1 Introduction and definitions

Let \(\mathcal {T}\) denote the class of functions of the form:

$$\begin{aligned} f(z)=z-\sum \limits _{n=2}^{\infty }a_{n}z^{n},\quad a_{n}\ge 0, \end{aligned}$$
(1.1)

which are analytic and univalent in the open unit disk \(\mathcal {U} =\{z:\vert z\vert <1\}.\) Let \(T^{*}(\alpha )\) and \(C(\alpha )\) denote the subclasses of \(\mathcal {T}\) consisting of starlike and convex functions of order \(\alpha (0\le \alpha <1),\) respectively [11].

A function f of the form (1.1) is in \(\mathcal {S}(k,\lambda )\) if it satisfies the condition

$$\begin{aligned} \left| \frac{\frac{zf^{\prime }(z)}{(1-\lambda )f(z)+\lambda zf^{\prime }(z)}-1}{\frac{zf^{\prime }(z)}{(1-\lambda )f(z)+\lambda zf^{\prime }(z)}+1} \right|<k,\;(0<k\le 1,\text { }0\le \lambda <1,z\in \mathcal {U}) \end{aligned}$$
(1.2)

and \(f\in \, \mathcal {C}(k,\lambda )\) if and only if \(zf^{\prime }\in \mathcal {S}(k,\lambda ).\)

We note that \(\mathcal {S}(k,0)=\mathcal {S}(k)\) and \(\mathcal {C}(k,0)= \mathcal {C}(k)\), where the classes \(\mathcal {S}(k)\) and \(\mathcal {C}(k)\) were introduced and studied by Padmanabhan [8] (see also, [4, 7]).

Let F(a, b; c; z) be the (Gaussian) hypergeometric function defined by

$$\begin{aligned} F(a,b;c;z)=\sum \limits _{n=0}^{\infty }\frac{(a)_{n}(b)_{n}}{(c)_{n}(1)_{n}} z^{n}, \end{aligned}$$
(1.3)

where \(c\ne 0,{-1},{-2},\ldots ,\) and \((x)_{n}\) is the Pochhammer symbol defined by

$$\begin{aligned} (x)_{n}:=\frac{\Gamma (x+n)}{\Gamma (x)}=\left\{ \begin{array}{ll} 1, &{} n=0, \\ x(x+1)(x+2)\cdots (x+n-1), &{} n\in \mathbb {N}:\{1,2,\ldots \}. \end{array} \right. \end{aligned}$$
(1.4)

We note that F(a, b; c; 1) converges for \(\mathrm {Re}(c-a-b)>0\) and is related to the Gamma function by

$$\begin{aligned} F(a,b;c;1)=\frac{\Gamma (c)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}. \end{aligned}$$
(1.5)

The Gauss hypergeometric function F(a, b; c; z) is one of the most important special functions in complex analysis. Due to its three defining parameters, its structure is extremely rich and contains almost all special functions as a particular or limiting case. Important orthogonal polynomials such as Chebyshev, Legendre, Gegenbauer, and Jacobi polynomials, can all be expressed with the F(a, b; c; z) function, see [9].

Silverman [12] gave necessary and sufficient conditions for zF(a, b; c; z) to be in the classes \(T^{*}{(\alpha )}\) and \(C {(\alpha )}\), and also examined a linear operator acting on hypergeometric functions. For more details, see the works done in [1,2,3,4,5,6, 10, 13].

In the present paper, we determine necessary and sufficient conditions for zF(a, b; c; z) to be in our new classes \(\mathcal {S}(k,\lambda )\) and \( \mathcal {C}(k,\lambda ).~\)Furthermore, we consider an integral operator related to hypergeometric functions.

The proof of Lemma 1.1 below is much akin to that of Theorem 1 in [7], so we choose to omit the details involved.

Lemma 1.1

  1. (i)

    A function f of the form (1.1) is in \(\mathcal {S} (k,\lambda )\) if and only if it satisfies

    $$\begin{aligned} \sum \limits _{n=2}^{\infty }[n((1-\lambda )+k(1+\lambda ))-(1-\lambda )(1-k)]a_{n}\le 2k \end{aligned}$$
    (1.6)

    where \(0<k\le 1\) and \(0\le \lambda <1.\) The result is sharp.

