Research article Special Issues

Exploring brain activity and transforming knowledge in visual and textual programming using neuroeducation approaches

  • Received: 05 May 2019 Accepted: 12 August 2019 Published: 02 September 2019
  • Eight (8) computer science students, novice programmers, who were in the first semester of their studies, participated in a field study in order to explore potential differences in their brain activity during programming with a visual programming language versus a textual programming language. The eight students were asked to develop two specific programs in both programming languages (a total of four tasks). The order of these programs was determined, while the order of languages in which they worked differed between the students. Measurement of cerebral activity was performed by the electroencephalography (EEG) imaging method. According to the analysis of the data it appears that the type of programming language did not affect the students' brain activity. Also, six students needed more time to successfully develop the programs they were asked with the first programming language versus the second one, regardless of the type of programming language that was first. In addition, it appears that six students did not show reducing or increasing brain activity as they spent their time on tasks and at the same time did not show a reduction or increase in the time they needed to develop the programs. Finally, the students showed higher average brain activity in the development of the fourth task than the third, and six of them showed higher average brain activity when developing the first versus the second program, regardless of the programming language. The results can contribute to: a) highlighting the need for a diverse educational approach for students when engaging in program development and b) identifying appropriate learning paths to enhance student education in programming.

    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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  • Eight (8) computer science students, novice programmers, who were in the first semester of their studies, participated in a field study in order to explore potential differences in their brain activity during programming with a visual programming language versus a textual programming language. The eight students were asked to develop two specific programs in both programming languages (a total of four tasks). The order of these programs was determined, while the order of languages in which they worked differed between the students. Measurement of cerebral activity was performed by the electroencephalography (EEG) imaging method. According to the analysis of the data it appears that the type of programming language did not affect the students' brain activity. Also, six students needed more time to successfully develop the programs they were asked with the first programming language versus the second one, regardless of the type of programming language that was first. In addition, it appears that six students did not show reducing or increasing brain activity as they spent their time on tasks and at the same time did not show a reduction or increase in the time they needed to develop the programs. Finally, the students showed higher average brain activity in the development of the fourth task than the third, and six of them showed higher average brain activity when developing the first versus the second program, regardless of the programming language. The results can contribute to: a) highlighting the need for a diverse educational approach for students when engaging in program development and b) identifying appropriate learning paths to enhance student education in programming.


    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,
    $
    (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).
    $

    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,
    $
    (1.2)

    analytic in $ \mathcal{U} $ such that

    $ 1+1b{p(z)1}1+Az1+Bz.
    $

    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))+(1b)(1+Bw(z))1+Bw(z).
    $

    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,
    $
    (1.3)

    in $ \mathcal{U} $ such that

    $ 1+1b{zg(z)g(z)1}1+Az1+Bz, 1B<A1,zU.
    $

    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)z0gα(t)p(t)tiβ1dt}1α+iβ,
    $

    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+iβ1)zf(z)f(z)=αzg(z)g(z)+zp(z)p(z)+iβ,
    $

    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+iβ1)zf(z)f(z)=αzg(z)g(z)+zp(z)p(z)+iβ.
    $

    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),
    $

    or

    $ z1iβf(z)f1(α+iβ)gα(z)=p(z), zU.
    $

    Since $ p\in \mathcal{P}[b, A, B], $ it follows that

    $ 1+1b{z1iβf(z)f1(α+iβ)gα(z)1}1+Az1+Bz,
    $

    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 }, $

    $ 12π2π0|p(reiθ)|2dθ1+[|b|2(AB)21]r21r2.
    $

    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)+(1b),˜pP[A,B].
    $

    Let $ \tilde{p}(z) = 1+\sum_{n = 1}^{\infty }c_{n}\, z^{n} $. Then

    $ 1+n=1pnzn=b(1+n=1cnzn)+(1b).
    $

    Comparing the coefficients of $ z^{n} $, we have

    $ pn=bcn.
    $

    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

    $ 12π2π0|p(reiθ)|2dθ=12π2π0|n=0pnrneinθ|2dθ=12π2π0(n=0|pn|2r2n)dθ=n=0|pn|2r2n1+|b|2(AB)2n=1r2n=1+|b|2(AB)2r21r2=1+(|b|2(AB)21)r21r2.
    $

