Citation: Spyridon Doukakis. Exploring brain activity and transforming knowledge in visual and textual programming using neuroeducation approaches[J]. AIMS Neuroscience, 2019, 6(3): 175-190. doi: 10.3934/Neuroscience.2019.3.175
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Let $ \mathfrak{A} $ denote the family of all functions $ f $ which are analytic in the open unit disc $ \mathcal{U} = \left \{ z:\left \vert z\right \vert < 1\right \} $ and satisfying the normalization
$ f(z)=z+∞∑n=2anzn, $
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(1.1) |
while by $ \mathcal{S} $ we mean the class of all functions in $ \mathfrak{A} $ which are univalent in $ \mathcal{U}. $ Also let $ \mathcal{S}^{\ast } $ and $ \mathcal{C} $ denote the familiar classes of starlike and convex functions, respectively. If $ f $ and $ g $ are analytic functions in $ \mathcal{U }, $ then we say that $ f $ is subordinate to $ g, $ denoted by $ f\prec g, $ if there exists an analytic Schwarz function $ w $ in $ \mathcal{U} $ with $ w\left(0\right) = 0 $ and $ \left \vert w(z)\right \vert < 1 $ such that $ f(z) = g\left(w(z)\right). $ Moreover if the function $ g $ is univalent in $ \mathcal{U} $, then
$ f(z)≺g(z)⇔f(0)=g(0) and f(U)⊂g(U). $
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For arbitrary fixed numbers $ A, $ $ B $ and $ b $ such that $ A $, $ B $ are real with $ -1\leq B < A\leq 1 $ and $ b\in \mathbb{C} \backslash \{0\} $, let $ \mathcal{P}[b, A, B] $ denote the family of functions
$ p(z)=1+∞∑n=1pnzn, $
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(1.2) |
analytic in $ \mathcal{U} $ such that
$ 1+1b{p(z)−1}≺1+Az1+Bz. $
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Then, $ p\in \mathcal{P}[b, A, B] $ can be written in terms of the Schwarz function $ w $ by
$ p(z)=b(1+Aw(z))+(1−b)(1+Bw(z))1+Bw(z). $
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By taking $ b = 1-\sigma $ with $ 0\leq \sigma < 1, $ the class $ \mathcal{P} [b, A, B] $ coincides with $ \mathcal{P}[\sigma, A, B], $ defined by Polatoğ lu [17,18] (see also [2]) and if we take $ b = 1, $ then $ \mathcal{ P}[b, A, B] $ reduces to the familiar class $ \mathcal{P}[A, B] $ defined by Janowski [10]. Also by taking $ A = 1, $ $ B = -1 $ and $ b = 1 $ in $ \mathcal{P} [b, A, B] $, we get the most valuable and familiar set $ \mathcal{P} $ of functions having positive real part. Let $ \mathcal{S}^{\ast }[A, B, b] $ denote the class of univalent functions $ g $ of the form
$ g(z)=z+∞∑n=2bnzn, $
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(1.3) |
in $ \mathcal{U} $ such that
$ 1+1b{zg′(z)g(z)−1}≺1+Az1+Bz, −1≤B<A≤1,z∈U. $
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Then $ S^* [A, B] : = S^* [A, B, 1] $ and the subclass $ \mathcal{S}^{\ast}[1,-1,1] $ coincides with the usual class of starlike functions.
The set of Bazilevič functions in $ \mathcal{U} $ was first introduced by Bazilevič [7] in 1955. He defined the Bazilevič function by the relation
$ f(z)={(α+iβ)z∫0gα(t)p(t)tiβ−1dt}1α+iβ, $
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where $ p\in \mathcal{P}, $ $ g\in \mathcal{S}^{\ast }, $ $ \beta $ is real and $ \alpha > 0 $. In 1979, Campbell and Pearce [8] generalized the Bazilevi č functions by means of the differential equation
$ 1+zf′′(z)f′(z)+(α+iβ−1)zf′(z)f(z)=αzg′(z)g(z)+zp′(z)p(z)+iβ, $
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where $ \alpha +i\beta \in \mathbb{C} -\left \{ \text{negative integers}\right \}. $ They associate each generalized Bazilevič functions with the quadruple $ (\alpha, \beta, g, p). $
Now we define the following subclass.
Definition 1.1. Let $ g $ be in the class $ \mathcal{S}^{\ast }[A, B] $ and let $ p\in \mathcal{P} [b, A, B] $. Then a function $ f $ of the form $ \left(1.1 \right) $ is said to belong to the class of generalized Bazilevič function associated with the quadruple $ (\alpha, \beta, g, p) $ if $ f $ satisfies the differential equation
$ 1+zf′′(z)f′(z)+(α+iβ−1)zf′(z)f(z)=αzg′(z)g(z)+zp′(z)p(z)+iβ. $
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where $ \alpha + i\beta \in \mathbb{C} - \{{\text {negative integers}} \} $.