  2. (ii)

    A function f of the form (1.1) is in \( \mathcal {C}(k,\lambda )\) if and only if it satisfies

    $$\begin{aligned} \sum \limits _{n=2}^{\infty }n[n((1-\lambda )+k(1+\lambda ))-(1-\lambda )(1-k)]a_{n}\le 2k \end{aligned}$$
    (1.7)

    where \(0<k\le 1\) and \(0\le \lambda <1.\) The result is sharp.

2 The necessary and sufficient conditions

Unless otherwise mentioned, we shall assume in this paper that \(0<k\le 1\) and \(\ 0\le \lambda <1.\)

Theorem 2.1

  1. (i)

    If \(a,b>-1\), \(c>0,\) and \(ab<0\), then zF(a, b; c; z) is in \( \mathcal {S}(k,\lambda )\) if and only if

    $$\begin{aligned} c\ge a+b+1-\frac{((1-\lambda )+k(1+\lambda ))ab}{2k}. \end{aligned}$$
    (2.1)
  2. (ii)

    If \(a,b>0\), \(c>a+b+1,\) then \( F_{1}(a,b;c;z)=z(2-F(a,b;c;z))\) is in \(\mathcal {S}(k,\lambda )\) if and only if

    $$\begin{aligned} \frac{\Gamma (c)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}\left( 1+\frac{ ((1-\lambda )+k(1+\lambda ))ab}{2k(c-a-b-1)}\right) \le 2. \end{aligned}$$
    (2.2)

Proof

  1. (i)

    The function zF(a, b; c; z) can be written as

    $$\begin{aligned} zF(a,b;c;z)= & {} z+\frac{ab}{c}\sum \limits _{n=2}^{\infty }\frac{ (a+1)_{n-2}(b+1)_{n-2}}{(c+1)_{n-2}(1)_{n-1}}z^{n} \nonumber \\= & {} z-\left| \frac{ab}{c}\right| \sum \limits _{n=2}^{\infty }\frac{ (a+1)_{n-2}(b+1)_{n-2}}{(c+1)_{n-2}(1)_{n-1}}z^{n}. \end{aligned}$$
    (2.3)

    According to (1.6) of Lemma 1.1, we must show that

    $$\begin{aligned} \sum \limits _{n=2}^{\infty }[n((1-\lambda )+k(1+\lambda ))+(1-\lambda )(k-1)] \frac{(a+1)_{n-2}(b+1)_{n-2}}{(c+1)_{n-2}(1)_{n-1}}\le \left| \frac{c}{ ab}\right| 2k. \end{aligned}$$
    (2.4)

    Noting that \((\tau )_{n}=\tau (\tau +1)_{n-1}\) and then applying (1.5), we have

    $$\begin{aligned}&\sum \limits _{n=0}^{\infty }[n((1-\lambda )+k(1+\lambda ))+(1-\lambda )(k-1)]\frac{(a+1)_{n}(b+1)_{n}}{(c+1)_{n}(1)_{n+1}} \\&\quad =\ (1-\lambda )+k(1+\lambda )\sum \limits _{n=0}^{\infty }\frac{ (a+1)_{n}(b+1)_{n}}{(c+1)_{n}(1)_{n}}+2k\frac{c}{ab}\sum \limits _{n=1}^{ \infty }\frac{(a)_{n}(b)_{n}}{(c)_{n}(1)_{n}}\ \\&\quad =(1-\lambda )+k(1+\lambda )\frac{\Gamma (c)\Gamma (c-a-b-1)}{\Gamma (c-a)\Gamma (c-b)}+2k\frac{c}{ab}\left( \frac{\Gamma (c)\Gamma (c-a-b)}{ \Gamma (c-a)\Gamma (c-b)}-1\right) \!. \end{aligned}$$

    Hence, (2.4) is equivalent to

    $$\begin{aligned}&\frac{\Gamma (c)\Gamma (c-a-b-1)}{\Gamma (c-a)\Gamma (c-b)}\left( (1-\lambda )+k(1+\lambda )+2k\frac{c-a-b-1}{ab}\right) \nonumber \\&\quad \le 2k\left( \frac{c}{ \left| ab\right| }+\frac{c}{ab}\right) =0. \end{aligned}$$
    (2.5)

    Thus, (2.5) is valid if and only if \((1-\lambda )+k(1+\lambda )+2k\frac{ c-a-b-1}{ab}\le 0\) or, \(c\ge a+b+1-\frac{((1-\lambda )+k(1+\lambda ))ab}{2k }.\)

  2. (ii)

