda + \phi\cos\lambda\right)}{p - \eta}$$

and:

$$D\left(z\right) = \frac{\eta\left(\phi\cos\lambda\right)}{p-\eta}.$$

$$\text{Now, if we apply Lemma 1 with } A = 0 \text{, we get: }$$

$$\frac{f'(z)}{\Psi'(z)} = h\left(z\right) \prec p\_k\left(z\right). \tag{20}$$

Furthermore, from (18), we have:

$$p\left(z\right) + zp'\left(z\right)q\left(z\right) \prec p\_k\left(z\right).$$

By using Lemma 2 on (20), we obtain the desired result. This completes the proof of Theorem 5.