    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+,(zU),
    $

    then for any complex number $ t $,

    $ |ω2tω21|max{1,|t|}.
    $

    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 $,

    $ |p2μp21||b|(AB)max{1,|μb(AB)+B|}={|b|(AB),if|μb(AB)+B|1,|b|(AB)|μb(AB)+B|,if|μb(AB)+B|1.
    $

    This result is sharp.

    Proof. Let $ p\in \mathcal{P}[b, A, B] $. Then we have

    $ 1+1b{p(z)1}1+Az1+Bz,
    $

    or, equivalently

    $ p(z)1+[bA+(1b)B]z1+Bz=1+b(AB)n=1(B)n1zn,
    $

    which would further give

    $ 1+p1z+p2z2+=1+b(AB)ω(z)+b(AB)(B)ω2(z)+=1+b(AB)(ω1z+ω2z2+) +b(AB)(B)(ω1z+ω2z2+)2+=1+b(AB)ω1z+b(AB){ω2Bω21}z2+.
    $

    Comparing the coefficients of $ z $ and $ z^{2} $, we obtain

    $ p1=b(AB)ω1p2=b(AB)ω2b(AB)Bω21.
    $

    By a simple computation,

    $ |p2μp21|=|b|(AB)|ω2(μb(AB)+B)ω21|.
    $

    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+(1b)B)z21+Bz2=1+b(AB)z2+b(AB)(B)z4+,p1(z)=1+(bA+(1b)B)z1+Bz=1+b(AB)z+b(AB)(B)z2+.
    $

    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].
    $

    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,
    $

    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).
    $

    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|=|L1A(t)D(t)dt|<d(r)L1|dD(t)|=d(r)D(z0).
    $

    This implies that

    $ |N(z0)D(z0)c(r)|<d(r).
    $

    Therefore

    $ N(z)D(z)P[b,A,B].
    $

    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.
    $

    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,zU,
    $

    has a univalent solution in $ \mathcal{U} $ given by

    $ q(z)={zβ1+γ(1+Bz)β1(AB)/Bβ1z0tβ1+γ1(1+Bt)β1(AB)/Bdtγβ1,B0,zβ1+γeβ1Azβ1z0tβ1+γ1eβ1Atdtγβ1,B=0.
    $

    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)+γ1+Az1+Bz,
    $

    then

    $ p(z)q(z)1+Az1+Bz,
    $

    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, $

    $ |b3μb22|(AB)2max{1,|2(AB)μ(A2B))|}.
    $

    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)AB2B=z+12(AB)z3+,B0,zeA2z2=z+A2z3+,B=0,
    $

    or

    $ g2(z)={z(1+Bz)ABB,B0,zeAz,B=0,={z+(AB)z2+12(AB)(A2B)z3+,B0,z+Az2+12A2z3+,B=0.
    $

    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+iβz0tc+iβ1gα(t)dt,
    $
    (3.1)

    is in $ \mathcal{S}^{\ast }[A, B] $. In addition

    $ RezG(z)G(z)>δ=min|z|=1Req(z),
    $

    where

    $ q(z)={1αα+iβ+c2F1(1;α(1AB);α+iβ+c+1;Bz1+Bz)(c+iβ),B0,1αα+iβ+c1F1(1;α+iβ+c+1;αAz)(c+iβ),B=0.
    $

    Proof. From $ \left(3.1\right), $ we have

    $ zc+iβGα(z)=(c+α+iβ)z0tc+iβ1gα(t)dt.
    $

    Differentiating and rearranging gives

    $ (c+α+iβ)gα(z)Gα(z)=(c+iβ)+αp(z),
    $
    (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).
    $