    Since

    $$\begin{aligned} F_{1}(a,b;c;z)=z-\sum \limits _{n=2}^{\infty }\frac{(a)_{n-1}(b)_{n-1}}{ (c)_{n-1}(1)_{n-1}}z^{n}, \end{aligned}$$

    from the condition (1.6), we need only to show that

    $$\begin{aligned} \sum \limits _{n=2}^{\infty }[n((1-\lambda )+k(1+\lambda ))-(1-\lambda )(1-k)] \frac{(a)_{n-1}(b)_{n-1}}{(c)_{n-1}(1)_{n-1}}\le 2k. \end{aligned}$$

    Now,

    $$\begin{aligned}&\sum \limits _{n=2}^{\infty }[n((1-\lambda )+k(1+\lambda ))-(1-\lambda )(1-k)]\frac{(a)_{n-1}(b)_{n-1}}{(c)_{n-1}(1)_{n-1}} \\&\quad =(1-\lambda )+k(1+\lambda )\sum \limits _{n=1}^{\infty }\frac{n(a)_{n}(b)_{n} }{(c)_{n}(1)_{n}}+2k\sum \limits _{n=1}^{\infty }\frac{(a)_{n}(b)_{n}}{ (c)_{n}(1)_{n}} \\&\quad =\frac{((1-\lambda )+k(1+\lambda ))ab}{c}\sum \limits _{n=1}^{\infty }\frac{ (a+1)_{n-1}(b+1)_{n-1}}{(c+1)_{n-1}(1)_{n-1}}+2k\sum \limits _{n=1}^{\infty } \frac{(a)_{n}(b)_{n}}{(c)_{n}(1)_{n}} \\&\quad =\frac{\Gamma (c)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}\left( \frac{ ((1-\lambda )+k(1+\lambda ))ab}{c-a-b-1}+2k\right) -2k. \end{aligned}$$

    But this last expression is bounded above by 2k if and only if (2.2) holds. \(\square \)

Theorem 2.2

  1. (i)

    If \(a,b>-1\), \(ab<0,\)\(c>a+b+2\), then zF(a, b; c; z) is in \( \mathcal {C}(k,\lambda )\) if and only if

    $$\begin{aligned} ((1-\lambda )+k(1+\lambda ))(a)_{2}(b)_{2}+2((1-\lambda )+k(2+\lambda ))ab(c-a-b-2)+2k(c-a-b-2)_{2}\ge 0. \end{aligned}$$
    (2.6)
  2. (ii)

    If \(a,b>0,\)\(c>a+b+2,\) then \( F_{1}(a,b;c;z)=z(2-F(a,b;c;z))\) is in \(\mathcal {C}(k,\lambda )\) if and only if

    $$\begin{aligned}&\frac{\Gamma (c)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}\left( \frac{ ((1-\lambda )+k(1+\lambda ))(a)_{2}(b)_{2}}{2k(c-a-b-2)_{2}}\nonumber \right. \\&\quad \left. +\left( \frac{ 1-\lambda +k(2+\lambda )}{k}\right) \left( \frac{ab}{c-a-b-1}\right) +1\right) \le 2. \end{aligned}$$
    (2.7)

Proof

  1. (i)

    From (2.3) and (1.7) it follows that

    $$\begin{aligned} \sum \limits _{n=2}^{\infty }n[n((1-\lambda )+k(1+\lambda ))-(1-\lambda )(1-k)] \frac{(a+1)_{n-2}(b+1)_{n-2}}{(c+1)_{n-2}(1)_{n-1}}\le \frac{2ck}{ \left| ab\right| }. \end{aligned}$$

    Writing

    $$\begin{aligned}&(n+2)[(n+2)((1-\lambda )+k(1+\lambda ))-(1-\lambda )(1-k)] \\&\quad =((1-\lambda )+k(1+\lambda ))(n+1)^{2}+((1-\lambda )+k(3+\lambda ))(n+1)+2k, \end{aligned}$$