The above differential equation can equivalently be written as
$ zf′(z)f(z)=(g(z)z)α(zf(z))α+iβp(z), $
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or
$ z1−iβf′(z)f1−(α+iβ)gα(z)=p(z), z∈U. $
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Since $ p\in \mathcal{P}[b, A, B], $ it follows that
$ 1+1b{z1−iβf′(z)f1−(α+iβ)gα(z)−1}≺1+Az1+Bz, $
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where $ g\in \mathcal{S}^{\ast }[A, B]. $
Several research papers have appeared recently on classes related to the Janowski functions, Bazilevič functions and their generalizations, see [3,4,5,13,16,21,22].
The following are some results that would be useful in proving the main results.
Lemma 2.1. Let $ p\in \mathcal{P}[b, A, B] $ with $ b\neq 0, $ $ -1\leq B < A\leq 1, $ and has the form $ \left(1.2\right). $ Then for $ z = re^{i\theta }, $
$ 12π∫2π0|p(reiθ)|2dθ≤1+[|b|2(A−B)2−1]r21−r2. $
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Proof. The proof of this lemma is straightforward but we include it for the sake of completeness. Since $ p\in \mathcal{P}[b, A, B] $, we have
$ p(z)=b˜p(z)+(1−b),˜p∈P[A,B]. $
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Let $ \tilde{p}(z) = 1+\sum_{n = 1}^{\infty }c_{n}\, z^{n} $. Then
$ 1+∞∑n=1pnzn=b(1+∞∑n=1cnzn)+(1−b). $
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Comparing the coefficients of $ z^{n} $, we have
$ pn=bcn. $
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Since $ \left \vert c_{n}\right \vert \leq A-B $ [20], it follows that $ \left \vert p_{n}\right \vert \leq \left \vert b\right \vert (A-B) $ and so
$ 12π∫2π0|p(reiθ)|2dθ=12π∫2π0|∞∑n=0pnrneinθ|2dθ=12π∫2π0(∞∑n=0|pn|2r2n)dθ=∞∑n=0|pn|2r2n≤1+|b|2(A−B)2∞∑n=1r2n=1+|b|2(A−B)2r21−r2=1+(|b|2(A−B)2−1)r21−r2. $
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Thus the proof is complete.
Lemma 2.2. [1] Let $ \Omega $ be the family of analytic functions $ \omega $ on $ \mathcal{U} $, normalized by $ \omega (0) = 0 $, satisfying the condition $ \left \vert \omega (z)\right \vert < 1 $. If $ \omega \in \Omega $ and
$ ω(z)=ω1z+ω2z2+⋯,(z∈U), $
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then for any complex number $ t $,
$ |ω2−tω21|≤max{1,|t|}. $
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The above inequality is sharp for $ \omega (z) = z $ or $ \omega (z) = z^{2} $.
Lemma 2.3. Let $ p(z) = 1+\sum \limits_{n = 1}^{\infty }p_{n}z^{n}\in \mathcal{P}[b, A, B], $ $ b\in \mathbb{C}\backslash \{0\}, $ $ -1\leq B < A\leq 1. $ Then for any complex number $ \mu $,
$ |p2−μp21|≤|b|(A−B)max{1,|μb(A−B)+B|}={|b|(A−B),if|μb(A−B)+B|≤1,|b|(A−B)|μb(A−B)+B|,if|μb(A−B)+B|≥1. $
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This result is sharp.
Proof. Let $ p\in \mathcal{P}[b, A, B] $. Then we have
$ 1+1b{p(z)−1}≺1+Az1+Bz, $
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or, equivalently
$ p(z)≺1+[bA+(1−b)B]z1+Bz=1+b(A−B)∞∑n=1(−B)n−1zn, $
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which would further give
$ 1+p1z+p2z2+⋯=1+b(A−B)ω(z)+b(A−B)(−B)ω2(z)+⋯=1+b(A−B)(ω1z+ω2z2+⋯) +b(A−B)(−B)(ω1z+ω2z2+⋯)2+⋯=1+b(A−B)ω1z+b(A−B){ω2−Bω21}z2+⋯. $
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Comparing the coefficients of $ z $ and $ z^{2} $, we obtain
$ p1=b(A−B)ω1p2=b(A−B)ω2−b(A−B)Bω21. $
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By a simple computation,
$ |p2−μp21|=|b|(A−B)|ω2−(μb(A−B)+B)ω21|. $
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Now by using Lemma 2.2 with $ t = \mu b(A-B)+B $, we get the required result. Equality holds for the functions
$ p∘(z)=1+(bA+(1−b)B)z21+Bz2=1+b(A−B)z2+b(A−B)(−B)z4+⋯,p1(z)=1+(bA+(1−b)B)z1+Bz=1+b(A−B)z+b(A−B)(−B)z2+⋯. $
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Now we prove the following result by using a method similar to the one in Libera [12].
