1 Introduction

Let \(\mathbb {D}=\{z{\text {:}}\,|z|<1\}.\) For \(a,\,b,\,c\in \mathbb {C}\) and \(c\ne 0,\,-1,\,-2,\ldots ,\) the Gaussian hypergeometric function is defined by

$$\begin{aligned} F(a,\,b;\,c;\,z):= _2F_1(a,\,b;\,c;\,z)=\sum _{k=0}^{\infty }\frac{(a)_k(b)_k}{(c)_kk!}z^k,\quad z\in \mathbb {D}, \end{aligned}$$

where \((a)_k\) is the Pochhammer symbol defined by

$$\begin{aligned} (a)_k=\frac{\varGamma {(a+k)}}{\varGamma {(a)}}, \end{aligned}$$

k is a positive integer and \(\varGamma \) denotes the Gamma function. Finding the condition on the triplet \((a,\,b,\,c)\) so that the normalized Gaussian hypergeometric function \(zF(a,\,b;\,c;\,z)\) is starlike and convex in \(\mathbb {D}\) has been discussed by several authors, but the complete range on the triplet \((a,\,b,\,c)\) for this problem is still open. For example, see [6, 9] and the references therein.

In the given context, a function is starlike if it maps \(\mathbb {D}\) onto a domain starlike with respect to origin and convex if it maps \(\mathbb {D}\) onto a convex domain. \(\mathcal {S}^{*}\) and \(\mathcal {C}\) denote, respectively, the class of all starlike functions and convex functions in \(\mathbb {D}.\) These two classes are well known subclasses of the class \(\mathcal {S}\) of univalent (one-to-one) functions in \(\mathbb {D},\) which itself is contained in the class \(\mathcal {A}\) of analytic functions f,  with the normalization \(f(0)=f^{\prime }(0)-1=0.\) Naturally, \(\mathcal {S}\) and its subclasses are also assumed to have this normalization.

These two subclasses \(\mathcal {S}^{*}\) and \(\mathcal {C}\) also have their respective generalizations \(\mathcal {S}^{*}(\alpha )\) and \(\mathcal {C}(\alpha ),\,0\le \alpha <1,\) as class of starlike function of order \(\alpha \) and class of convex function of order \(\alpha ,\) respectively, with the analytic characterization.

$$\begin{aligned} f\in \mathcal {S}^{*}(\alpha ) \Leftrightarrow \mathrm{Re}\left( \frac{zf^{\prime }}{f}\right)>\alpha , \quad f\in \mathcal {C}(\alpha ) \Leftrightarrow \mathrm{Re}\left( 1+\frac{zf^{\prime \prime }}{f^{\prime }}\right) >\alpha , \end{aligned}$$

\(0\le \alpha <1\) and \(z\in \mathbb {D}.\) Note that \(\mathcal {S}^{*}(0)=\mathcal {S}^{*}\) and \(\mathcal {C}(0)=\mathcal {C}.\) The relation between these two classes is given by famous Alexander transform,

$$\begin{aligned} f\in \mathcal {C}(\alpha ) \Longleftrightarrow zf^{\prime }(z)\in \mathcal {S}^{*}(\alpha ),\quad \alpha \in [0,\,1). \end{aligned}$$

Another important subclass of \(\mathcal {S}\) is the class of close-to-convex functions of order \(\alpha ,\, \alpha \in [0,\,1),\) with respect to a starlike function \(g\in \mathcal {S}^{*}\) given by the analytic characterization,

$$\begin{aligned} \mathrm{Re}\left( \mathrm{e}^{i\gamma }\frac{zf^{\prime }(z)}{g(z)}\right) >\alpha ,\quad \hbox {for some real}\,\gamma \in (-\pi /2,\,\pi /2). \end{aligned}$$

In the sequel, we consider only its particular case where \(\gamma =0.\) Several interesting properties of this and other subclasses can be found in [4, 5, 14] and the references therein.

A function \(f\in \mathcal {A}\) satisfying \(\mathrm{Im}(z)\mathrm{Im}(f(z))>0\) where \(z\in \mathbb {D}\) is said to be typically real function. A function \(f\in \mathcal {A}\) is convex in the direction of imaginary axis if every line parallel to imaginary axis either intersect \(f(\mathbb {D})\) in a line segment or has an empty intersection. In [15] it is proved that \(f\in \mathcal {A}\) with real coefficient is convex in the direction of imaginary axis if \(zf^{\prime }\) is typically real which is equivalent to \(\mathrm{Re}[(1-z^2)f^{\prime }(z)]>0.\)

Ruscheweyh [16] proved that if for \(f\in \mathcal {A},\,f^{\prime }\) is typically real and \(\mathrm{Re}(f^{\prime }(z))>0\) in \(\mathbb {D}\) then f is necessarily starlike and this result was extended to starlike of order \(\alpha \) in [12]. We state the extension as a lemma.

Lemma 1.1

([12], Theorem 3.1) Let \(0\le \alpha <1\) and \(f\in \mathcal {A}\) be such that \(f^{\prime }(z)\) and \(f^{\prime }(z)-\alpha \frac{f(z)}{z}\) are typically real in \(\mathbb {D}.\) Further if \(\mathrm{Re}f^{\prime }(z)>0\) and \(\mathrm{Re}\left( f^{\prime }(z)-\alpha \frac{f(z)}{z}\right) >0,\) then \(f\in \mathcal {S}^{*}(\alpha ).\)

Various techniques are used to obtain the geometric and analytic properties of these subclasses of \(\mathcal {S}.\) Among these, the positivity of trigonometric sum is of recent interest. One of the remarkable results for the positivity of sine and cosine sum is given by Vietoris [17].