    we see that

    $$\begin{aligned}&\sum \limits _{n=0}^{\infty }(n+2)[(n+2)((1-\lambda )+k(1+\lambda ))-(1-\lambda )(1-k)]\frac{(a+1)_{n}(b+1)_{n}}{(c+1)_{n}(1)_{n+1}}\\&\quad =((1-\lambda )+k(1+\lambda ))\sum \limits _{n=0}^{\infty }(n+1)\frac{ (a+1)_{n}(b+1)_{n}}{(c+1)_{n}(1)_{n}} \\&\qquad +\,((1-\lambda )+k(3-\lambda ))\sum \limits _{n=0}^{\infty }\frac{ (a+1)_{n}(b+1)_{n}}{(c+1)_{n}(1)_{n}}+2k\sum \limits _{n=0}^{\infty }\frac{ (a+1)_{n}(b+1)_{n}}{(c+1)_{n}(1)_{n+1}}\ \ \\&\quad =\ \frac{((1-\lambda )+k(1+\lambda ))(a+1)(b+1)}{c+1} \sum \limits _{n=0}^{\infty }\frac{(a+2)_{n}(b+2)_{n}}{(c+2)_{n}(1)_{n}} \\&\qquad +\,2[(1-\lambda )+k(2+\lambda )]\sum \limits _{n=0}^{\infty }\frac{ (a+1)_{n}(b+1)_{n}}{(c+1)_{n}(1)_{n}}+\frac{2kc}{ab}\sum \limits _{n=0}^{ \infty }\frac{(a)_{n}(b)_{n}}{(c)_{n}(1)_{n}}\ \\&\quad =\frac{\Gamma (c+1)\Gamma (c-a-b-2)}{\Gamma (c-a)\Gamma (c-b)}\left( \phantom {\frac{2k}{ab}} ((1-\lambda )+k(1+\lambda ))(a+1)(b+1)\right. \\&\qquad \left. +\,2[(1-\lambda )+k(2+\lambda )]\left( c-a-b-2\right) +\frac{2k}{ab} \left( c-a-b-2\right) _{2}\right) -\frac{2kc}{ab}. \end{aligned}$$

    This last expression is bounded above by \(2ck/\left| ab\right| ~\)if and only if 

    $$\begin{aligned}&((1-\lambda )+k(1+\lambda ))(a+1)(b+1)+2((1-\lambda )+k(2+\lambda ))\left( c-a-b-2\right) \\&\quad +\,\frac{2k}{ab}\left( c-a-b-2\right) _{2} \\&\quad \le 0, \end{aligned}$$

    which is equivalent to (2.6).

  2. (ii)

    In view of (1.7), we need only to show that

    $$\begin{aligned} \sum \limits _{n=2}^{\infty }n[n((1-\lambda )+k(1+\lambda ))-(1-\lambda )(1-k)] \frac{(a)_{n-1}(b)_{n-1}}{(c)_{n-1}(1)_{n-1}}\le 2k. \end{aligned}$$

    Now,

    $$\begin{aligned}&\sum \limits _{n=0}^{\infty }(n+2)[(n+2)((1-\lambda )+k(1+\lambda ))-(1-\lambda )(1-k)]\frac{(a)_{n+1}(b)_{n+1}}{(c)_{n+1}(1)_{n+1}}\nonumber \\&\quad =((1-\lambda )+k(1+\lambda ))\sum \limits _{n=0}^{\infty }(n+2)^{2}\frac{ (a)_{n+1}(b)_{n+1}}{(c)_{n+1}(1)_{n+1}}\nonumber \\&\qquad -\,(1-\lambda )(1-k)\sum \limits _{n=0}^{\infty }(n+2)\frac{(a)_{n+1}(b)_{n+1}}{ (c)_{n+1}(1)_{n+1}}. \end{aligned}$$
    (2.8)

    Writing \(n+2=(n+1)+1,\) we have

    $$\begin{aligned}&\sum \limits _{n=0}^{\infty }(n+2)\frac{(a)_{n+1}(b)_{n+1}}{(c)_{n+1}(1)_{n+1}} =\sum \limits _{n=0}^{\infty }\frac{(a)_{n+1}(b)_{n+1}}{(c)_{n+1}(1)_{n}} +\sum \limits _{n=0}^{\infty }\frac{(a)_{n+1}(b)_{n+1}}{(c)_{n+1}(1)_{n+1}},\nonumber \\&\sum \limits _{n=0}^{\infty }(n+2)^{2}\frac{(a)_{n+1}(b)_{n+1}}{ (c)_{n+1}(1)_{n+1}}\nonumber \\&\quad =\sum \limits _{n=0}^{\infty }(n+1)\frac{(a)_{n+1}(b)_{n+1}}{(c)_{n+1}(1)_{n} }+2\sum \limits _{n=0}^{\infty }\frac{(a)_{n+1}(b)_{n+1}}{(c)_{n+1}(1)_{n}} +\sum \limits _{n=0}^{\infty }\frac{(a)_{n+1}(b)_{n+1}}{(c)_{n+1}(1)_{n+1}} \nonumber \\&\quad =\sum \limits _{n=1}^{\infty }\frac{(a)_{n+1}(b)_{n+1}}{(c)_{n+1}(1)_{n-1}} +3\sum \limits _{n=0}^{\infty }\frac{(a)_{n+1}(b)_{n+1}}{(c)_{n+1}(1)_{n}} +\sum \limits _{n=1}^{\infty }\frac{(a)_{n}(b)_{n}}{(c)_{n}(1)_{n}}. \end{aligned}$$
    (2.9)

