1 Introduction

On a simply connected domain \(\varOmega\subset \mathbb{C}\) a complex-valued harmonic mapping f can be written as \(f=h+\overline{g}\), where h and g are analytic mappings. By Lewy [8], it is locally univalent sense-preserving if and only if its Jacobian \(\mathcal{J}_{f}=|h'|^{2}-|g'|^{2}\) is positive or, equivalently, its dilatation \(\omega_{f}:=g'/h'\) lies in \(\mathbb{D}:=\{ z\in\mathbb{C}:|z|<1\}\). Let \(\mathcal{H}\) denote the class of all locally univalent sense-preserving harmonic mappings \(f=h+\overline {g}\) defined on \(\mathbb{D}\). Also, let \(\mathcal{S}_{H}\) denote the subclass of \(\mathcal{H}\) consisting of univalent mappings with normalization \(f(0)=0=f_{z}(0)-1\). Moreover, let \(\mathcal{S}_{H}^{0}\) be the subclass of \(\mathcal{S}_{H}\) that contains all mappings \(f=h+\overline{g}\) such that \(f_{\overline {z}}(0)=0\). For \(0\leq\nu<\pi\), a mapping φ is called convex in the direction ν if \(\varphi(\mathbb{D})\) has connected intersection with every line that is parallel to the line joining \(e^{ i \nu}\) to the origin. Such a mapping is also called a directional convex mapping. If \(\nu=0\) (or \(\pi/2\)), then φ is known as convex in the real (or imaginary) direction. A harmonic mapping \(f=h+\overline{g}\in\mathcal{S}_{H}^{0}\) is said to be a right half-plane or a vertical strip mapping if it maps \(\mathbb{D}\) onto the right half-plane

$$R=\bigl\{ w\in\mathbb{C}: \operatorname {Re}(w)>-1/2\bigr\} $$

or the vertical strip

$$V_{\alpha}:= \biggl\{ w\in\mathbb{C}: \frac{\alpha-\pi}{2\sin \alpha} < \operatorname {Re}w < \frac{\alpha}{2\sin\alpha} \biggr\} , \quad \frac{\pi}{2}< \alpha< \pi, $$

respectively. It is well known [1, 4] that if \(f=h+\overline{g}\) is a right half-plane harmonic mapping then \(h'(z)+g'(z)=(1-z)^{-2}\), and if it is a vertical strip harmonic mapping then \(h'(z)+g'(z)=(1+2z \cos{\alpha}+z^{2})^{-1}\). In this article, we find some sufficient conditions for the convex combination of the right half-plane mappings, the vertical strip mappings, their rotations, and some other harmonic mappings to be univalent and convex in a particular direction. Generally, the convex combination of two analytic/harmonic mappings does not carry the univalency or other geometric properties of individual mappings. One can refer to the survey article by Campbell [2] and the references therein for the univalency and other geometric properties of the convex combination of analytic mappings. However, recently, a convex combination of some harmonic mappings has been studied in [5, 7, 11–13]. In particular, Wang et al. [13] and Kumar et al. [7] respectively studied the directional convexity of convex combination of harmonic mappings, which are shears of the analytic mappings \(z/(1-z)\) and \(z(1-\alpha z)/(1-z^{2})\), \(-1\leq\alpha\leq1\). Motivated by the work carried out in [7, 13], we study the convex combination of harmonic mappings which are shears of the analytic mapping \(\psi_{\mu,\nu}p_{k}\), where \(p_{k}\) is analytic with positive real part on \(\mathbb{D}\) and

$$ \psi_{\mu,\nu}(z)=\frac{1}{1-2z e^{ -i \mu}\cos\nu+z^{2} e^{-2 i \mu}},\quad\mu, \nu\in[0, 2\pi). $$
(1.1)

In particular, we show that the combination \(f=tf_{1}+(1-t)f_{2}\), \(0\leq t\leq1\) of the mappings \(f_{k}=h_{k}+\overline{g_{k}}\in\mathcal{S}_{H}\), \(k=1,2\), satisfying \(h_{k}'-e^{2i\mu}g_{k}'=\psi_{\mu,\nu}p_{k}\) is univalent and convex in the direction μ for some specific dilatations of \(f_{1}\) and \(f_{2}\). The following result by Royster and Ziegler [10] is used to check the convexity in a particular direction of analytic mappings.

Lemma 1.1

Letϕbe a non-constant analytic mapping in\(\mathbb{D}\). Thenϕmaps\(\mathbb{D}\)onto a domain convex in the directionγ (\(0\leq\gamma<\pi\)) if and only if there are real numbersμandν (\(0\leq\nu<2\pi\)) such that

$$ \operatorname {Re}\bigl( e^{ i (\mu-\gamma)}\bigl(1-2z e^{- i \mu}\cos \nu+z^{2} e^{-2 i \mu}\bigr)\phi'(z) \bigr) \geq0,\quad z\in\mathbb{D}. $$
(1.2)

Remark 1.2

By taking γ or \(\gamma+\pi\) equal to μ in Lemma 1.1, we see that a non-constant analytic mapping ϕ is convex in the direction μ if, for some real number ν (\(0\leq \nu<2\pi\)), the real part of the mapping \(\phi'/\psi_{\mu,\nu}\), where \(\psi_{\mu,\nu}\) is given by (1.1), is either non-negative or non-positive on \(\mathbb{D}\).

Lemma 1.1 along with the following result due to Clunie and Sheil-Small [3], known as shear construction, is used to check the convexity in a particular direction of harmonic mappings.

Lemma 1.3

A locally univalent and sense-preserving harmonic mapping\(f=h+\overline{g}\)on\(\mathbb{D}\)is univalent and maps\(\mathbb {D}\)onto a domain convex in the directionγ (\(0\leq\gamma <\pi\)) if and only if the analytic mapping\(h-e^{2 i \gamma}g\)is univalent and maps\(\mathbb{D}\)onto a domain convex in the direction γ.