Lemma 2.4. Suppose that $ N $ and $ D $ are analytic in $ \mathcal{U} $ with $ N(0) = D(0) = 0 $ and $ D $ maps $ \mathcal{U} $ onto a many sheeted region which is starlike with respect to the origin. If $ \frac{N^{\prime }(z)}{D^{\prime }(z) }\in \mathcal{P}[b, A, B] $, then
$ N(z)D(z)∈P[b,A,B]. $
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Proof. Let $ \frac{N^{\prime }(z)}{D^{\prime }(z)}\in \mathcal{P}[b, A, B]. $ Then by using a result due to Attiya [6], we have
$ |N′(z)D′(z)−c(r)|≤d(r),|z|<r,0<r<1, $
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where $ c(r) = \frac{1-B\left[B+b\left(A-B\right) \right] r^{2}}{1-B^{2}r^{2}} $ and $ d\left(r\right) = \frac{\left \vert b\right \vert \left(A-B\right) r^{2}}{1-B^{2}r^{2}}. $ We choose $ A\left(z\right) $ such that $ \left \vert A\left(z\right) \right \vert < d\left(r\right) $ and
$ A(z)D′(z)=N′(z)−c(r)D′(z). $
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Now for a fixed $ z_{0} $ in $ \mathcal{U} $, consider the line segment $ L $ joining $ 0 $ and $ D\left(z_{0}\right) $ which remains in one sheet of the starlike image of $ \mathcal{U} $ by $ D. $ Suppose that $ L^{-1} $ is the pre-image of $ L $ under $ D $. Then
$ |N(z0)−c(r)D(z0)|=|∫z00(N′(t)−c(r)D′(t))dt|=|∫L−1A(t)D′(t)dt|<d(r)∫L−1|dD(t)|=d(r)D(z0). $
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This implies that
$ |N(z0)D(z0)−c(r)|<d(r). $
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Therefore
$ N(z)D(z)∈P[b,A,B]. $
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For $ A = -B = b = 1 $, we have the following result due to Libera [12].
Lemma 2.5. If $ N $ and $ D $ are analytic in $ \mathcal{U} $ with $ N(0) = D(0) = 0 $ and $ D $ maps $ \mathcal{U} $ onto a many sheeted region which is starlike with respect to the origin, then
$ N′(z)D′(z)∈P implies N(z)D(z)∈P. $
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Lemma 2.6. [14] If $ -1\leq B < A\leq 1, \beta _{1} > 0 $ and the complex number $ \gamma $ satisfies $ {Re}\left \{ \gamma \right \} \geq -\beta_{1}\left(1-A\right) /\left(1-B\right), $ then the differential equation
$ q(z)+zq′(z)β1q(z)+γ=1+Az1+Bz,z∈U, $
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has a univalent solution in $ \mathcal{U} $ given by
$ q(z)={zβ1+γ(1+Bz)β1(A−B)/Bβ1∫z0tβ1+γ−1(1+Bt)β1(A−B)/Bdt−γβ1,B≠0,zβ1+γeβ1Azβ1∫z0tβ1+γ−1eβ1Atdt−γβ1,B=0. $
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If $ p\left(z\right) = 1+p_{1}z+p_{2}z^{2}+\cdots $ is analytic in $ \mathcal{U} $ and satisfies
$ p(z)+zp′(z)β1p(z)+γ≺1+Az1+Bz, $
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then
$ p(z)≺q(z)≺1+Az1+Bz, $
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and $ q\left(z\right) $ is the best dominant.
Before proving the results for the generalized Bazilevič functions, let us discuss a few results related to the function $ g \in S^*[A, B] $.