Theorem 1.1

[17] If \(a_0\ge a_1 \ge a_2 \cdots \ge a_n >0\) and \(2k a_{2k}\le (2k-1) a_{2k-1},\,k\ge 1,\) then for \(n\ge 1\) and \(\theta \in (0,\,\pi ),\)

$$\begin{aligned} \sum _{k=1}^n a_k\sin {k\theta }>0 \quad \hbox {and}\quad \sum _{k=0}^n a_k\cos {k\theta }>0. \end{aligned}$$

Among various results involving the coefficients obtained in [17], the inequalities for the specific case in which \(a_k=c_k\) is important with reference to this work, where the sequence \(c_k\) is defined as

$$\begin{aligned} c_0=c_1=1\quad \mathrm{and}\quad c_{2k}=c_{2k+1}=\frac{1\cdot 3\cdot 5\cdots (2k-1)}{2\cdot 4\cdot 6\cdots 2k} =\dfrac{(1/2)_k}{k!}, \quad k=1,\,2,\ldots . \end{aligned}$$

Theorem 1.1 has several interesting applications in geometric function theory. Using Theorem 1.1, Ruscheweyh [16] obtained coefficient conditions for \(f\!\in \!\mathcal {A}\) to be starlike.

Theorem 1.2

[16] Let \(a_1=1\) and \(a_k\ge 0\) satisfy

$$\begin{aligned}&ka_k \ge (k+1)a_{k+1},\\&(2k-1)a_{2k} \ge (2k+1)a_{2k+1}. \end{aligned}$$

Then \(f(z)=\displaystyle \sum \nolimits _{k=1}^{\infty }a_kz^k\) is starlike.

Among several generalizations of Theorem 1.1, we are interested in the following result given in [13].

Theorem 1.3

[13] Let \(\beta \ge 0,\,\mu \in \mathbb {R}\) such that \(0<\mu +\beta <1\) and \(n\in \mathbb {N}.\) If \(d_0=d_1=1\) and \(d_{2k}=d_{2k+1}=\frac{(1+\beta )_{n-k}n!}{(n-k)!(1+\beta )_n}. \frac{(\mu +\beta )_k}{k!}\) for \(1\le k\le n.\) Then

  1. (1)

    \(\displaystyle \sum \nolimits _{k=0}^{n}d_k\cos k\theta >0 \Longleftrightarrow \mu +\beta \le \mu ^{*}(1/2)=0.691556,\ldots ,\)

  2. (2)

    \(\displaystyle \sum \nolimits _{k=1}^{2n+1}d_k\sin k\theta >0 \Longleftrightarrow \mu +\beta \le \mu ^{*}(1/2),\)

  3. (3)

    \(\displaystyle \sum \nolimits _{k=1}^{2n}d_k\sin k\theta >0\) for \(\mu +\beta \le \dfrac{1+\beta }{2},\)

where \(\mu ^{*}(\lambda ),\,\lambda \in (0,\,1]\) is the unique solution in \((0,\,1)\) of

$$\begin{aligned} \int _{0}^{(\lambda +1)\pi }\dfrac{\sin (t-\lambda \pi )}{t^{1-\mu }}\mathrm{d}t=0. \end{aligned}$$

Note that this \(\mu ^{*}(\lambda )\) was first obtained by Koumandos and Ruscheweyh [8]. The particular case \(\mu ^{*}(1/2)=\mu _0^{*}\) is used in this work extensively.

This work is organized as follows: in the next section, certain conditions on the starlikeness of \(f\in \mathcal {A}\) by means of its Taylor coefficients are obtained using Theorem 1.3. Conditions on the coefficient \(\{a_k\}\) are obtained that ensure that the function \(f(z)=\sum _{k=1}^{\infty }a_kz^k\) is starlike and also the partial sum \(f_n(z)\) to be close-to-convex or convex in the direction of imaginary axis. In the last section, conditions on the triplet \((a,\,b,\,c)\) are obtained for which the normalized Gaussian hypergeometric function \(zF(a,\,b;\,c;\,z)\) is starlike of certain order.

2 Main Results

The positivity of the cosine sum and sine sum given in Theorem 1.3 [13] can be improved as given below by direct application of Abel’s summation formula.

Lemma 2.1

Let \(\beta \ge 0,\,\mu \in \mathbb {R}\) such that \(0<\mu +\beta <1\) and \(n\in \mathbb {N}.\) If \(\{a_k\}\) be decreasing sequence of non- negative numbers satisfying \(a_0>0\) and

$$\begin{aligned} k(n-k+1+\beta )a_{2k}\le (n-k+1)(\mu +\beta +k-1)a_{2k-1},\quad \hbox {for}\quad 1\le k\le n, \end{aligned}$$
(2.1)

then for all \(0<\theta <\pi ,\)

$$\begin{aligned} \sum _{k=0}^{n}a_k\cos k\theta >0 \Longleftrightarrow \mu +\beta \le \mu _0^{*}. \end{aligned}$$

Proof

Let the sequence \(\{d_k\}\) be as given in Theorem 1.3. Then Theorem 1.3 guarantees that \(\displaystyle \sum \nolimits _{k=0}^{n}d_k\cos k\theta >0\) by the hypothesis given in the statement. Using summation by parts, if \(\displaystyle \sum \nolimits _{k=0}^{n}a_k\cos k\theta \) is rewritten as

$$\begin{aligned} \sum _{k=0}^{n}a_k\cos k\theta =\sum _{k=0}^{n-1}\left( \frac{a_k}{d_k}- \frac{a_{k+1}}{d_{k+1}}\right) \sum _{j=0}^k d_j\cos j\theta + \frac{a_n}{d_n}\sum _{j=0}^{n}d_j\cos j\theta , \end{aligned}$$

then

$$\begin{aligned}&\frac{a_{2k-1}}{d_{2k-1}}-\frac{a_{2k}}{d_{2k}}\\&\quad =\frac{(1+\beta )_n}{n!} \frac{(n-k)!(k-1)!}{(\mu +\beta )_{k-1} (1+\beta )_{n-k}} \left[ \frac{n-k+1}{n-k+1+\beta }a_{2k-1}-\frac{k}{\mu +\beta +k-1}a_{2k}\right] \\&\quad \ge 0. \end{aligned}$$

which gives the desired result. \(\square \)

Following the same procedure as in Lemma 2.1 the following result for sine sums can be obtained.