2 Main results

Theorem 2.1

For\(k=1,2\), let\(f_{k}=h_{k}+\overline{g_{k}}\in\mathcal{S}_{H}\)such that

$$ h_{k}(z)-e^{2i\mu}g_{k}(z)= \int_{0}^{z} \psi_{\mu,\nu}(\xi)p_{k}( \xi)\, {d}\xi,\quad \mu, \nu\in[0,2\pi), $$
(2.1)

where\(p_{k}\)is an analytic mapping with\(\operatorname {Re}p_{k}>0\)on\(\mathbb{D}\)and\(\psi_{\mu, \nu}\)is given by (1.1). Then the mapping\(f=tf_{1}+(1-t)f_{2}\)is univalent and is convex in the directionμfor\(0\leq t \leq1\)if it is locally univalent and sense-preserving.

Proof

Let \(f=h+\overline{g}\), then

$$h=th_{1}+(1-t)h_{2}\quad\text{and}\quad g= tg_{1}+(1-t)g_{2}, $$

and thus

$$h-e^{2i\mu}g=t\bigl(h_{1}-e^{2i\mu}g_{1} \bigr)+(1-t) \bigl(h_{2}-e^{2i\mu}g_{2}\bigr). $$

Therefore, in view of (2.1), it follows that

$$\operatorname {Re}\biggl(\frac{h'-e^{2i\mu}g'}{\psi_{\mu,\nu}} \biggr)=t\operatorname {Re}p_{1}+(1-t)\operatorname {Re}p_{2}>0 $$

on \(\mathbb{D}\) for \(0\leq t\leq1\). Hence, by Lemma 1.1, it follows that the mapping \(h-e^{2i\mu}g\) is convex in the direction μ. The result now follows by Lemma 1.3. □

Theorem 2.1 has the following obvious extension to n mappings.

Theorem 2.2

For\(k=1,2,\dots,n\), let\(f_{k}=h_{k}+\overline{g_{k}}\in\mathcal{S}_{H}\)satisfy (2.1), where\(p_{k}\)is an analytic mapping with\(\operatorname {Re}p_{k}>0\)on\(\mathbb{D}\)and\(\psi_{\mu, \nu}\)is given by (1.1). If\(\sum_{t=1}^{n}t_{k}=1\), \(0\leq t_{k}\leq1\), then the mapping\(f=\sum_{t=1}^{n}t_{k}f_{k}\)is univalent and is convex in the directionμprovided it is locally univalent and sense-preserving.

In Theorems 2.1 and 2.2 we assumed f to be locally univalent and sense-preserving on \(\mathbb{D}\). Next, we will study some cases where this assumption can be relaxed.

Theorem 2.3

For\(k=1,2\), let\(f_{k}=h_{k}+\overline{g_{k}}\in\mathcal{S}_{H}\)satisfy (2.1), where\(p_{k}\)is an analytic mapping with\(\operatorname {Re}p_{k}>0\)on\(\mathbb{D}\)and\(\psi_{\mu, \nu}\)is given by (1.1). Let\(\omega_{f_{k}}\)be the dilatation of\(f_{k}\), then the mapping\(f=tf_{1}+(1-t)f_{2}\)is univalent and is convex in the directionμfor\(0\leq t\leq1\)if\(\omega_{f_{k}}\)and\(p_{k}\)satisfy one of the following:

  1. (i)

    \(\omega_{f_{1}}=\omega_{f_{2}}\),

  2. (ii)

    \(p_{1}/(1-e^{2i\mu}\omega_{f_{1}}) = p_{2}/(1-e^{2i\mu}\omega _{f_{2}})\),

  3. (iii)

    \(p_{1}=p_{2}\),

  4. (iv)

    \(\omega_{f_{2}}=-\omega_{f_{1}}\)and\(\operatorname {Re}(p_{2}(1-e^{2i\mu}\omega_{f_{1}})/(p_{1}(1+e^{2i\mu}\omega _{f_{1}})) )>0\).

Proof

In view of Theorem 2.1, it is enough to show that f is locally univalent and sense-preserving or, equivalently, \(|\omega _{f}|<1\) on \(\mathbb{D}\), where \(\omega_{f}\) is the dilatation of f. Since for \(t=0\) and 1 the result is obvious, we consider \(0< t<1\). On differentiation (2.1) gives

$$h_{k}'-e^{2i\mu}g_{k}'= \psi_{\mu,\nu}p_{k}. $$

The above equation along with \(g_{k}'=\omega_{f_{k}}h_{k}'\) gives

$$ h_{k}'=\frac{\psi_{\mu,\nu} p_{k}}{1- e^{2i\mu}\omega_{f_{k}}}. $$
(2.2)

Since \(f=h+\overline{g}:=th_{1}+(1-t)h_{2}+\overline{tg_{1}+(1-t)g_{2}}\), in view of (2.2), \(\omega_{f}\) is given by

$$\begin{aligned} \omega_{f}&=\frac{g'}{h'}=\frac{tg_{1}'+(1-t)g_{2}'}{th_{1}'+(1-t)h_{2}'} \\ &=\frac{t\omega_{f_{1}}h_{1}'+(1-t)\omega _{f_{2}}h_{2}'}{th_{1}'+(1-t)h_{2}'} \\ &=\frac{t\omega_{f_{1}}(1- e^{2i\mu}\omega_{f_{2}})p_{1}+(1-t)\omega _{f_{2}}(1- e^{2i\mu}\omega_{f_{1}})p_{2}}{t(1- e^{2i\mu}\omega _{f_{2}})p_{1}+(1-t)(1- e^{2i\mu}\omega_{f_{1}})p_{2}}. \end{aligned}$$
(2.3)

Let \(\omega_{f_{1}}=\omega_{f_{2}}\), then (2.3) gives that \(\omega_{f}=\omega_{f_{1}}\) and hence \(|\omega_{f}|<1\). Also, let \(p_{k}\) and \(\omega_{f_{k}}\) be given by (ii), then (2.3) gives that \(\omega_{f}=t\omega_{f_{1}}+(1-t)\omega_{f_{2}}\). Hence, \(|\omega_{f_{k}}|<1\) follows that \(|\omega_{f}|<1\). Moreover, let \(p_{1}=p_{2}\), then (2.3) shows that

$$\omega_{f}=\frac{t\omega_{f_{1}}(1- e^{2i\mu}\omega _{f_{2}})+(1-t)\omega_{f_{2}}(1- e^{2i\mu}\omega_{f_{1}})}{t(1- e^{2i\mu }\omega_{f_{2}})+(1-t)(1- e^{2i\mu}\omega_{f_{1}})}. $$