    Since $ g\in \mathcal{S}^{\ast }[A, B] $, it follows that

    $ p(z)+zp(z)αp(z)+(c+iβ)1+Az1+Bz.
    $

    Now by using Lemma 2.6, for $ \beta _{1} = \alpha $ and $ \gamma = c+i\beta, $ we obtain

    $ p(z)q(z)1+Az1+Bz,
    $

    where

    $ q(z)={zc+α+iβ(1+Bz)α(AB)/Bαz0tc+α+iβ1(1+Bt)α(AB)/Bdtc+iβα,B0,zc+α+iβeαAzαz0tc+α+iβ1eαAtdtc+iβα,B=0.
    $

    Now by using the properties of the familiar hypergeometric functions proved in [15], we have

    $ q(z)={1αα+iβ+c2F1(1;α(1AB);α+iβ+c+1;Bz1+Bz)(c+iβ),B0,α+iβ+c1F1(1;α+iβ+c+1;αAz)(c+iβ),B=0.
    $

    This implies that

    $ p(z)q(z)={1αα+iβ+c2F1(1;α(1AB);α+iβ+c+1;Bz1+Bz)(c+iβ),B0,1αα+iβ+c1F1(1;α+iβ+c+1;αAz)(c+iβ),B=0,
    $

    and

    $ RezG(z)G(z)=Rep(z)>δ=min|z|=1Req(z).
    $

    Theorem 3.3. Let $ g\in \mathcal{S}^{\ast }[A, B]. $ Then

    $ S(z)=z0tc+iβ1gα(t)dt,
    $

    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

    $ RezD1(z)D1(z)=Re{(c+iβ)+αzg(z)g(z)}=c+αRezg(z)g(z).
    $

    Since $ g\in \mathcal{S}^{\ast }[A, B]\subset \mathcal{S}^{\ast }\left(\frac{ 1-A}{1-B}\right) $, see [10], it follows that

    $ RezD1(z)D1(z)>c+α(1A1B)>0.
    $

    Also

    $ ReD1(z)N1(z)=Re{(c+iβ)+αzg(z)g(z)}>c+α(1A1B)>0.
    $

    Now by using Lemma 2.5$, $ we have

    $ ReD1(z)N1(z)>0 or RezS(z)S(z)>0.
    $

    By the mean value theorem for harmonic functions,

    $ RezS(z)S(z)|z=0=12π2π0RereiθS(reiθ)S(reiθ)dθ.
    $

    Therefore

    $ 2π0RereiθS(reiθ)S(reiθ)dθ=2πRe{c+iβ+αzg(z)g(z)}|z=0=2π(c+α).
    $

    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+iβzcz0tc1fα+iβ(t)dt]1α+iβ
    $
    (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+iβzcz0tc1(f(t))α+iβdt.
    $

    This implies that

    $ zcFα+iβ(z)=(c+α+iβ)z0tc1(f(t))α+iβdt.
    $

    Differentiate both sides and rearrange, we get

    $ czc1Fα+iβ(z)+(α+iβ)zcFα+iβ1(z)F(z)=(c+α+iβ)zc1(f(z))α+iβ,
    $

    and

    $ z1iβF(z)F1(α+iβ)(z)=1α+iβ{(c+α+iβ)ziβfα+iβ(z)cziβFα+iβ(z)}.
    $

    Now from $ (3.1) $, we have

    $ z1iβF(z)F1(α+iβ)Gα(z)=1α+iβ{(c+α+iβ)ziβfα+iβ(z)cziβc(c+α+iβ)z0tc1(f(t))α+iβdt}(c+α+iβ)zc+iβz0tc+iβ1gα(t)dt=1α+iβ{(zcfα+iβ(z)cz0tc1(f(t))α+iβdt}z0tc+iβ1gα(t)dt:=N(z)D(z).
    $