Lemma 2.2

Let \(0\le \beta \le 2\mu _0^{*}-1,\, \mu \in \mathbb {R}\) such that \(0<\mu +\beta <1\) and \(n\in \mathbb {N}.\) If \(\{a_k\}\) be decreasing sequence of non-negative numbers satisfying \(a_0>0\) and (2.1), then for all \(0<\theta <\pi ,\)

$$\begin{aligned} \sum _{k=0}^{n}a_k\sin k\theta >0 \Longleftrightarrow \mu +\beta \le \frac{1+\beta }{2}. \end{aligned}$$

The above two lemmas can be used to find a geometric property of an arbitrary but normalized analytic function, using the positivity behaviour related to its Taylor coefficients.

Theorem 2.1

Let \(0\le \beta \le 2\mu _0^{*}-1\) and \(-\beta <\mu \le \frac{1-\beta }{2},\, a_1=1,\,a_k\ge 0\) satisfy

$$\begin{aligned}&[k(1+\beta )(1-\mu -\beta )-1+2\mu +\beta ]a_k\nonumber \\&\quad \ge [(k+1)(1+\beta )(1-\mu -\beta )-1+2\mu +\beta ]a_{k+1}, \end{aligned}$$
(2.2)
$$\begin{aligned}&(n-k+1)(k+\mu +\beta -1)[2k(1+\beta )(1-\mu -\beta )-1+2\mu +\beta ]a_{2k}\nonumber \\&\quad \ge k(n-k+1+\beta ) [(2k+1)(1+\beta )(1-\mu -\beta )-1+2\mu +\beta ]a_{2k+1}, \end{aligned}$$
(2.3)

for \(1\le k \le n.\) Then \(f_n(z)=\displaystyle \sum \nolimits _{k=1}^{n}a_kz^k\) is starlike of order \(\frac{1-2\mu -\beta }{(1+\beta )(1-\mu -\beta )}.\) Moreover in the limiting case, \(f(z)=\displaystyle \lim \nolimits _{n\rightarrow \infty }f_n(z)=\displaystyle \sum \nolimits _{k=1}^{\infty }a_kz^k\) is starlike of the same order if \(\{a_k\}\) satisfy (2.2) and in addition

$$\begin{aligned}&(k+\mu +\beta -1)[2k(1+\beta )(1-\mu -\beta )-1+2\mu +\beta ]a_{2k}\nonumber \\&\quad \ge k [(2k+1)(1+\beta )(1-\mu -\beta )-1+2\mu +\beta ]a_{2k+1}, \quad \hbox {for}\quad k\ge 1.\nonumber \\ \end{aligned}$$
(2.4)

Proof

Let \(f_{n}(z)=z+\displaystyle \sum \nolimits _{k=2}^{n}a_kz^k\) and \(\gamma :=\dfrac{1-2\mu -\beta }{(1+\beta )(1-\mu -\beta )}.\) Then

$$\begin{aligned} g_{n}(z)=f^{\prime }_{n}(z)-\dfrac{\gamma f_n(z)}{z}=b_0+\sum _{k=1}^{n-1}b_kz^k, \end{aligned}$$

where \(b_0=(1-\gamma )\) and \(b_k=(k+1-\gamma )a_{k+1}\) for \(1\le k\le n-1.\) Clearly \(b_0>0.\) Further, we claim that these \(\{b_k\}\) are decreasing and satisfy (2.1) for \(0\le k\le n-1.\) To prove that the sequence \(\{b_k\}\) is decreasing it is enough to establish that \((k-\gamma )a_k\ge (k+1-\gamma )a_{k+1}\) or equivalently

$$\begin{aligned} \left( k-\dfrac{1-2\mu -\beta }{(1+\beta )(1-\mu -\beta )}\right) a_k \ge \left( k+1-\dfrac{1-2\mu -\beta }{(1+\beta )(1-\mu -\beta )}\right) a_{k+1}, \end{aligned}$$

which by rearranging the terms, leads to (2.2). To verify that \(\{b_k\}\) satisfy (2.1), we write

$$\begin{aligned}&(n-k+1)(\mu +\beta +k-1)(2k-\gamma )a_{2k}\ge k(n-k+1+\beta )(2k+1-\gamma )a_{2k+1},\\&(n-k+1)(\mu +\beta +k-1)\left( 2k-\dfrac{1-2\mu -\beta }{(1+\beta )(1-\mu -\beta )}\right) a_{2k}\\&\quad \ge k(n-k+1+\beta )\left( 2k+1-\dfrac{1-2\mu -\beta }{(1+\beta )(1-\mu -\beta )}\right) a_{2k+1}, \end{aligned}$$

which is nothing but (2.3) given in the hypothesis.