Therefore, \(|\omega_{f_{k}}|<1\) implies that

$$\operatorname {Re}\biggl(\frac{1+e^{2i\mu}\omega_{f}}{1-e^{2i\mu}\omega_{f}} \biggr)=t\operatorname {Re}\biggl(\frac{1+ e^{2i\mu}\omega_{f_{1}}}{1- e^{2i\mu}\omega _{f_{1}}} \biggr)+(1-t)\operatorname {Re}\biggl(\frac{1+ e^{2i\mu}\omega_{f_{2}}}{1- e^{2i\mu}\omega_{f_{2}}} \biggr)>0. $$

Hence, \(|\omega_{f}|<1\). Lastly, let \(\omega_{f_{2}}=-\omega_{f_{1}}\), then from (2.3) we have

$$\omega_{f}=\omega_{f_{1}}\frac{t(1+ e^{2i\mu}\omega_{f_{1}})p_{1}-(1-t)(1- e^{2i\mu}\omega_{f_{1}})p_{2}}{t(1+ e^{2i\mu}\omega_{f_{1}})p_{1}+(1-t)(1- e^{2i\mu}\omega_{f_{1}})p_{2}}=: \omega_{f_{1}}\varphi. $$

Therefore, \(|\omega_{f}|<1\) if \(|\varphi|<1\). Now, by the assumption in (iv), we have

$$\operatorname {Re}\biggl(\frac{1+\varphi}{1-\varphi} \biggr)=\operatorname {Re}\biggl(\frac {t(1+ e^{2i\mu}\omega_{f_{1}})p_{1}}{(1-t)(1- e^{2i\mu}\omega _{f_{1}})p_{2}} \biggr)>0. $$

Hence, \(|\varphi|<1\). This proves the result when \(\omega_{f_{k}}\) and \(p_{k}\) satisfy condition (iv). This completes the proof. □

From its proof, it is easily seen that Theorem 2.3, except case (iv), has a natural extension to n mappings as follows.

Theorem 2.4

For\(k=1,2,\dots,n\), let\(f_{k}=h_{k}+\overline{g_{k}}\in\mathcal{S}_{H}\)have dilatation\(\omega_{f_{k}}\)and satisfy (2.1), where\(p_{k}\)is an analytic mapping with\(\operatorname {Re}p_{k}>0\)on\(\mathbb{D}\)and\(\psi_{\mu, \nu}\)is given by (1.1). If\(\sum_{t=1}^{n}t_{k}=1\), \(0\leq t_{k}\leq1\), then the mapping\(f=\sum_{t=1}^{n}t_{k}f_{k}\)is univalent and is convex in the directionμprovided\(\omega _{f_{k}}\)and\(p_{k}\)satisfy one of the following:

  1. (i)

    \(\omega_{f_{1}}=\omega_{f_{2}}=\cdots=\omega_{f_{n}}\),

  2. (ii)

    \(p_{1}/(1-e^{2i\mu}\omega_{f_{1}}) = p_{2}/(1-e^{2i\mu}\omega _{f_{2}}) = \cdots= p_{k}/(1-e^{2i\mu}\omega_{f_{k}})\),

  3. (iii)

    \(p_{1}=p_{2}=\cdots=p_{n}\).

The following example gives an illustration of Theorem 2.3.

Example 2.5

For \(k=1, 2\), let \(f_{k}=h_{k}+\overline{g_{k}}\in\mathcal{S}_{H}\) be given by

$$\begin{aligned} f_{1}(z)= h_{1}(z)+\overline{g_{1}(z)}= \frac{1}{2}\log \biggl(\frac {1+z}{1-z} \biggr)+\overline{z- \frac{1}{2}\log \biggl(\frac {1+z}{1-z} \biggr)} \end{aligned}$$

and

$$\begin{aligned} f_{2}(z)= h_{2}(z)+\overline{g_{2}(z)}= \frac{1}{2}\log \biggl(\frac {1+z}{1-z} \biggr)+\overline{ \frac{1}{2}\log\frac{1}{1-z^{2}}}. \end{aligned}$$

Then, \(\omega_{f_{k}}\), the dilatation of \(f_{k}\), is given by \(\omega _{f_{1}}(z)=-z^{2}\) and \(\omega_{f_{2}}(z)=z\). Also, we can see that

$$h_{k}'(z)+g_{k}'(z)= \frac{1+\omega_{f_{k}}(z)}{1-z^{2}}. $$

Thus, \(f_{k}\) satisfies (2.1) with \(\mu=\pi/2\), \(\nu=\pi/2\) and \(p_{k}=1+\omega_{f_{k}}\), where \(\operatorname {Re}p_{k} >0\) on \(\mathbb{D}\). Therefore, it follows from Theorem 2.3 that the mapping \(f=tf_{1}+(1-t)f_{2}\) is univalent and convex in the imaginary direction for \(0\leq t\leq1\). Images of \(\mathbb{D}\) under f at \(t=1\), \(t=0\), and \(t=1/3\) are shown in Fig. 1.

Figure 1
figure 1

Images of \(\mathbb{D}\) under f at different values of t

Full size image

We will use the following lemma to prove our next results.

Lemma 2.6

For\(n\in\mathbb{N}\)and\(k=1,2\), let\(f_{k}=h_{k}+\overline{g_{k}}\in \mathcal{S}_{H}\)such that

$$ h_{k}(z)-g_{k}(z)=\bigl(1+(-1)^{k}a \bigr) \int_{0}^{z} \frac {q(\xi)\,{d}\xi}{\psi_{\mu,\nu_{k}}(\xi^{n})},\quad\mu, \nu_{k}\in [0,2\pi), $$
(2.4)

whereqis an analytic mapping and\(\psi_{\mu,\nu}\)is defined by (1.1). Let\(\omega_{f_{k}}\)be the dilatation of\(f_{k}\). If

$$ \omega_{f_{1}}(z)=-\omega_{f_{2}}(z)= \frac{a+e^{i(\theta-\mu )}z^{n}}{1+ae^{i(\theta-\mu)}z^{n}},\quad a\in(-1,1), \theta\in[0,2\pi), $$
(2.5)

then the mapping\(f=tf_{1}+(1-t)f_{2}\)is locally univalent and sense-preserving for\(0\leq t \leq1\)provided:

  1. (i)

    \(\cos\theta> \max\{\cos\nu_{1},-\cos\nu_{2}\}\)and\(\cos \nu_{1}>\cos\nu_{2}\), or

  2. (ii)

    \(\cos\theta< \min\{\cos\nu_{1}, -\cos\nu_{2}\}\)and\(\cos\nu_{2}>\cos\nu_{1}\).