    With this, note that

    $ N(z)D(z)=1α+iβ{(czc1fα+iβ(z)+(α+iβ)zcfα+iβ1(z)f(z)czc1(f(z))α+iβ}zc+iβ1gα(z)=z1iβf(z)f1(α+iβ)(z)gα(z),
    $

    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

    $ z1iβF(z)F1(α+iβ)(z)Gα(z)P[b,A,B].
    $

    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)zcz0tc1gα(t)dt
    $

    belong to $ S^{\ast }\left(\delta _{1}\right) $, where

    $ δ1=(1+2c)+(1+2c)2+8α4α,(see [16]).
    $

    Hence $ G $ is starlike when $ g\in \mathcal{S}^{\ast }, $ and

    $ Fα(z)=(α+c)zcz0tc1gα(t)dt
    $

    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

    $ |a33+α+iβ2(2+α+iβ)a22|AB2|2+α+iβ|[α+|b|max{2,|b(AB)+2B|}].
    $

    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+iβ1)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+(6a34a22)z2+,zf(z)f(z)=1+a2z+(2a3a22)z2+,zg(z)g(z)=1+b2z+(2b3b22)z2+,zp(z)p(z)=p1z+(2p2p21)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(2b3b22)+2p2p21+a22(3+α+iβ).
    $
    (4.4)

    From $ \left(4.4\right) $, we have

    $ |a33+α+iβ2(2+α+iβ)a22|=|α(b312b22)+(p212p21)2+α+iβ|α|b312b22||2+α+iβ|+|p212p21||2+α+iβ|.
    $

    Now by using Theorem 3.1 and Lemma 2.3, both with $ \mu = \frac{1}{2} $, we obtain

    $ |b312b22|AB2max{1,|B|}=AB2,
    $

    and

    $ |p212p21||b|(AB)max{1,12|b(AB)+2B|}.
    $

    Therefore, we have

    $ |a33+α+iβ2(2+α+iβ)a22|AB2|2+α+iβ|[α+2|b|max{1,12|b(AB)+2B|}].
    $

    The equality

    $ |b312b22|=AB2
    $

    for $ B\neq 0 $ can be obtained for

    $ g(z)={z(1+Bz)ABB=z+(AB)z2+12(AB)(A2B)z3+,z(1+Bz2)AB2B=z+12(AB)z3+.
    $

    Similarly, the equality

    $ |b312b22|=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+(1b)B)z1+Bz or p1(z)=1+(bA+(1b)B)z21+Bz2.
    $

    Corollary 4.4. For $ A = 1, $ $ B = -1 $ and $ b = 1 $, we have the result proved in [8]:

    $ |a33+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

    $ |a323a22|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|(AB)(α+|b|)|1+α+iβ|.
    $

    (ii) If $ \alpha = 0, $ then

    $ |a3||b|(AB)|2+iβ|max{1,|b(AB)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|.
    $

    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|AB and |p1||b|(AB).
    $

    This implies that

    $ |a2|(α+|b|)(AB)|1+α+iβ|.
    $

    Equality can be obtained by the functions

    $ g(z)=z(1+Bz)ABB, B0 and p(z)=1+[bA+(1b)B]z1+Bz.
    $

    (ⅱ) Let $ \alpha = 0. $ Then from $ \left(4.3\right) $ and $ \left(4.4 \right) $, we have

    $ (2+iβ)a3=p212p21+(3+iβ)p212(1+iβ)2=p212(1(3+iβ)(1+iβ)2)p21.
    $

    This implies

    $ |a3|=1|2+iβ||p2μp21|,
    $

    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|(AB)|2+iβ|max{1,|b(AB)2(2β2+iβ)(1+iβ)2)+B|}.
    $

    Sharpness can be attained by the functions

    $ p0(z)=1+(bA+(1b)B)z21+Bz2=1+b(AB)z2+b(AB)(B)z4+,p1(z)=1+(bA+(1b)B)z1+Bz=1+b(AB)z+b(AB)(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)M1α(r)[1(1r)α(ABB)],0<α1,C(α,b,A,B)m1α(r)[1(1r)α(ABB)],α>1,
    $

    where

    $ C(α,b,A,B)=2π|b|B[(AB)+1α].
    $

    Proof. As $ f $ is a generalized Bazilevič function associated by the quadruple $ (\alpha, 0, g, p) $, we have