Thus \(\{b_k\}\) satisfy the hypothesis of Lemmas 2.1 and 2.2. Hence for \(0<\mu +\beta \le (1+\beta )/2,\, \displaystyle \sum \nolimits _{k=0}^{n}b_k\cos k\theta >0\) and \(\displaystyle \sum \nolimits _{k=1}^{n}b_k\sin k\theta >0\) for \(\theta \in (0,\,\pi ).\) By the minimum principle of harmonic functions we obtain that

$$\begin{aligned} \mathrm {Re}\left( g_n(z)\right) =\sum _{k=0}^{n}b_kr^k\cos k\theta >0,\quad \hbox {for}\quad \theta \in (0,\,\pi )\quad \hbox {and}\quad r\in [0,\,1). \end{aligned}$$

Moreover \(\mathrm {Im}(g_n(z))=\displaystyle \sum \nolimits _{k=1}^{n}b_kr^k\sin k\theta \equiv 0\) if \(-1<z=x+iy<1\) and \(\mathrm {Im}(g_n(z))>0\) in \(\mathbb {D}\cap \{z{\text {:}}\,\mathrm {Im}(z)>0\}.\) The reflection principle yields \(\mathrm {Im}(g_n(z))<0\) in \(\mathbb {D}\cap \{z{\text {:}}\,\mathrm {Im}(z)<0\}.\) Thus \(g_n(z)\) is typically real.

Now to complete the proof of the theorem, it remains to justify that \(f^{\prime }_n(z)\) is typically real having real part. Now,

$$\begin{aligned} f^{\prime }_n(z)=1+\sum _{k=1}^{n-1}(k+1)a_{k+1}z^{k}. \end{aligned}$$

By the same argument as above, we have that \(\mathrm {Re}(f^{\prime }_n(z))>0\) and typically real function. So by Lemma 1.1, we get that \(f_n(z)\) is starlike of order \(\dfrac{1-2\mu -\beta }{(1+\beta )(1-\mu -\beta )}.\) Clearly in the limiting case as \(n\rightarrow \infty \) (2.3) becomes (2.4) and since the family of starlike functions of a fixed order is normal, we get \(f(z)=\displaystyle \lim \nolimits _{n\rightarrow \infty }f_n(z)\) is starlike of order \(\dfrac{1-2\mu -\beta }{(1+\beta )(1-\mu -\beta )}\) if \(\{a_k\}\) satisfy (2.2) and (2.4). \(\square \)

Remark 2.1

By using \(\beta =2\mu _0^{*}-1\) and \(\mu =1-\mu _0^{*}\) in Theorem 2.1, inequalities (2.2) and (2.4) reduce to

$$\begin{aligned}&ka_k \ge (k+1)a_{k+1},\\&\left( k-1+\mu _0^{*}\right) a_{2k} \ge \mu _0^{*}(2k+1)a_{2k+1}. \end{aligned}$$

Clearly, these inequalities provide a better lower bound for the ratio \(\frac{a_{2k}}{a_{2k+1_{}}}\) compared to the one given in Theorem 1.2. This is due to the fact that, Theorem 1.2 is obtained using Theorem 1.1, whereas Theorem 2.1 is obtained using Theorem 1.3.

Using Lemma 2.1 and minimum principle for harmonic functions, the following result for \(f(z)=z+\displaystyle \sum \nolimits _{k=2}^na_kz^k\) to be close-to-convex with respect to starlike function \(g(z)=z\) is obtained.

Theorem 2.2

Let \(\mu \in \mathbb {R}\) and \(\beta \ge 0\) such that \(0<\mu +\beta <1\) and let \(a_1=1\) and \(a_k\ge 0\) satisfy

$$\begin{aligned} 0\le na_n\le \cdots \le (k+1)a_{k+1}\le ka_k \le \cdots \le 3a_3\le 2a_2\le \frac{\mu +\beta }{\mu _0^{*}},\quad \mu +\beta \in \left( 0,\,\mu _0^{*}\right] , \end{aligned}$$
(2.5)

and

$$\begin{aligned} 2(n-k+1)(\mu +\beta +k-1)a_{2k}\ge (2k+1)(n-k+1+\beta )a_{2k+1},\quad 1\le k\le \left[ \frac{n}{2}\right] . \end{aligned}$$
(2.6)

Then \(f_n(z)=z+\displaystyle \sum \nolimits _{k=2}^n a_k z^k\) satisfies \(\mathrm{Re}(f_n^{\prime }(z))>1-\frac{\mu +\beta }{\mu _0^{*}}.\)

Proof

Let \(\delta =1-\frac{\mu +\beta }{\mu _0^{*}}\) and \(f_n(z) =z+\displaystyle \sum \nolimits _{k=2}^n a_k z^k.\) Then

$$\begin{aligned} \frac{f_n^{\prime }(z)-\delta }{1-\delta }=\sum _{k=0}^{n-1} b_k z^k, \end{aligned}$$

where \(b_0=1\) and \(b_k=\frac{(k+1)a_{k+1}}{(1-\delta )}\) for \(1\le k\le n-1.\) Clearly \(a_k\ge 0\) implies \(b_k\ge 0\) for \(k\ge 1.\) To prove the theorem it is required that \(\{b_k\}\) are decreasing and satisfy (2.1). Now \(\{b_k\}\) are decreasing if \(\{a_k\}\) satisfy

$$\begin{aligned} (k+2)a_{k+2}\le (k+1)a_{k+1},\quad \hbox {for}\quad 1\le k\le n-2, \end{aligned}$$

and \(b_1\le b_0\) implies \(2a_2\le 1-\delta \) which is nothing but (2.5). Moreover \(\{b_k\}\) satisfy (2.1) if \(\{a_k\}\) satisfy

$$\begin{aligned}&k(n-k+1+\beta )(2k+1)a_{2k+1}&\le (n-k+1)(\mu +\beta +k-1)2k a_{2k}, \quad \hbox {for}\\&\quad 1\le k\le \left[ \frac{n}{2}\right] , \end{aligned}$$

which implies (2.6). Similar arguments using minimum principle for harmonic function yield the required result. \(\square \)