Theorem 3.1. Let $ g\in \mathcal{S}^{\ast }[A, B] $ and of the form $ \left(1.3\right) $. Then for any complex number $ \mu, $
$ |b3−μb22|≤(A−B)2max{1,|2(A−B)μ−(A−2B))|}. $
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Proof. The proof of the result is the same as of Lemma 2.3. The result is sharp and equality holds for the function defined by
$ g1(z)={z(1+Bz2)A−B2B=z+12(A−B)z3+⋯,B≠0,zeA2z2=z+A2z3+⋯,B=0, $
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or
$ g2(z)={z(1+Bz)A−BB,B≠0,zeAz,B=0,={z+(A−B)z2+12(A−B)(A−2B)z3+⋯,B≠0,z+Az2+12A2z3+⋯,B=0. $
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Theorem 3.2. Let $ g\in \mathcal{S}^{\ast }[A, B]. $ Then for $ c > 0, $ $ \alpha > 0 $ and $ \beta $ any real number,
$ Gα(z)=c+α+iβzc+iβ∫z0tc+iβ−1gα(t)dt, $
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(3.1) |
is in $ \mathcal{S}^{\ast }[A, B] $. In addition
$ RezG′(z)G(z)>δ=min|z|=1Req(z), $
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where
$ q(z)={1αα+iβ+c2F1(1;α(1−AB);α+iβ+c+1;Bz1+Bz)−(c+iβ),B≠0,1αα+iβ+c1F1(1;α+iβ+c+1;−αAz)−(c+iβ),B=0. $
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Proof. From $ \left(3.1\right), $ we have
$ zc+iβGα(z)=(c+α+iβ)∫z0tc+iβ−1gα(t)dt. $
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Differentiating and rearranging gives
$ (c+α+iβ)gα(z)Gα(z)=(c+iβ)+αp(z), $
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(3.2) |
where $ p(z) = \frac{zG^{\prime }(z)}{G(z)} $. Then differentiating $ \left(3.2\right) $ logarithmically, we have
$ zg′(z)g(z)=p(z)+zp′(z)αp(z)+(c+iβ). $
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Since $ g\in \mathcal{S}^{\ast }[A, B] $, it follows that
$ p(z)+zp′(z)αp(z)+(c+iβ)≺1+Az1+Bz. $
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Now by using Lemma 2.6, for $ \beta _{1} = \alpha $ and $ \gamma = c+i\beta, $ we obtain
$ p(z)≺q(z)≺1+Az1+Bz, $
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where
$ q(z)={zc+α+iβ(1+Bz)α(A−B)/Bα∫z0tc+α+iβ−1(1+Bt)α(A−B)/Bdt−c+iβα,B≠0,zc+α+iβeαAzα∫z0tc+α+iβ−1eαAtdt−c+iβα,B=0. $
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Now by using the properties of the familiar hypergeometric functions proved in [15], we have
$ q(z)={1αα+iβ+c2F1(1;α(1−AB);α+iβ+c+1;Bz1+Bz)−(c+iβ),B≠0,α+iβ+c1F1(1;α+iβ+c+1;−αAz)−(c+iβ),B=0. $
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This implies that
$ p(z)≺q(z)={1αα+iβ+c2F1(1;α(1−AB);α+iβ+c+1;Bz1+Bz)−(c+iβ),B≠0,1αα+iβ+c1F1(1;α+iβ+c+1;−αAz)−(c+iβ),B=0, $
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and
$ RezG′(z)G(z)=Rep(z)>δ=min|z|=1Req(z). $
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Theorem 3.3. Let $ g\in \mathcal{S}^{\ast }[A, B]. $ Then
$ S(z)=∫z0tc+iβ−1gα(t)dt, $
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is $ (\alpha +c) $-valent starlike, where $ \alpha > 0, $ $ c > 0 $ and $ \beta $ is a real number.
Proof. Let $ D_{1}(z) = zS^{\prime }(z) = z^{c+i\beta }g^{\alpha }(z) $ and $ N_{1}(z) = S(z). $ Then
$ RezD′1(z)D1(z)=Re{(c+iβ)+αzg′(z)g(z)}=c+αRezg′(z)g(z). $
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Since $ g\in \mathcal{S}^{\ast }[A, B]\subset \mathcal{S}^{\ast }\left(\frac{ 1-A}{1-B}\right) $, see [10], it follows that
$ RezD′1(z)D1(z)>c+α(1−A1−B)>0. $
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Also
$ ReD′1(z)N′1(z)=Re{(c+iβ)+αzg′(z)g(z)}>c+α(1−A1−B)>0. $
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Now by using Lemma 2.5$, $ we have
$ ReD1(z)N1(z)>0 or RezS′(z)S(z)>0. $
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By the mean value theorem for harmonic functions,
$ RezS′(z)S(z)|z=0=12π∫2π0RereiθS′(reiθ)S(reiθ)dθ. $
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Therefore
$ ∫2π0RereiθS′(reiθ)S(reiθ)dθ=2πRe{c+iβ+αzg′(z)g(z)}|z=0=2π(c+α). $
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Now by using a result due to [9,p 212], we have that $ S $ is $ (c+\alpha) $-valent starlike function.