To prove the above lemma, we will use the following result commonly known as Cohn’s rule [9].

Theorem 2.7

Let\(r(z)=a_{0}+a_{1}z+\cdots+a_{n}z^{n}\)be a polynomial of degreenand

$$r^{*}(z)=z^{n}\overline{r(1/\overline{z})}=\overline{a}_{n}+ \overline {a}_{n-1}z+\cdots+\overline{a}_{0}z^{n}. $$

Letsand\(s_{1}\)be the number of zeros ofrinside and on the unit circle\(|z|=1\), respectively. If\(|a_{0}| < |a_{n}|\), then

$$r_{1}(z)=\frac{\overline{a}_{n}r(z)-a_{0}r^{*}(z)}{z} $$

is a polynomial of degree\(n-1\)and has\(s-1\)and\(s_{1}\)number of zeros inside and on the unit circle\(|z|=1\), respectively.

Proof of Theorem 2.1

Since \(f_{k}\in\mathcal{S}_{H}\), we need to prove the result only for \(0< t<1\). First of all, we will show that both conditions (i) and (ii) imply

$$ \bigl\vert 1-(\cos\nu_{1}-\cos\nu_{2})e^{-i\theta} \bigr\vert < 1 $$
(2.6)

and

$$ \biggl\vert \frac{\cos\nu_{1}+\cos\nu_{2}}{\cos\nu_{1}-\cos\nu_{2}+2\cos \theta} \biggr\vert < 1. $$
(2.7)

We see that \((\cos\nu_{1}-\cos\nu_{2})(\cos\nu_{1}-\cos\nu_{2}-2\cos \theta)<0\) if condition (i) or (ii) is satisfied. Therefore,

$$\begin{aligned} \bigl\vert 1-(\cos\nu_{1}-\cos\nu_{2})e^{-i\theta} \bigr\vert ^{2}-1&=\bigl(1-(\cos\nu_{1}-\cos \nu_{2})\cos\theta\bigr)^{2}+\bigl((\cos\nu_{1}- \cos\nu_{2})\sin\theta\bigr)^{2} \\ &=(\cos\nu_{1}-\cos\nu_{2}) (\cos\nu_{1}-\cos \nu_{2}-2\cos\theta)< 0. \end{aligned}$$

Hence, both (i) and (ii) imply (2.6). Next, let condition (i) be satisfied. Then \(\cos\theta>\cos\nu_{1}\) and \(\cos\theta>-\cos\nu_{2}\), and hence

$$ \cos\nu_{1}-\cos\nu_{2}-2\cos\theta< \cos \nu_{1}+\cos\nu_{2}< -\cos\nu _{1}+\cos \nu_{2}+2\cos\theta. $$
(2.8)

Similarly, if condition (ii) is satisfied, then

$$ -\cos\nu_{1}+\cos\nu_{2}+2\cos\theta < \cos \nu_{1}+\cos\nu_{2} < \cos\nu_{1}-\cos \nu_{2}-2\cos\theta. $$
(2.9)

Therefore, (2.8) and (2.9) show that both (i) and (ii) imply (2.7).

Now, differentiating (2.4), we have

$$\bigl(h_{k}'(z)-g_{k}'(z)\bigr) \psi_{\mu,\nu_{k}}\bigl(z^{n}\bigr)=\bigl(1+(-1)^{k}a\bigr) q(z). $$

The above equation along with \(g_{k}'=\omega_{f_{k}} h_{k}'\) gives

$$h_{k}'(z)=\frac{(1+(-1)^{k}a) q(z)}{\psi_{\mu,\nu_{k}}(z^{n})(1- \omega_{f_{k}}(z))}. $$

Therefore \(\omega_{f}\), the dilatation of \(f=tf_{1}+(1-t)f_{2}\), is given by

$$\begin{aligned} \omega_{f}(z)&=\frac {tg_{1}'(z)+(1-t)g_{2}'(z)}{th_{1}'(z)+(1-t)h_{2}'(z)} \\ &=\frac{t\omega_{f_{1}}(z)h_{1}'(z)+(1-t)\omega _{f_{2}}(z)h_{2}'(z)}{th'_{1}(z)+(1-t)h_{2}'(z)} \\ &=\frac{t\omega_{f_{1}}(z)\psi_{\mu,\nu_{2}}(z^{n})(1- \omega _{f_{2}}(z))(1-a)+(1-t)\omega_{f_{2}}(z)\psi_{\mu,\nu_{1}}(z^{n})(1-\omega _{f_{1}}(z))(1+a)}{t\psi_{\mu,\nu_{2}}(z^{n})(1- \omega _{f_{2}}(z))(1-a)+(1-t)\psi_{\mu,\nu_{1}}(z^{n})(1-\omega _{f_{1}}(z))(1+a)}. \end{aligned}$$
(2.10)

Now, on substituting the values of \(\omega_{f_{k}}\), given by (2.5), in (2.10), we obtain

$$\begin{aligned} \selectfont\begin{aligned} \omega_{f}(z) &=\omega_{f_{1}}(z)\\&\quad\times \biggl(\frac{t(1+ ae^{i(\theta-\mu )}z^{n}+a+e^{i(\theta-\mu)}z^{n})\psi_{\mu,\nu_{2}}(z^{n})(1-a)-(1-t)(1+ ae^{i(\theta-\mu)}z^{n}-a-e^{i(\theta-\mu)}z^{n})\psi_{\mu,\nu _{1}}(z^{n})(1+a)}{t(1+ ae^{i(\theta-\mu)}z^{n}+a+e^{i(\theta-\mu )}z^{n})\psi_{\mu,\nu_{2}}(z^{n})(1-a)+(1-t) (1+ ae^{i(\theta-\mu )}z^{n}-a-e^{i(\theta-\mu)}z^{n})\psi_{\mu,\nu_{1}}(z^{n})(1+a)} \biggr) \\ &=\omega_{f_{1}}(z)\frac{t(1+e^{i(\theta-\mu)}z^{n})\psi_{\mu,\nu _{2}}(z^{n})-(1-t)(1-e^{i(\theta-\mu)}z^{n})\psi_{\mu,\nu _{1}}(z^{n})}{t(1+e^{i(\theta-\mu)}z^{n})\psi_{\mu,\nu_{2}}(z^{n})+(1-t) (1-e^{i(\theta-\mu)}z^{n})\psi_{\mu,\nu_{1}}(z^{n})}.\end{aligned} \end{aligned}$$