The next result gives the condition for which \(f_n(z)\) is convex in the direction of imaginary axis which is equivalent to the condition that \(f_n(z)\) is close-to-convex with respect to the starlike function \(g(z)=z/(1-z^2).\)

Theorem 2.3

Let \(0\le \beta \le 2\mu _0^{*}-1,\,a_1=1,\,a_k\ge 0\) satisfy

$$\begin{aligned}&k a_k \ge (k+1)a_{k+1}, \quad k=1,\,2,\ldots ,n-1, \nonumber \\&(n-k+1)(k+\mu +\beta -1)(2k-1)a_{2k-1} \ge (n-k+1+\beta )2k^2a_{2k},\nonumber \\&\quad k=1,\,2,\ldots ,[n/2], \end{aligned}$$
(2.7)

for \(k\ge 1,\,-\beta <\mu \le \frac{1-\beta }{2}.\) Then \(f_n(z)=\displaystyle \sum \nolimits _{k=1}^{n}a_kz^k\) is convex in the direction of imaginary axis.

Proof

\(f_n(z)\) is convex in the direction of imaginary axis which is equivalent to \(zf_n^{\prime }(z)\) that is typically a real function. Also \(f_n(z)\) has real coefficients.

$$\begin{aligned} zf_n^{\prime }(z)=z+\sum _{k=2}^n b_k z^k, \end{aligned}$$

where \(b_k=ka_k.\) To obtain the result, it is required that \(\{b_k\}\) satisfy the conditions given in Lemma 2.2. Then for \(1\le k\le n,\)

$$\begin{aligned} b_k-b_{k+1}=ka_k-(k+1)a_{k+1}\ge 0, \end{aligned}$$

and for \(k\in \{1,\,2\ldots ,[n/2]\},\)

$$\begin{aligned}&(n-k+1)(k+\mu +\beta -1)b_{2k-1}-k(n-k+1+\beta )b_{2k}\\&\quad =(n-k+1)(k+\mu +\beta -1)(2k-1)a_{2k-1}-(n-k+1+\beta )2k^2a_{2k}\ge 0. \end{aligned}$$

Thus \(\{b_k\}\) satisfy the conditions of Lemma 2.2. Then for \(\mu +\beta \in \left( 0,\,\frac{1+\beta }{2}\right] ,\) using minimum principle for harmonic functions \(\mathrm{Im}(zf_n^{\prime }(z))=\displaystyle \sum \nolimits _{k=1}^nb_kr^k\sin {k\theta }>0\) where \(\theta \in (0,\,\pi )\) and \(r\in (0,\,1)\) and \(\mathrm{Im}(zf_n^{\prime }(z))\equiv 0\) for \(z\in \mathbb {D}\cap \{z=x+iy{\text {:}}\,-1<x<1,\,y=0\}.\) Schwarz reflection principle yields that \(\mathrm{Im}(zf_n^{\prime }(z))<0\) for \(\theta \in (\pi ,\,2\pi ).\) So \(zf_n^{\prime }(z)\) is typically real function or equivalently \(f_n(z)\) is convex in the direction of imaginary axis. \(\square \)

Theorem 2.3 generalizes various earlier known results. For example, \(\beta =0\) gives the following corollary due to Ali et al. [3], which for \(\mu =1/2,\) further reduces to a result obtained by Acharya [1].

Corollary 2.1

[3] Let \(\{a_k\}\) be a sequence of non-negative real numbers such that \(a_1=1\) and if it satisfies (2.7) and

$$\begin{aligned} 2k^2 a_{2k} \le (k+\mu -1)(2k-1)a_{2k-1}, \quad (k=1,\ldots ,n/2), \end{aligned}$$

then the function \(\displaystyle \sum \nolimits _{k=1}^na_kz^k\) is convex in the direction of imaginary axis whenever \(\mu \in (0,\,1/2].\)

Note that, Theorem 2.3 has a lower bound for \(\frac{a_{2k}}{a_{2k-1}},\) which is lesser than the lower bound given in [11, Theorem 2.3]. The following example supports this observation. However both these results are not sharp.

Example 2.1

Consider the sequence \(\{a_k\}\) such that \(a_1=1\) and

$$\begin{aligned} 2ka_{2k}=(2k+1)a_{2k+1}=\frac{(\mu +\beta )_k}{(1+\beta )^k k!}. \end{aligned}$$

Then by Theorem 2.3 we conclude that \(f(z)=z+\displaystyle \sum \nolimits _{k=2}^na_kz^k\) is convex in the direction of imaginary axis.

3 Starlikeness of Gaussian Hypergeometric Function

Among various special functions, Gaussian hypergeometric functions play an important role in the study of univalent function. Hence the starlikeness of Gaussian hypergeometric function is of considerable interest. Note that in [9], the order of starlikeness is analyzed for various ranges of the parameters \(a,\,b,\,c\) of \(zF(a,\,b;\,c;\,z).\) The starlikeness of \(zF(a,\,b;\,c;\,z)\) using duality technique was studied by several authors including [2]. For other results in this direction, we refer the [3, 6, 7, 10] and the references therein. In this section, we use Theorem 2.1 in finding the starlikeness of the Gaussian hypergeometric function \(zF(a,\,b;\,c;\,z).\) Examples are also given to compare with the range available in the literature. Sufficient conditions for odd Gaussian hypergeometric function \(zF(a,\,b;\,c;\,z^2)\) to be starlike are also obtained.