Now we are ready to discuss some results related to the defined generalized Bazilevič functions.
Theorem 4.1. Let $ f $ be a generalized Bazilevič function associated by the quadruple $ \left(\alpha, \beta, g, p\right), $ where $ g\in \mathcal{S} ^{\ast }[A, B] $ of the form $ \left(1.3\right) $ and $ p\in \mathcal{P} [b, A, B] $ of the form $ \left(1.2\right) $. Then for $ c > 0, $
$ F(z)=[c+α+iβzc∫z0tc−1fα+iβ(t)dt]1α+iβ $
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(4.1) |
is a generalized Bazilevič function associated by the quadruple $ (\alpha, \beta, G, p), $ where $ G\in \mathcal{S}^{\ast }[A, B, \delta] $, as defined by $ \left(3.1\right). $
Proof. From $ \left(4.1\right) $, we have
$ Fα+iβ(z)=c+α+iβzc∫z0tc−1(f(t))α+iβdt. $
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This implies that
$ zcFα+iβ(z)=(c+α+iβ)∫z0tc−1(f(t))α+iβdt. $
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Differentiate both sides and rearrange, we get
$ czc−1Fα+iβ(z)+(α+iβ)zcFα+iβ−1(z)F′(z)=(c+α+iβ)zc−1(f(z))α+iβ, $
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and
$ z1−iβF′(z)F1−(α+iβ)(z)=1α+iβ{(c+α+iβ)z−iβfα+iβ(z)−cz−iβFα+iβ(z)}. $
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Now from $ (3.1) $, we have
$ z1−iβF′(z)F1−(α+iβ)Gα(z)=1α+iβ{(c+α+iβ)z−iβfα+iβ(z)−cz−iβ−c(c+α+iβ)∫z0tc−1(f(t))α+iβdt}(c+α+iβ)zc+iβ∫z0tc+iβ−1gα(t)dt=1α+iβ{(zcfα+iβ(z)−c∫z0tc−1(f(t))α+iβdt}∫z0tc+iβ−1gα(t)dt:=N(z)D(z). $
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With this, note that
$ N′(z)D′(z)=1α+iβ{(czc−1fα+iβ(z)+(α+iβ)zcfα+iβ−1(z)f′(z)−czc−1(f(z))α+iβ}zc+iβ−1gα(z)=z1−iβf′(z)f1−(α+iβ)(z)gα(z), $
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which implies $ \frac{N^{\prime }(z)}{D^{\prime }(z)} \in \mathcal{P}[b, A, B] $. By Theorem 3.3, we know that $ D(z) = \int_{0}^{z}t^{c+i\beta -1}g^{\alpha }(t)dt $ is $ (\alpha +c) $-valent starlike. Therefore by using Lemma 2.4, we obtain
$ z1−iβF′(z)F1−(α+iβ)(z)Gα(z)∈P[b,A,B]. $
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This is the equivalent form of Definition 1.1. Hence the result follows.
Corollary 4.2. Let $ A = 1, B = -1 $ and $ \beta = 0 $ in Theorem 4.1. Then
$ Gα(z)=(α+c)zc∫z0tc−1gα(t)dt $
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belong to $ S^{\ast }\left(\delta _{1}\right) $, where
$ δ1=−(1+2c)+√(1+2c)2+8α4α,(see [16]). $
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Hence $ G $ is starlike when $ g\in \mathcal{S}^{\ast }, $ and
$ Fα(z)=(α+c)zc∫z0tc−1gα(t)dt $
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belongs to the class of Bazilevič functions associated by the quadruple $ (\alpha, 0, G, p). $
Theorem 4.3. Let $ f $ of the form $ \left(1.1\right) $ be a generalized Bazilevič function associated by the quadruple $ (\alpha, \beta, g, p) $, with $ g\in \mathcal{S}^{\ast }[A, B] $ of the form $ \left(1.3\right) $ and $ p\in \mathcal{P}[b, A, B] $ of the form $ \left(1.2\right) $. Then
$ |a3−3+α+iβ2(2+α+iβ)a22|≤A−B2|2+α+iβ|[α+|b|max{2,|b(A−B)+2B|}]. $
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This inequality is sharp.