The above equation, after substituting the values of \(\psi_{\mu,\nu _{k}}\) and then putting \(e^{-i\mu}z^{n}=w\), is equivalent to

$$\begin{aligned} \begin{aligned}[b] \omega_{f}\bigl(\bigl( e^{i\mu}w\bigr)^{1/n}\bigr) & =\omega_{f_{1}}\bigl( \bigl( e^{i\mu}w\bigr)^{1/n} \bigr) \\&\quad\times\biggl(\frac{t(1+e^{i\theta}w)(1-2w\cos\nu _{1}+w^{2})-(1-t)(1-e^{i\theta}w)(1-2w\cos\nu_{2} +w^{2})}{t(1+e^{i\theta }w)(1-2w\cos\nu_{1} +w^{2})+(1-t)(1-e^{i\theta}w)(1-2w\cos\nu_{2} +w^{2})} \biggr) \\ &=:\omega_{f_{1}}\bigl( \bigl( e^{i\mu}w\bigr)^{1/n} \bigr)W(w).\end{aligned}\hspace{-36pt} \end{aligned}$$
(2.11)

To prove our result, we have to show \(|\omega_{f}|<1\) on \(\mathbb{D}\). Since \(|\omega_{f_{1}}|<1\), in view of (2.11), it is enough to show that \(|W|<1\) on \(\mathbb{D}\). Let

$$\begin{aligned} W(w)= e^{-i\theta}\frac{\mathfrak{p}(w)}{\mathfrak{q}(w)}, \end{aligned}$$

where, after a simplification,

$$\begin{aligned} \mathfrak{p}(w)={}&e^{i\theta}w^{3}+\bigl(2t-1-2te^{i\theta} \cos\nu _{1}-2(1-t)e^{i\theta}\cos\nu_{2} \bigr)w^{2} \\ &+\bigl(e^{i\theta}-2t\cos\nu _{1}+2(1-t)\cos\nu_{2} \bigr)w+2t-1 \end{aligned}$$

and

$$\begin{aligned} \mathfrak{q}(w)={}&(2t-1)w^{3}+\bigl(e^{-i\theta}-2t\cos \nu_{1}+2(1-t)\cos\nu _{2}\bigr)w^{2} \\ &+\bigl(2t-1-2te^{-i\theta}\cos\nu_{1}-2(1-t)e^{-i\theta}\cos \nu _{2}\bigr)w+e^{-i\theta}. \end{aligned}$$

Clearly \(\mathfrak{q}(w)=w^{3}\overline{\mathfrak{p}(1/\overline {w})}\). Hence, we can write W as follows:

$$W(w)=\frac{\mathfrak{p}(w)}{w^{3}\overline{\mathfrak{p}(1/\overline {w})}}=e^{i\theta}\prod_{i=1}^{3} \frac{w-w_{i}}{1-\overline{w_{i}}w}, $$

where \(w_{1}\), \(w_{2}\), and \(w_{3}\) are the zeros of k. Thus, to show \(|W|<1\), it is enough to show \(w_{1},w_{2},w_{3}\in\mathbb{D}\). We will discuss it for the cases \(t=1/2\) and \(t\neq1/2\) separately. For \(t\neq1/2\), we have \(0<|2t-1|<|e^{i\theta}|=1\). Define a polynomial \(\mathfrak{p}_{1}\) by

$$\begin{aligned} \mathfrak{p}_{1}(w)&=\frac{e^{-i\theta}\mathfrak {p}(w)-(2t-1)\mathfrak{q}(w)}{w}. \end{aligned}$$

A calculation gives

$$\begin{aligned} \mathfrak{p}_{1}(w)&=4t(1-t)w^{2}-4t(1-t) (\cos \nu_{1}+\cos\nu _{2})w+4t(1-t) \bigl(1-(\cos \nu_{1}-\cos\nu_{2})e^{-i\theta}\bigr) \\ &=4t(1-t)\tilde{\mathfrak{p}}_{1}(w), \end{aligned}$$

where

$$\tilde{\mathfrak{p}}_{1}(w)=w^{2}-(\cos\nu_{1}+ \cos\nu_{2})w+1-(\cos\nu _{1}-\cos\nu_{2})e^{-i\theta}. $$

Recall that inequality (2.6) holds. Again, define a polynomial \(\mathfrak{p}_{2}\) by

$$\mathfrak{p}_{2}(w)=\frac{\tilde{\mathfrak{p}}_{1}(w)-(1-(\cos\nu _{1}-\cos\nu_{2})e^{-i\theta})\tilde{\mathfrak{p}}_{1}^{*}(w)}{w}, $$

where \(\tilde{\mathfrak{p}}_{1}^{*}(w)=w^{2} \overline{ \tilde{\mathfrak {p}}_{1} (1/\overline{w}) }\). Furthermore, we see that

$$\begin{aligned} \mathfrak{p}_{2}(w)&=\bigl(1- \bigl\vert 1-(\cos\nu_{1}- \cos\nu_{2})e^{-i\theta } \bigr\vert ^{2}\bigr)w-\bigl( \cos^{2}\nu_{1}-\cos^{2}\nu_{2} \bigr)e^{-i\theta} \\ &=-(\cos\nu_{1}-\cos\nu_{2}) \bigl((\cos\nu_{1}- \cos\nu_{2}-2\cos \theta)w+(\cos\nu_{1}+\cos \nu_{2})e^{-i\theta} \bigr). \end{aligned}$$

Since \(\cos\nu_{1} \neq \cos\nu_{2}\), it follows from (2.7) that the only zero of \(\mathfrak{p}_{2}\) lies in \(\mathbb {D}\). Thus, by Theorem 2.7 both the zeros of \(\mathfrak {p}_{1}\) and hence all the three zeros of \(\mathfrak{p}\) lie in \(\mathbb {D}\). This completes the proof for \(t\neq1/2\). Now, for \(t=1/2\), we have

$$ \mathfrak{p}(w) = e^{i \theta} w \mathfrak{p}_{1}(w). $$
(2.12)