    Substituting (2.9) into the right-hand side of (2.8), we obtain

    $$\begin{aligned}&((1-\lambda )+k(1+\lambda ))\sum \limits _{n=0}^{\infty }\frac{ (a)_{n+2}(b)_{n+2}}{(c)_{n+2}(1)_{n}}+2((1-\lambda )+k(2+\lambda ))\sum \limits _{n=0}^{\infty }\frac{(a)_{n+1}(b)_{n+1}}{(c)_{n+1}(1)_{n}} \nonumber \\&\quad +\,2k\sum \limits _{n=0}^{\infty }\frac{(a)_{n+1}(b)_{n+1}}{(c)_{n+1}(1)_{n+1}} . \end{aligned}$$
    (2.10)

    Since \((a)_{n+j}=(a)_{j}(a+k)_{n},\ \)we write (2.10) as

    $$\begin{aligned}&\frac{((1-\lambda )+k(1+\lambda ))(a)_{2}(b)_{2}}{(c)_{2}}\frac{\Gamma (c+2)\Gamma (c-a-b-2)}{\Gamma (c-a)\Gamma (c-b)} \\&\quad +\,2((1-\lambda )+k(2+\lambda )){\frac{ab}{c}}{\frac{\Gamma (c+1)\Gamma (c-a-b-1)}{\Gamma (c-a)\Gamma (c-b)}} \\&\quad +\,2k\left( \frac{\Gamma (c)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)} -1\right) . \end{aligned}$$

    By simplification, we see that the last expression is bounded above by \( (1-\alpha )\) if and only if (3.6) holds. \(\square \)

3 An integral operator

In the next theorems, we obtain similar-type in connections with a particular integral operator G(a, b; c; z) acting on F(a, b; c; z) as follows:

$$\begin{aligned} G(a,b;c;z)=\int \limits _{0}^{z}F(a,b;c;t)dt. \end{aligned}$$
(3.1)

Theorem 3.1

Let \(a,b>-1\), \(ab<0,\) and \(c>\max \{0,a+b\}.\) Then, G(a, b; c; z) defined by (3.1) is in \(\mathcal {S}(k,\lambda )\) if and only if

$$\begin{aligned}&\frac{\Gamma (c+1)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}\left( \frac{ ((1-\lambda )+k(1+\lambda ))}{ab}-\frac{(1-\lambda )(1-k)(c-a-b)}{ (a-1)_{2}(b-1)_{2}}\right) \nonumber \\&\quad +\,\frac{(1-\lambda )(1-k)(c-1)_{2}}{(a-1)_{2}(b-1)_{2}}\le 0. \end{aligned}$$
(3.2)

Proof

Since

$$\begin{aligned} G(a,b;c;z)=z-\frac{\left| ab\right| }{c}\sum \limits _{n=2}^{\infty } \frac{(a+1)_{n-2}(b+1)_{n-2}}{(c+1)_{n-2}(1)_{n}}z^{n}, \end{aligned}$$

by (1.6), we need to show that

$$\begin{aligned} \sum \limits _{n=2}^{\infty }[n((1-\lambda )+k(1+\lambda ))-(1-\lambda )(1-k)] \frac{(a+1)_{n-2}(b+1)_{n-2}}{(c+1)_{n-2}(1)_{n}}\le \frac{2kc}{\left| ab\right| }. \end{aligned}$$