Proof. Since $ f $ is a generalized Bazilevič function associated by the quadruple $ (\alpha, \beta, g, p) $, we have
$ 1+zf′′(z)f′(z)+(α+iβ−1)zf′(z)f(z)=αzg′(z)g(z)+zp′(z)p(z)+iβ. $
|
(4.2) |
As $ f, \ g $ and $ p $ respectively have the form $ \left(1.1\right), \left(1.3\right) $ and $ \left(1.2\right), $ it is easy to get
$ 1+zf′′(z)f′(z)=1+2a2z+(6a3−4a22)z2+⋯,zf′(z)f(z)=1+a2z+(2a3−a22)z2+⋯,zg′(z)g(z)=1+b2z+(2b3−b22)z2+⋯,zp′(z)p(z)=p1z+(2p2−p21)z2+⋯. $
|
Putting these values in $ \left(4.2\right) $ and comparing the coefficients of $ z, $ we obtain
$ (1+α+iβ)a2=αb2+p1. $
|
(4.3) |
Similarly by comparing the coefficients of $ z^{2} $ and rearranging, we have
$ 2(2+α+iβ)a3=α(2b3−b22)+2p2−p21+a22(3+α+iβ). $
|
(4.4) |
From $ \left(4.4\right) $, we have
$ |a3−3+α+iβ2(2+α+iβ)a22|=|α(b3−12b22)+(p2−12p21)2+α+iβ|≤α|b3−12b22||2+α+iβ|+|p2−12p21||2+α+iβ|. $
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Now by using Theorem 3.1 and Lemma 2.3, both with $ \mu = \frac{1}{2} $, we obtain
$ |b3−12b22|≤A−B2max{1,|B|}=A−B2, $
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and
$ |p2−12p21|≤|b|(A−B)max{1,12|b(A−B)+2B|}. $
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Therefore, we have
$ |a3−3+α+iβ2(2+α+iβ)a22|≤A−B2|2+α+iβ|[α+2|b|max{1,12|b(A−B)+2B|}]. $
|
The equality
$ |b3−12b22|=A−B2 $
|
for $ B\neq 0 $ can be obtained for
$ g(z)={z(1+Bz)A−BB=z+(A−B)z2+12(A−B)(A−2B)z3+⋯,z(1+Bz2)A−B2B=z+12(A−B)z3+⋯. $
|
Similarly, the equality
$ |b3−12b22|=A2 $
|
for $ B = 0 $ can be obtained for the function $ g_{\ast }(z) = $ $ ze^{\frac{A}{2} z^{2}} = z+\frac{A}{2}z^{3}+\cdots. $ Also equality for the functional $ \left \vert p_{2}-\frac{1}{2}p_{1}^{2}\right \vert $ can be obtained by the functions
$ p∘(z)=1+(bA+(1−b)B)z1+Bz or p1(z)=1+(bA+(1−b)B)z21+Bz2. $
|
Corollary 4.4. For $ A = 1, $ $ B = -1 $ and $ b = 1 $, we have the result proved in [8]:
$ |a3−3+α+iβ2(2+α+iβ)a22|≤α+2|2+α+iβ|. $
|
For $ \alpha = 1 $, $ \beta = 0 $, we have $ f\in \mathcal{K} $, the class of close-to-convex functions, and
$ |a3−23a22|≤1. $
|
The latter result has been proved in [11].
Theorem 4.5. Let $ f $ of the form $ \left(1.1\right) $ be a generalized Bazilevič function associated by the quadruple $ (\alpha, \beta, g, p) $, with $ g\in \mathcal{S}^{\ast }[A, B] $ and of the form $ \left(1.3\right) $ and $ p\in \mathcal{P}[b, A, B] $ of the form $ \left(1.2\right) $. Then
(i)
$ |a2|≤(A−B)(α+|b|)|1+α+iβ|. $
|
(ii) If $ \alpha = 0, $ then
$ |a3|≤|b|(A−B)|2+iβ|max{1,|b(A−B)2(1−(3+iβ)(1+iβ)2)+B|}. $
|
Both the above inequalities are sharp.