Theorem 3.1

Let \(0\le \beta \le 2\mu _0^{*}-1\) and \(-\beta <\mu \le \frac{1-\beta }{2}\) and \(a,\,b\le \mu +\beta -1\) satisfy \((a)_k(b)_k\ge 0\) for \(k\ge 2.\) If \(Mc\ge ab\) where \(M=\frac{\beta (1-\mu -\beta )+\mu }{2\beta (1-\mu -\beta )+1-\beta }>0,\) then \(zF(a,\,b;\,c;\,z)\) is starlike of order \(\frac{1-2\mu -\beta }{(1+\beta )(1-\mu -\beta )}.\)

Proof

Clearly \(z F(a,\,b;\,c;\,z) =\displaystyle \sum \nolimits _{k=1}^{\infty }b_k z^k\) implies \(b_1=1\) and \(b_k=\frac{(a)_{k-1}(b)_{k-1}}{(c)_{k-1}(k-1)!}\) for \(k\ge 2.\) Note that \(b_k\) and \(b_{k+1}\) can be related by

$$\begin{aligned} b_{k+1}=\frac{(a+k-1)(b+k-1)}{(c+k-1)k}b_k,\quad \hbox {for}\quad k\ge 1. \end{aligned}$$

To establish the theorem it is enough to prove that \(\{b_k\}\) satisfy the conditions (2.2) and (2.4) of Theorem 2.1. Clearly, a simple computation using the above relation between \(b_k\) and \(b_{k+1}\) provides.

$$\begin{aligned}&[k(1+\beta )(1-\mu -\beta )-1+2\mu +\beta ]b_k-[(k+1)(1+\beta )(1-\mu -\beta )-1+2\mu +\beta ]b_{k+1}\\&\quad =\frac{b_k}{(c+k-1)k} h(k), \end{aligned}$$

where h(k) is defined as

$$\begin{aligned} h(k):&=[k(1+\beta )(1-\mu -\beta )-1+2\mu +\beta ]\left[ k^2+(c-1)k\right] -[k(1+\beta )(1-\mu -\beta )\\&\quad +(1+\beta )(1-\mu -\beta )-1+2\mu +\beta ]\left[ k^2+(a+b-2)k+(a-1)(b-1)\right] . \end{aligned}$$

Let

$$\begin{aligned} A_1&=(c-a-b)(1+\beta )(1-\mu -\beta ),\\ A_2&={-}(a+b)[2\beta (1-\mu -\beta )+1-\beta ]+ab[\beta (1-\mu -\beta )+\mu ]+\beta (1-\mu -\beta )\\&\quad +\mu +(c-ab)[2\beta (1-\mu -\beta )+1-\beta ],\\ A_3&=c[\beta (1-\mu -\beta )+\mu ]-ab[2\beta (1-\mu -\beta )+1-\beta ], \end{aligned}$$

such that

$$\begin{aligned} h(k)=A_1(k-1)^2+A_2(k-1)+A_3. \end{aligned}$$

The conditions \(a\le \mu +\beta -1\le \frac{\beta -1}{2},\, b\le \mu +\beta -1\le \frac{\beta -1}{2}\) and \(Mc \ge ab>0,\) where

$$\begin{aligned} M=\frac{\beta (1-\mu -\beta )+\mu }{2\beta (1-\mu -\beta )+1-\beta }, \end{aligned}$$

show that \(A_1\ge 0,\,A_3\ge 0.\) Note that M can be written as \(M=\frac{(1-\beta )(\mu +\beta )}{2\beta (1-\mu -\beta )+1-\beta }\) so that \(M\ge 0.\) Since \(2\beta (1-\mu -\beta )+1-\beta \ge 0,\) we consider,

$$\begin{aligned} \frac{A_2}{2\beta (1-\mu -\beta )+1-\beta }={-}(a+b)+(ab+1)M+(c-ab). \end{aligned}$$

Now, using \(Mc\ge ab>0,\) we get

$$\begin{aligned} c-ab\ge \left( \frac{1}{M}-1\right) ab, \end{aligned}$$

which gives

$$\begin{aligned} c-ab \ge \frac{(1+\beta )(1-\mu -\beta )}{(1-\beta )(\mu +\beta )}ab>0. \end{aligned}$$

This means

$$\begin{aligned} \frac{A_2}{2\beta (1-\mu -\beta )+1-\beta }={-}a-b+(ab+1)M+(c-ab)\ge 0. \end{aligned}$$

Thus \(A_1\ge 0,\,A_2\ge 0\) and \(A_3\ge 0\) contribute to \(h(k)\ge 0\) for \(k\ge 1.\)

It remains to verify (2.4)

$$\begin{aligned}&(k+\mu +\beta -1)[2k(1+\beta )(1-\mu -\beta )-1+2\mu +\beta ]b_{2k}\\&\quad \ge k[(2k+1)(1+\beta )(1-\mu -\beta )-1+2\mu +\beta ]b_{2k+1}. \end{aligned}$$

Clearly,

$$\begin{aligned}&(k+\mu +\beta -1)[2k(1+\beta )(1-\mu -\beta )-1+2\mu +\beta ]b_{2k}\\&\quad - k[(2k+1)(1+\beta )(1-\mu -\beta )-1+2\mu +\beta ]b_{2k+1}=\frac{b_{2k}}{2(c+2k-1)}g(k), \end{aligned}$$

where

$$\begin{aligned} g(k):&=(k+\mu +\beta -1)(c+2k-1)2(2k(1+\beta )(1-\mu -\beta )-1+2\mu +\beta )\\&\quad -(a+2k-1)(b+2k-1)((2k+1)(1+\beta )(1-\mu -\beta )-1+2\mu +\beta ). \end{aligned}$$