    $ zf(z)=f1α(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|f1α(z)gα(z)p(z)|dθ,M1α(r)2π0r0|αg(z)gα1(z)p(z)+gα(z)p(z)|dsdθ,M1α(r){2π0r0α|gα(z)|s|h(z)p(z)|dsdθ+2π0r0|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))M1α(r)×r0sα1(1|B|s)αBAB{α2π0|h(z)|2dθ2π0|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|(AB)r1B2r2,
    $

    for $ p\in \mathcal{P}[b, A, B] $ (see [19]), we can write

    $ Lr(f(z))2πM1α(r)×r0sα1(1|B|s)αBAB{α1+[(AB)21]s21s21+[|b|2(AB)21]s21s2+|b|(AB)s1B2s2}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))2πM1α(r)[|b|(AB)2+|b|(AB)]r01(1s)αBAB+1ds=C(α,b,A,B)M1α(r)[1(1r)α(ABB)],
    $

    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)m1α(r)[1(1r)α(ABB)].
    $

    Corollary 4.8. For $ g\in \mathcal{S}^{\ast } $ and $ p\in \mathcal{P}(b), $ we have

    $ Lr(f(z)){2π|b|(2+1α)M1α(r)[1(1r)2α1],0<α1,2π|b|(2+1α)m1α(r)[1(1r)2α1],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(AB+1α)limr1M1α(r),0<α1,1n|b|B(AB+1α)limr1m1α(r),α>1.
    $

    Proof. By Cauchy's theorem for $ z = re^{i\theta }, $ $ n\geq 2, $ we have

    $ nan=12πrn2π0zf(z)einθdθ.
    $

    Therefore

    $ n|an|12πrn2π0|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(AB+1α)M1α(r)[1(1r)αABB]).
    $

    Hence, by taking $ r $ approaches $ 1^{-} $,

    $ |an|1n|b|B(AB+1α)limr1M1α(r).
    $

    For $ \alpha > 1, $ we have

    $ |an|1n|b|B(AB+1α)limr1m1α(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)|αα(1B2)+(AB)(|b|BRe(b))1Brα2F1(α(1AB)+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)f1α(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)=αz0t1gα(t)p(t)dt,
    $

    and so

    $ |f(z)|αα|z|0|t1||gα(t)||p(t)|d|t|,=αr0s1|gα(t)||p(t)|ds.
    $

    Now by using the results

    $ |g(z)|r(1|B|r)ABB,B0, (see [10]), 
    $

    and

    $ |p(z)|1+|b|(AB)rB[(AB)Re{b}+B]r21|B|2r2, (see [6]), 
    $

    we have

    $ |f(z)|ααr0s1sα(1|B|s)α(1AB)1+|b|(AB)sB[(AB)Re{b}+B]s21|B|2s2dsα(1B2)+(AB)(|b|BRe{b})1+|B|r0sα1(1|B|s)α(1AB)1ds.
    $

    Putting $ s = ru, $ we have

    $ |f(z)|αα(1B2)+(AB)(|b|BRe{b})1Brα10uα1(1|B|ru)α(1AB)1du=(1B2)+(AB)(|b|BRe{b})1Brα 2F1(α(1AB)+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)|α2αrα2F1(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.


    Acknowledgments



    This research was partially supported by Fulbright Foundation–Greece, The American College of Greece, The Institute of Educational Policy, Greece, BiHeLab, Ionian University, Greece and Villanova University, USA. I would like to thank Dr Plerou for her support in EEG analysis, Professor Papalaskari and Mrs Giannopoulou for their comments.

    Conflict of interest



    The authors declare no conflicts of interest.

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    沈阳化工大学材料科学与工程学院 沈阳 110142

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