Proof. (ⅰ) From $ \left(4.3\right) $, we have
$ (1+α+iβ)a2=αb2+p1. $
|
This implies that
$ |a2|≤α|b2|+|p1||1+α+iβ|. $
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By using the coefficient bound for $ \mathcal{S}^{\ast }[A, B] $ along with the coefficient bound of $ \mathcal{P}[b, A, B] $, we have
$ |b2|≤A−B and |p1|≤|b|(A−B). $
|
This implies that
$ |a2|≤(α+|b|)(A−B)|1+α+iβ|. $
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Equality can be obtained by the functions
$ g∘(z)=z(1+Bz)A−BB, B≠0 and p∘(z)=1+[bA+(1−b)B]z1+Bz. $
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(ⅱ) Let $ \alpha = 0. $ Then from $ \left(4.3\right) $ and $ \left(4.4 \right) $, we have
$ (2+iβ)a3=p2−12p21+(3+iβ)p212(1+iβ)2=p2−12(1−(3+iβ)(1+iβ)2)p21. $
|
This implies
$ |a3|=1|2+iβ||p2−μp21|, $
|
where $ \mu = \frac{1}{2}-\frac{(3+i\beta)}{2\left(1+i\beta \right) ^{2}}. $ Now by using Lemma 2.3, we obtain
$ |a3|≤|b|(A−B)|2+iβ|max{1,|b(A−B)2(−2−β2+iβ)(1+iβ)2)+B|}. $
|
Sharpness can be attained by the functions
$ p0(z)=1+(bA+(1−b)B)z21+Bz2=1+b(A−B)z2+b(A−B)(−B)z4+⋯,p1(z)=1+(bA+(1−b)B)z1+Bz=1+b(A−B)z+b(A−B)(−B)z2+⋯. $
|
Corollary 4.6. For $ A = 1, $ $ B = -1 $ and $ b = 1 $, we have
$ |a2|≤2(α+1)|1+α+iβ|, $
|
and
$ |a3|=2|2+iβ|max{1,|(3+iβ)(1+iβ)2|}. $
|
In the final part of this paper, we look at some results for the generalized Bazilevič functions associated with $ \beta = 0 $.
Let $ C_{r} $ denote the closed curve which is the image of the circle $ \left \vert z\right \vert = r < 1 $ under the mapping $ w = $ $ f(z) $, and $ L_{r}\left(f(z)\right) $ denote the length of $ C_{r} $. Also let $ M(r) = {\max} _{\left \vert z\right \vert = r} \left \vert f(z)\right \vert \ $and $ m(r) = { \min}_{\left \vert z\right \vert = r} \left \vert f(z)\right \vert $. We now prove the following result.
Theorem 4.7. Let $ f $ be a generalized Bazilevič function associated by the quadruple $ (\alpha, 0, g, p) $. Then for $ B\neq 0 $,
$ Lr(f(z))≤{C(α,b,A,B)M1−α(r)[1−(1−r)α(A−BB)],0<α≤1,C(α,b,A,B)m1−α(r)[1−(1−r)α(A−BB)],α>1, $
|
where
$ C(α,b,A,B)=2π|b|B[(A−B)+1α]. $
|
Proof. As $ f $ is a generalized Bazilevič function associated by the quadruple $ (\alpha, 0, g, p) $, we have
$ zf′(z)=f1−α(z)gα(z)p(z), $
|
where $ g\in \mathcal{S}^{\ast }[A, B] $ and $ p\in \mathcal{P}[b, A, B] $. Since for $ z = re^{i\theta } $, $ 0 < r < 1 $,
$ Lr(f(z))=∫2π0|zf′(z)|dθ, $
|
we have for $ 0 < \alpha \leq 1 $,
$ Lr(f(z))=∫2π0|f1−α(z)gα(z)p(z)|dθ,≤M1−α(r)∫2π0∫r0|αg′(z)gα−1(z)p(z)+gα(z)p′(z)|dsdθ,≤M1−α(r){∫2π0∫r0α|gα(z)|s|h(z)p(z)|dsdθ+∫2π0∫r0|gα(z)|s|zp′(z)|dsdθ}, $
|
where $ \frac{zg^{\prime }(z)}{g(z)} = h\left(z\right) \in \mathcal{P}[A, B] $. Now by using the distortion theorem for Janowski starlike functions when $ B\neq 0 $ (see [10]) and the Cauchy-Schwarz inequality, we have
$ Lr(f(z))≤M1−α(r)×∫r0sα−1(1−|B|s)αB−AB{α√∫2π0|h(z)|2dθ√∫2π0|p(z)|2dθ+∫2π0|zp′(z)|dθ}ds. $
|
Now by using Lemma 2.1 for both the classes $ \mathcal{P}[b, A, B] $ and $ \mathcal{P}[A, B] $, along with the result
$ ∫2π0|zp′(z)|≤|b|(A−B)r1−B2r2, $
|
for $ p\in \mathcal{P}[b, A, B] $ (see [19]), we can write
$ Lr(f(z))≤2πM1−α(r)×∫r0sα−1(1−|B|s)αB−AB{α√1+[(A−B)2−1]s21−s2√1+[|b|2(A−B)2−1]s21−s2+|b|(A−B)s1−B2s2}ds. $
|
Since $ 1-\left \vert B\right \vert r\geq 1-r $ and $ 1- B^2 r^2\geq 1-r^2 $,
$ Lr(f(z))≤2πM1−α(r)[|b|(A−B)2+|b|(A−B)]∫r01(1−s)αB−AB+1ds=C(α,b,A,B)M1−α(r)[1−(1−r)α(A−BB)], $
|
where $ C(\alpha, b, A, B) = 2\pi |b| [(A-B) + (1/ \alpha)] B $.