Let

$$\begin{aligned} B_1&=4\left[ -2(1+\beta )(1-\mu -\beta )^2-(a+b)(1+\beta )(1-\mu -\beta )+c(1+\beta )(1-\mu -\beta )\right] ,\\ B_2&=2\left[ -4(1+\beta )+(7+9\beta )(\mu +\beta )-(2+6\beta )(\mu +\beta )^2\right] \\&\quad -2(a+b)[3(1+\beta )(1-\mu -\beta )\\&\quad +\beta (1-\mu -\beta )+\mu ]+2c[2\beta (1-\mu -\beta )+(1-\beta )\\&\quad +2(\mu +\beta )(1+\beta )(1-\mu -\beta )]+2ab(\mu +\beta ),\\ B_3&=[-1-(a+b)]\cdot [2(1+\beta )(1-\mu -\beta )+\beta (1-\mu -\beta )+\mu ]\\&\quad +2(\mu +\beta )[2\beta (1-\mu -\beta )+1-\beta ]+2c(\mu +\beta )[2\beta (1-\mu -\beta )+1-\beta ]\\&\quad -ab[2(1+\beta )(1-\mu -\beta )+\beta (1-\mu -\beta )+\mu ], \end{aligned}$$

so that \(g(k)=B_1(k-1)^2+B_2(k-1)+B_3.\) We claim that \(B_1,\,B_2,\,B_3\) are non-negative.

For \(a\le \mu +\beta -1\) and \(b\le \mu +\beta -1,\) we have

$$\begin{aligned} B_1\ge 4\left[ -2(1+\beta )(1-\mu -\beta )^2-2(\mu +\beta -1)(1+\beta )(1-\mu -\beta )\right] =0, \end{aligned}$$

and

$$\begin{aligned} B_2&>2\left[ -4(1+\beta )+(7+9\beta )(\mu +\beta )-(2+6\beta )(\mu +\beta )^2\right] \\&\quad -2(a+b)[3(1+\beta )(1-\mu -\beta )+\beta (1-\mu -\beta )+\mu ]\\&=2\left[ 2(1+\beta )-(3+5\beta )(\mu +\beta )+2(1+\beta )(\mu +\beta )^2\right] =:\mathcal {N}, \end{aligned}$$

where \(\mathcal {N}:=\mathcal {N}(\mu +\beta )\) is a second-degree polynomial in \((\mu +\beta )\) and \(0<\mu +\beta \le \frac{1+\beta }{2}.\)

Now

$$\begin{aligned} \mathcal {N}(0)&=2(1+\beta )>0 \quad \hbox {and}\\ \mathcal {N}\left( \frac{1+\beta }{2}\right)&=2(1+\beta )\frac{(1+\beta )^2}{4}-(3+5\beta )\frac{(1+\beta )}{2}+2(1+\beta )\\&=\left( \frac{1+\beta }{2}\right) (\beta -1)(\beta -2)>0. \end{aligned}$$

Clearly \(\mathcal {N}\) has no zeros in \(\left( 0,\,\frac{1+\beta }{2}\right] .\) So \(\mathcal {N}>0\) implies \(B_2>0.\) \(B_3\) can also be rewritten as

$$\begin{aligned} B_3=B_{31}+2A_3+B_{32}, \end{aligned}$$

where

$$\begin{aligned} B_{31}&=[-1-(a+b)][2(1+\beta )(1-\mu -\beta )+\beta (1-\mu -\beta )+\mu ]\\&\quad +2(\mu +\beta )[2\beta (1-\mu -\beta )+1-\beta ],\\ B_{32}&=2c(\mu +\beta )[2\beta (1-\mu -\beta )+1-\beta ]\\&\quad -ab[2(1+\beta )(1-\mu -\beta )+\beta (1-\mu -\beta )+\mu ]\\&\quad -2c[\beta (1-\mu -\beta )+\mu ]+2ab[2\beta (1-\mu -\beta )+1-\beta ]\\&=4c\beta (\mu +\beta )(1-\mu -\beta )+ab[\beta (1-\mu -\beta )+\mu ]>0. \end{aligned}$$

To prove our claim it remains to prove that \(B_{31}>0.\) However,

$$\begin{aligned} B_{31}&\ge {-}[2\beta (1-\mu -\beta )+2-(1+\beta )(\mu +\beta )]-2(\mu +\beta -1)[2\beta (1-\mu -\beta )\\&\quad +2-(1+\beta )(\mu +\beta )]+2(\mu +\beta )[2\beta (1-\mu -\beta )+1-\beta ]\\&=2(1+\beta )(\mu +\beta )^2-(3+5\beta )(\mu +\beta )+2(1+\beta )=\mathcal {N}>0. \end{aligned}$$

This means \(B_3>0.\) Hence inequality (2.4) holds good. The desired result follows from Theorem 2.1. \(\square \)

Note that Küstner [9] obtained the order of starlikeness of normalize Gaussian hypergeometric function \(z\rightarrow zF(a,\,b;\,c;\,z)\) for all parameters \(a,\,b,\,c\in \mathbb {R},\, c>0\) except for the following ranges:

  1. (i)

    \(a<-1,\,0<b<c<b-a+1,\)

  2. (ii)

    \(0<a<c<b-a+1,\,c+1<b,\)

  3. (iii)

    \(a\le b<0<c,\)

  4. (iv)

    \(0<c<a\le b,\)

  5. (v)

    \(a<0<c<b.\)

Theorem 3.1 gives the answer for the range (iii). Since a and b are negative real numbers in Theorem 3.1, and have no relation between them, our results cover the case \(b\le a\) also.

The result in Theorem 3.1 is not sharp. However, this result is better than the earlier results available in the literature for certain ranges of \(a,\,b\) and c. If we take \(\mu =1-\mu _0^{*}\) and \(\beta =2\mu _0^{*}-1\) Theorem 3.1 leads to the following example.

Example 3.1

Let \(a,\,b\le \mu _0^{*}-1\) satisfy \((a)_k(b)_k\ge 0\) for \(k\ge 2\) and if \(c\ge 2ab,\) then \(z F(a,\,b;\,c;\,z)\) is starlike.