Since \(\mathfrak{p}_{1}\) has two zeros and both of them lie in \(\mathbb {D}\), by (2.12), all the three zeros of \(\mathfrak{p}\) lie in \(\mathbb{D}\). This completes the proof of Theorem 2.1. □

Corollary 2.8

For\(n\in\mathbb{N}\)and\(k=1,2\), let\(f_{k}=h_{k}+\overline{g_{k}}\in \mathcal{S}_{H}\)such that

$$ h_{k}(z)-g_{k}(z)=\bigl(1+(-1)^{k}a \bigr) \int_{0}^{z} q(\xi )\psi_{\mu,\nu_{k}}\bigl( \xi^{n}\bigr)\,{d}\xi,\quad\mu, \nu\in[0,2\pi), $$
(2.13)

whereqis an analytic mapping and\(\psi_{\mu,\nu}\)is defined by (1.1). Let\(\omega_{f_{k}}\)be the dilatation of\(f_{k}\). If

$$\omega_{f_{1}}(z)=-\omega_{f_{2}}(z)=\frac{a+e^{i(\theta-\mu )}z^{n}}{1+ae^{i(\theta-\mu)}z^{n}},\quad a \in(-1,1), \theta\in[0,2\pi ), $$

then the mapping\(f=tf_{1}+(1-t)f_{2}\)is locally univalent and sense-preserving for\(0\leq t\leq1\)provided:

  1. (i)

    \(\cos\theta> \max\{\cos\nu_{2},-\cos\nu_{1}\}\)and\(\cos \nu_{2}>\cos\nu_{1}\), or

  2. (ii)

    \(\cos\theta< \min\{\cos\nu_{2}, -\cos\nu_{1}\}\)and\(\cos\nu_{1} >\cos\nu_{2}\).

Proof

Following similarly as in Lemma 2.6, we find the expression for the dilatation \(\omega_{f}\) of \(f=tf_{1}+(1-t)f_{2}\) as follows:

$$\omega_{f}(z)=\frac{t\omega_{f_{1}}(z)\psi_{\mu,\nu_{1}}(z^{n})(1- \omega _{f_{2}}(z))(1-a)+(1-t)\omega_{f_{2}}(z)\psi_{\mu,\nu_{2}}(z^{n})(1-\omega _{f_{1}}(z))(1+a)}{t\psi_{\mu,\nu_{1}}(z^{n})(1- \omega _{f_{2}}(z))(1-a)+(1-t)\psi_{\mu,\nu_{2}}(z^{n})(1-\omega_{f_{1}}(z))(1+a)}. $$

The above equation is identical with (2.10) except that \(\cos \nu_{1}\) and \(\cos\nu_{2} \) are interchanged. Hence, the result follows by Lemma 2.6. □

By using Lemma 2.6, we now examine the local univalence of f in Theorem 2.1 for some specific values of \(p_{k}\).

Theorem 2.9

For\(k=1,2\), let\(f_{k}=h_{k}+\overline{g_{k}}\in\mathcal{S}_{H}\)such that

$$ h_{k}(z)+e^{2i\mu}g_{k}(z)= \bigl(1+(-1)^{k}a\bigr) \int_{0}^{z} \psi_{\mu,\nu_{k}}(\xi)\, {d}\xi, \quad-1< a< 1, $$
(2.14)

where\(\psi_{\mu,\nu_{k}}\)is defined by (1.1). Let\(\omega _{f_{k}}\)be the dilatation of\(f_{k}\). If

$$ \omega_{f_{1}}(z)=-\omega_{f_{2}}(z)=-e^{-2i\mu} \frac{a+e^{i(\theta -\mu)}z}{1+ae^{i(\theta-\mu)}z},\quad0\leq\theta< 2\pi, $$
(2.15)

then the mapping\(f=tf_{1}+(1-t)f_{2}\)is univalent and convex in the direction\(\mu+\pi/2\)for\(0 \leq t \leq1\)providedθand\(\nu_{k}\)are given as in Corollary2.8.

Proof

Let \(F_{k}=H_{k}+\overline{G_{k}}\), where \(H_{k}=h_{k}\) and \(G_{k}=-e^{2i\mu }g_{k}\). Then, in view of (2.14) and (2.15), we have

$$H_{k}(z)-G_{k}(z)=\bigl(1+(-1)^{k}a\bigr) \int_{0}^{z} \psi_{\mu,\nu_{k}}(\xi)\,{d}\xi, $$

and the dilatation of \(\omega_{F_{k}}\) of \(F_{k}\) is given by

$$\omega_{F_{1}}(z)=-\omega_{F_{2}}(z)=\frac{a+e^{i(\theta-\mu )}z}{1+ae^{i(\theta-\mu)}z}. $$

Therefore, by Corollary 2.8, the mapping \(F:=tF_{1}+(1-t)F_{2}\) is locally univalent and sense-preserving. Thus,

$$\biggl\vert \frac{tG_{1}'+(1-t)G_{2}'}{tH_{1}'+(1-t)H_{2}'} \biggr\vert < 1 \quad\text{on } \mathbb{D}. $$

Equivalently,

$$\biggl\vert \frac{tg_{1}'+(1-t)g_{2}'}{th_{1}'+(1-t)h_{2}'} \biggr\vert < 1 \quad\text{on } \mathbb{D}. $$

Hence, f is locally univalent and sense-preserving. Now, we can write (2.14) as

$$ h_{k}(z)-e^{2i(\mu+\pi/2)}g_{k}(z)= \int_{0}^{z} \psi_{\mu+\pi/2,\pi /2}(\xi)p_{k}( \xi)\,{d}\xi, $$
(2.16)

where \(\psi_{\mu+\pi/2,\pi/2}\) is defined by (1.1) and

$$p_{k}(z)=\frac{(1+(-1)^{k}a)\psi_{\mu,\nu_{k}}(z)}{\psi_{\mu+\pi/2,\pi/2}(z)} =: \bigl(1+(-1)^{k}a\bigr) \tilde{p}_{k}\bigl( e^{-i \mu} z\bigr). $$

Therefore, in view of (2.16), Theorem 2.1 follows the result once we show that \(\operatorname {Re}p_{k}\) or, equivalently, \(\operatorname {Re}\tilde {p}_{k}\) is positive on \(\mathbb{D}\). Since

$$ \tilde{p}_{k}(z) = \frac{1-z^{2}}{ 1- 2z\cos\nu_{k} +z^{2}}, $$
(2.17)

we see that

$$\biggl\vert \frac{\tilde{p}_{k}(z)-1}{\tilde{p}_{k}(z)+1} \biggr\vert = \biggl\vert \frac{z (\cos\nu_{k} - z) }{ 1- z\cos\nu_{k} } \biggr\vert < 1, $$

and hence \(\operatorname {Re}\tilde{p}_{k}(z)>0\) on \(\mathbb{D}\). This completes the proof. □

Next, we give an illustration of Theorem 2.9 through an example.