Now,

$$\begin{aligned}&\sum \limits _{n=0}^{\infty }[n((1-\lambda )+k(1+\lambda ))-(1-\lambda )(1-k)] \frac{(a+1)_{n}(b+1)_{n}}{(c+1)_{n}(1)_{n+2}}\\&\quad =((1-\lambda )+k(1+\lambda ))\sum \limits _{n=0}^{\infty }\frac{ (a+1)_{n}(b+1)_{n}}{(c+1)_{n}(1)_{n+1}}-(1-\lambda )(1-k)\frac{c}{ab} \sum \limits _{n=1}^{\infty }\frac{(a)_{n}(b)_{n}}{(c)_{n}(1)_{n+1}} \\&\quad =\frac{\Gamma (c+1)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}\left( \frac{ (1-\lambda )+k(1+\lambda )}{ab}-\frac{(1-\lambda )(1-k)(c-a-b)}{ (a-1)_{2}(b-1)_{2}}\right) \\&\qquad +\,\frac{(1-\lambda )(1-k)(c-1)_{2}}{(a-1)_{2}(b-1)_{2}}-\frac{2kc}{ab}\le \frac{2kc}{\left| ab\right| }, \end{aligned}$$

which is equivalent to (3.2). \(\square \)

Now, we observe that \(G(a,b;c;z)\in \mathcal {C}(k,\lambda )\) if and only if \( zF(a,b;c;z)\in \mathcal {S}(k,\lambda ).\) Thus, any result of functions belonging to the class \(\mathcal {S}(k,\lambda )\) about zF leads to that of functions belonging to the class \(\mathcal {C}(k,\lambda )\). Hence, we obtain the following analogues to Theorem 2.1.

Theorem 3.2

Let \(a,b>-1\), \(ab<0,\) and \(c>a+b+2.\) Then, G(a, b; c; z) defined by (3.1) is in \(\mathcal {C}(k,\lambda )\) if and only if

$$\begin{aligned} c\ge a+b+1-\frac{((1-\lambda )+k(1+\lambda ))ab}{2k}. \end{aligned}$$

Letting \(\lambda =0\) in Theorems 2.1, 2.2, 3.1 and , we obtain the following corollaries.

Corollary 3.3

  1. (i)

    If \(a,b>-1\), \(c>0,\) and \(ab<0,\) then zF(a, b; c; z) is in \(\mathcal {S}(k)\) if and only if

    $$\begin{aligned} c\ge a+b+1-\frac{(1+k)ab}{2k}. \end{aligned}$$
    (3.3)
  2. (ii)

     If \(a,b>0\), \(c>a+b+1,\) then \( F_{1}(a,b;c;z)=z(2-F(a,b;c;z))\) is in \(\mathcal {S}(k)\) if and only if

    $$\begin{aligned} \frac{\Gamma (c)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}\left( 1+\frac{ (1+k)ab}{2k(c-a-b-1)}\right) \le 2. \end{aligned}$$
    (3.4)

Corollary 3.4

  1. (i)

    If \(a,b>-1\), \(ab<0,\)\(c>a+b+2,\) then zF(a, b; c; z) is in \(\mathcal {C}(k)\) if and only if

    $$\begin{aligned} (1+k)(a)_{2}(b)_{2}+2(1+2k)ab(c-a-b-2)+2k(c-a-b-2)_{2}\ge 0. \end{aligned}$$
    (3.5)
  2. (ii)

     If \(a,b>0\), \(c>a+b+2,\) then \( F_{1}(a,b;c;z)=z(2-F(a,b;c;z))\) is in \(\mathcal {C}(k,\lambda )\) if and only if

    $$\begin{aligned} \frac{\Gamma (c)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}\left( \frac{ (1+k)(a)_{2}(b)_{2}}{2k(c-a-b-2)_{2}}+\left( \frac{1+2k}{k}\right) \left( \frac{ab}{c-a-b-1}\right) +1\right) \le 2. \end{aligned}$$
    (3.6)

Corollary 3.5

Let \(a,b>-1\), \(ab<0,\) and \(c>\max \{0,a+b\}.\) Then, G(a, b; c; z) defined by (3.1) is in \(\mathcal {S}(k)\) if and only if

$$\begin{aligned} \frac{\Gamma (c+1)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}\left( \frac{(1+k) }{ab}-\frac{(1-k)(c-a-b)}{(a-1)_{2}(b-1)_{2}}\right) +\frac{(1-k)(c-1)_{2}}{ (a-1)_{2}(b-1)_{2}}\le 0. \end{aligned}$$
(3.7)

Let \(a,b>-1\), \(ab<0,\) and \(c>a+b+2.\) Then, G(a, b; c; z) defined by (3.1 ) is in \(\mathcal {C}(k)\) if and only if

$$\begin{aligned} c\ge a+b+1-\frac{(1+k)ab}{2k}. \end{aligned}$$