Remark 3.1

If we substitute \(\mu +\beta =\frac{1+\beta }{3-\beta }\) in Theorem 3.1, then \(M=1/3\) and \(c\ge 3ab,\) which lead to constructing the following case.

If \(a,\,b\in \left( -1,\,\frac{2(\mu _0^{*}-1)}{2-\mu _0^{*}}\right] \) then \(zF(a,\,b;\,c;\,z)\) is starlike of order 1/2. Note that for the given range of a and b,  we get \(c\ge 3ab\) which is better than the range \(c\ge 1+2ab\) provided by Hästö et al. [6, Corollary 1.7]. However, Corollary 1.7 of [6] is valid for all \(a,\,b\) non-zero real, which is not applicable in our case.

Using Alexander transform and

$$\begin{aligned} F^{\prime }(a,\,b;\,c;\,z)=\frac{ab}{c}F(a+1,\,b+1;\,c+1;\,z), \end{aligned}$$

the following consequence of Theorem 3.1 can be deduced which is of considerable interest.

Corollary 3.1

Let \(0\le \beta < 2\mu _0^{*}-1\) and \(-\beta <\mu \le \frac{1-\beta }{2}\) and \(a,\,b\le \mu +\beta -2\) satisfy \((a+1)_k(b+1)_k\ge 0\) for \(k\ge 2.\) If \(M(c+1)\ge (a+1)(b+1),\) then the function \(\frac{c}{ab}[F(a,\,b;\,c,\,z)-1]\) is convex of order \((1-2\mu -\beta )/(1+\beta )(1-\mu -\beta ).\)

Proof

If we replace \(a,\,b,\,c\) by \(a+1,\,b+1,\,c+1\), respectively, in Theorem 3.1 then we get that for \(a,\,b\le \mu +\beta -2,\,(a+1)_k(b+1)_k\ge 0\) for \(k\ge 2\) and if \(M(c+1)\ge (a+1)(b+1)\) then \(F(a+1,\,b+1;\,c+1;\,z)\) is starlike of order \((1-2\mu -\beta )/(1+\beta )(1-\mu -\beta ).\) Since

$$\begin{aligned} \int _0^z \frac{F(a+1,\,b+1;\,c+1;\,t)}{t}\mathrm{d}t=\frac{c}{ab}[F(a,\,b;\,c,\,z)-1], \end{aligned}$$

by Alexander transformation, \(\int _0^z F(a+1,\,b+1;\,c+1;\,t)\frac{\mathrm{d}t}{t}\) is convex of order \((1-2\mu -\beta )/(1+\beta )(1-\mu -\beta )\) for the same conditions. \(\square \)

The next result gives the condition on the triplet \((a,\,b,\,c)\) for the starlikeness of odd Gaussian hypergeometric function.

Theorem 3.2

Let \(0\le \beta < 2\mu _0^{*}-1\) and \(\mu \in \left( -\beta ,\,\frac{(1-\beta )^2}{(3-\beta )}\right] \) and \(a,\,b\le \mu +\beta -1\) satisfy \((a)_k(b)_k\ge 0\) for \(k\ge 2.\) If \(Mc\ge ab,\) then the function \(z F(a,\,b;\,c,\,z^2)\) is starlike of order \(\frac{(1+\beta -(3+\beta )(\mu +\beta ))}{(1+\beta )(1-\mu -\beta )}.\)

Proof

Let \(f(z)=z F(a,\,b;\,c,\,z)\) and \(zg(z)=f(z^2).\) Then \(g(z)=z F(a,\,b;\,c;\,z^2)\) and from Theorem 3.1 \(f\in \mathcal {S}^{*}\left( \frac{1-2\mu -\beta }{(1+\beta )(1-\mu -\beta )}\right) .\) Taking logarithmic derivative of g(z) and considering the real part, we get

$$\begin{aligned} \mathrm{{Re}}\frac{zg^{\prime }(z)}{g(z)}&=2\mathrm{{Re}}\frac{z^2f^{\prime }(z^2)}{f(z^2)}-1 >\frac{(1+\beta )-(3+\beta )(\mu +\beta )}{(1+\beta )(1-\mu -\beta )} \end{aligned}$$

\(\Longrightarrow g\in \mathcal {S}^{*}\left( \frac{(1+\beta )-(3+\beta )(\mu +\beta )}{(1+\beta )(1-\mu -\beta )} \right) .\) \(\square \)

For \(\beta =2\mu _0^{*}-1\) and \(\mu +\beta =\frac{1+\beta }{3-\beta },\) we obtain the following corollary of Theorem 3.2.

Corollary 3.2

Let \(a,\,b\le \frac{2(\mu _0^{*}-1)}{2-\mu _0^{*}}\) satisfy \((a)_k(b)_k\ge 0,\,k\ge 2.\) Then for \(c\ge 3ab,\,z F(a,\,b;\,c;\,z^2)\in \mathcal {S}^{*}.\)

Remark 3.2

  1. (i)

    Corollary 3.2 is better than the result [3, Corollary 3.10] that states that \(z F(a,\,b;\,c;\,z^2)\) is starlike provided \(a,\,b\le -2/3\) and \(c\ge 3ab.\) Corollary 3.2 gives a better range for a and b.

  1. (ii)

    When \(a,\,b\in \left( -1,\, \frac{2(\mu _0^{*}-1)}{2-\mu _0^{*}}\right] \) then \(zF(a,\,b;\,c;\,z^2)\in \mathcal {S}^{*}\) for \(c\ge 3ab\) which is better than the condition \(c\ge 1+2ab\) provided in [6], but for a smaller range of a and b.