Example 2.10

For \(k=1,2\), let \(f_{k}=h_{k}+\overline{g_{k}}\in\mathcal{S}_{H}\) be such that

$$\begin{gathered} h_{1}(z)=\frac{1}{4}\tan^{-1}z+\frac{3}{8} \log\frac{1+z^{2}}{(1-z)^{2}}, \\ g_{1}(z)=\frac{1}{4}\tan^{-1}z-\frac{3}{8} \log\frac{1+z^{2}}{(1-z)^{2}}, \\ h_{2}(z)=\frac{3}{2\sqrt{2}}\tan^{-1}(\sqrt{2} z-1)+ \frac{1}{8+4\sqrt {2}}\log\frac{(1+z)^{2}}{1-\sqrt{2} z+z^{2}}+\frac{3\pi}{8\sqrt{2}},\end{gathered} $$

and

$$g_{2}(z)=\frac{3}{2\sqrt{2}}\tan^{-1}(\sqrt{2} z-1)- \frac{1}{8+4\sqrt{2}}\log\frac{(1+z)^{2}}{1-\sqrt{2} z+z^{2}}+\frac {3\pi}{8\sqrt{2}}. $$

Then we have

$$\begin{gathered} h_{1}(z)+g_{1}(z)=\frac{1}{2}\tan^{-1}z= \int_{0}^{z} \frac{1/2}{1+\xi ^{2}}\,d\xi, \\ h_{2}(z)+g_{2}(z)=-\frac{3i}{2\sqrt{2}}\log\frac{\sqrt {2}-(1-i)z}{\sqrt{2}-(1+i)z}= \int_{0}^{z} \frac{3/2 }{1-\sqrt{2}\xi +\xi^{2}}\,d\xi,\end{gathered} $$

and

$$\omega_{f_{2}}(z)=\frac{g_{2}'(z)}{h_{2}'(z)}=\frac{1/2+z}{1+z/2} = - \omega_{f_{1}}(z), $$

where \(\omega_{f_{k}}\) is the dilatation of \(f_{k}\). Thus, it is seen that \(f_{k}\) satisfy (2.14) and (2.15) with \(\mu=0\), \(\nu_{1}=\pi/2\), \(\nu_{2}=\pi/4\), \(\theta=0\), and \(a=1/2\). Moreover, since \(\cos\nu_{2} =1/ \sqrt{2} > 0 = \cos\nu _{1}\) and

$$\cos\theta=1 > \frac{1}{\sqrt{2}}= \max\{\cos\nu_{2}, - \cos\nu _{1} \}, $$

condition (i) in Corollary 2.8 holds. Hence, by Theorem 2.9, the mapping \(f=tf_{1}+(1-t)f_{2}\) is univalent and convex in the imaginary direction for \(0\leq t\leq1\). Images of \(\mathbb{D}\) under f at \(t=0\), \(t=1\), and \(t=1/3\) are shown in Fig. 2.

Figure 2
figure 2

Images of \(\mathbb{D}\) under f at different values of t

Full size image

Theorem 2.11

For\(k=1,2\), let\(f=h_{k}+\overline{g_{k}}\in\mathcal{S}_{H}\)such that

$$ h_{k}(z)+e^{2i\mu}g_{k}(z)= \bigl(1+(-1)^{k}a\bigr)\frac{z(1-z e^{-i\mu}\cos\nu _{k})}{1-z^{2} e^{-2i\mu}},\quad -1< a< 1, $$
(2.18)

for\(\mu,\nu_{k}\in[0, 2\pi)\). If\(\omega_{f_{k}}\), the dilatation of\(f_{k}\)is given by

$$ \omega_{f_{1}}(z)=-\omega_{f_{2}}(z)=-e^{-2i\mu} \frac{a+e^{i(\theta -\mu)}z}{1+ae^{i(\theta-\mu)}z},\quad0\leq\theta< 2\pi, $$
(2.19)

then the mapping\(f=tf_{1}+(1-t)f_{2}\)is univalent and convex in the direction\(\mu+\pi/2\)for\(0\leq t\leq1\)providedθand\(\nu_{k}\)are given as in Lemma2.6.

Proof

Differentiating (2.18), we get

$$h_{k}'(z)+e^{2i\mu}g_{k}'(z)= \frac{(1+(-1)^{k}a)(1-2 z e^{-i\mu}\cos\nu _{k}+z^{2} e^{-2i\mu})}{(1-z^{2} e^{-2i\mu})^{2}}. $$

The above equation can be written as

$$ h_{k}(z)+e^{2i\mu}g_{k}(z)= \bigl(1+(-1)^{k}a\bigr) \int_{0}^{z}\frac{q(\xi)}{\psi_{\mu ,\nu_{k}}(\xi)}\,{d}\xi, $$
(2.20)

where \(q(z)=(1-z^{2} e^{-2i\mu})^{-2}\). Similar to the proof of Theorem 2.9, by Lemma 2.6, we obtain that f is locally univalent and sense-preserving. Also, we can write (2.20) as

$$ h_{k}(z)-e^{2i(\mu+\pi/2)}g_{k}(z)= \int_{0}^{z} p_{k}(\xi)\psi_{\mu+\pi /2,\pi/2}( \xi)\,{d}\xi, $$
(2.21)

where

$$p_{k}(z)=\frac{(1+(-1)^{k}a)(1-2 z e^{-i\mu}\cos\nu_{k}+z^{2} e^{-2i\mu })}{1-z^{2} e^{-2i\mu}}. $$

Note that \(p_{k}(z) = (1+(-1)^{k}a)/\tilde{p}_{k}(e^{-i\mu}z)\), where \(\tilde{p}_{k}\) is defined by (2.17) and thus \(\operatorname {Re}\tilde {p}_{k}\) or equivalently \(\operatorname {Re}p_{k}\) is positive on \(\mathbb{D}\). Therefore, in view of (2.21), Theorem 2.1 follows the result. □

Remark 2.12

If we put \(a=\theta=\mu=0\) in Theorem 2.11, we get Theorem 7 of Kumar et al. [7].