When $ \alpha > 1, $ we can prove similarly as above to get
$ Lr(f(z))≤C(α,b,A,B)m1−α(r)[1−(1−r)α(A−BB)]. $
|
Corollary 4.8. For $ g\in \mathcal{S}^{\ast } $ and $ p\in \mathcal{P}(b), $ we have
$ Lr(f(z))≤{2π|b|(2+1α)M1−α(r)[1(1−r)2α−1],0<α≤1,2π|b|(2+1α)m1−α(r)[1(1−r)2α−1],α>1. $
|
Theorem 4.9. Let $ f $ be a generalized Bazilevič function associated by the quadruple $ (\alpha, 0, g, p) $, where $ g\in \mathcal{S}^{\ast }[A, B] $ and $ p\in \mathcal{P} [b, A, B] $. Then for $ B\neq 0 $,
$ |an|≤{1n|b|B(A−B+1α)limr→1−M1−α(r),0<α≤1,1n|b|B(A−B+1α)limr→1−m1−α(r),α>1. $
|
Proof. By Cauchy's theorem for $ z = re^{i\theta }, $ $ n\geq 2, $ we have
$ nan=12πrn∫2π0zf′(z)e−inθdθ. $
|
Therefore
$ n|an|≤12πrn∫2π0|zf′(z)|dθ,=12πrnLr(f(z)). $
|
By using Theorem 4.7 for the case $ 0 < \alpha \leq 1, $ we have
$ n|an|≤12πrn(2π|b|B(A−B+1α)M1−α(r)[1−(1−r)αA−BB]). $
|
Hence, by taking $ r $ approaches $ 1^{-} $,
$ |an|≤1n|b|B(A−B+1α)limr→1−M1−α(r). $
|
For $ \alpha > 1, $ we have
$ |an|≤1n|b|B(A−B+1α)limr→1−m1−α(r). $
|
Theorem 4.10. Let $ f $ be a generalized Bazilevič function represented by the quadruple $ (\alpha, 0, g, p), $ where $ g\in \mathcal{S}^{\ast }[A, B] $ and $ p\in \mathcal{P }[b, A, B]. $ Then for $ B\neq 0 $,
$ |f(z)|α≤α(1−B2)+(A−B)(|b|−BRe(b))1−Brα2F1(α(1−AB)+1;α;α+1;|B|r). $
|
Proof. Since $ f $ is a generalized Bazilevič function associated by the quadruple $ (\alpha, 0, g, p), $ by definition, we have
$ zf′(z)f1−α(z)gα(z)=p(z), $
|
where $ g\in \mathcal{S}^{\ast }[A, B] $ and $ p\in \mathcal{P}[b, A, B]. $ This implies that
$ fα(z)=α∫z0t−1gα(t)p(t)dt, $
|
and so
$ |f(z)|α≤α∫|z|0|t−1||gα(t)||p(t)|d|t|,=α∫r0s−1|gα(t)||p(t)|ds. $
|
Now by using the results
$ |g(z)|≤r(1−|B|r)A−BB,B≠0, (see [10]), $
|
and
$ |p(z)|≤1+|b|(A−B)r−B[(A−B)Re{b}+B]r21−|B|2r2, (see [6]), $
|
we have
$ |f(z)|α≤α∫r0s−1sα(1−|B|s)α(1−AB)1+|b|(A−B)s−B[(A−B)Re{b}+B]s21−|B|2s2ds≤α(1−B2)+(A−B)(|b|−BRe{b})1+|B|∫r0sα−1(1−|B|s)−α(1−AB)−1ds. $
|
Putting $ s = ru, $ we have
$ |f(z)|α≤α(1−B2)+(A−B)(|b|−BRe{b})1−Brα∫10uα−1(1−|B|ru)−α(1−AB)−1du=(1−B2)+(A−B)(|b|−BRe{b})1−Brα 2F1(α(1−AB)+1;α;α+1;|B|r), $
|
where $ {_{2}}F_{1} \left(a; b;c; z\right) $ is the hypergeometric function.
Corollary 4.11. For $ g\in \mathcal{S}^{\ast \mathit{\text{}}} $and $ p\in \mathcal{P}, $ we have
$ |f(z)|α≤2αrα2F1(2α+1;α;α+1;r). $
|
The research for the fourth author is supported by the USM Research University Individual Grant (RUI) 1001/PMATHS/8011038.
The authors declare that they have no conflict of interests.
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