For \(\mu,\nu\in[0, 2\pi)\), define \(\varPhi_{\mu,\nu}\) by

$$ \varPhi_{\mu,\nu}(z)=\frac{1-\cos\nu}{4 e^{-i\mu}}\log \biggl( \frac{1+ e^{-i\mu} z}{1- e^{-i\mu} z} \biggr)+\frac{(1+\cos\nu )z}{2(1+ e^{-2i\mu} z^{2})}. $$
(2.22)

The mapping \(\varPhi_{0,\nu}\) maps \(\mathbb{D}\) onto a domain with parallel slits along the real direction and its harmonic shears along the real direction were studied in [6]. In the next result we find sufficient conditions for the directional convexity of the convex combination of harmonic shears of \(\varPhi_{\mu,\nu}\).

Theorem 2.13

For\(k=1,2\), let\(f_{k}=h_{k}+\overline{g_{k}}\in\mathcal{S}_{H}\)such that

$$ h_{k}(z)-e^{2i\mu}g_{k}(z)= \bigl(1+(-1)^{k}a\bigr)\varPhi_{\mu,\nu_{k}}(z), \quad a\in(-1,1), \mu, \nu_{k}\in[0, 2\pi), $$
(2.23)

where\(\varPhi_{\mu,\nu_{k}}\)is defined by (2.22). If\(\omega_{f_{k}}\), the dilatation of\(f_{k}\)is given by

$$\omega_{f_{1}}(z)=-\omega_{f_{2}}(z)=e^{-2i\mu} \frac{a+e^{i(\theta -2\mu)}z^{2}}{1+ae^{i(\theta-2\mu)}z^{2}},\quad0\leq\theta< 2\pi, $$

then the mapping\(f=tf_{1}+(1-t)f_{2}\)is univalent and convex in the directionμfor\(0\leq t\leq1\)providedθand\(\nu_{k}\)are given as in Lemma2.6.

Proof

On differentiating (2.23), we have

$$\begin{aligned} h'_{k}(z)-e^{2i\mu}g'_{k}(z)&= \bigl(1+(-1)^{k}a\bigr) \biggl(\frac{1-\cos\nu_{k}}{2(1- e^{-2i\mu}z^{2})}+\frac{(1+\cos\nu_{k})(1- e^{-2i\mu} z^{2})}{2(1+ e^{-2i\mu} z^{2})^{2}} \biggr) \\ &=\bigl(1+(-1)^{k}a\bigr)\frac{1-2\cos\nu_{k} e^{-2i\mu}z^{2}+e^{-4i\mu}z^{4}}{(1- e^{-2i\mu} z^{2})(1+ e^{-2i\mu} z^{2})^{2}}. \end{aligned}$$

Therefore,

$$ h_{k}(z)-e^{2i\mu}g_{k}(z)= \bigl(1+(-1)^{k}a\bigr) \int_{0}^{z}\frac{q(\xi)}{\psi_{2\mu ,\nu_{k}}(\xi^{2})}\,d\xi, $$
(2.24)

where

$$q(z)=\frac{1}{(1- e^{-2i\mu} z^{2})(1+ e^{-2i\mu} z^{2})^{2}}. $$

Hence, following similarly as in the proof of Theorem 2.9, we see by using Lemma 2.6 that f is locally univalent and sense-preserving. Moreover, (2.24) can also be written as

$$ h_{k}(z)-e^{2i\mu}g_{k}(z)= \int_{0}^{z}p_{k}(\xi)\psi_{\mu,\pi/2}( \xi)\,d\xi, $$
(2.25)

where

$$p_{k}(z)=\bigl(1+(-1)^{k}a\bigr)\frac{1-2\cos\nu_{k} e^{-2i\mu}z^{2}+e^{-4i\mu }z^{4}}{1- e^{-4i\mu} z^{4}}. $$

Since \(p_{k}(z) = (1+(-1)^{k}a)/\tilde{p}_{k}(e^{-2i\mu}z^{2})\), where \(\tilde{p}_{k}\) is defined by (2.17) and thus \(\operatorname {Re}\tilde {p}_{k}\) or equivalently \(\operatorname {Re}p_{k}\) is positive on \(\mathbb{D}\). Therefore, in view of (2.25), the result follows from Theorem 2.1. □

Theorem 2.14

For\(k=1,2\), let\(f_{k}=h_{k}+\overline{g_{k}}\in\mathcal{S}_{H}\)such that

$$h_{k}(z)-e^{2i\mu}g_{k}(z)=\bigl(1+(-1)^{k}a \bigr) \int_{0}^{z}\varPsi_{k}(\xi)\,d\xi, \quad a \in(-1,1), $$

where

$$\varPsi_{k}(z)=\frac{1-2\cos\nu_{k} e^{-i n\mu}z^{n}+e^{-2i n\mu }z^{2n}}{(1-e^{-2i n\mu}z^{2n})(1-2\cos\nu e^{-i\mu}z+e^{-2i\mu }z^{2})},\quad n\in\mathbb{N}, \mu, \nu, \nu_{k}\in[0, 2\pi). $$

If\(\omega_{f_{k}}\), the dilatation of\(f_{k}\)is given by

$$\omega_{f_{1}}(z)=-\omega_{f_{2}}(z)=e^{-2i\mu} \frac{a+e^{i(\theta -n\mu)}z^{n}}{1+ae^{i(\theta-n\mu)}z^{n}},\quad0\leq\theta< 2\pi, $$

then the mapping\(f=tf_{1}+(1-t)f_{2}\)is univalent and convex in the directionμfor\(0\leq t\leq1\)providedθand\(\nu_{k}\)are given as in Lemma2.6.

The proof of the above theorem is similar to that of Theorem 2.13 and is thus omitted here.