On some geometric properties of multivalent functions

Lemma 2.1. [5] Let $f(z) = z + a_2 z^2 + \cdots$ be analytic in the unit disc and suppose that $$|f''(z)| < 1, \quad |z| < 1,$$ (2.1) then f(z) is univalent in |z| < 1.

Lemma 2.2. [6] Let p(z) be an analytic function in |z| < 1 with p(0) = 1, $p(z) \neq 0$ . If there exists a point $z_0$ , $|z_0| < 1$ , such that $$\left|\arg\{p(z)\}\right| < \frac{\pi\alpha}{2} \quad for |z| < |z_0|$$ and $$\left|\arg\{p(z_0)\}\right| = \frac{\pi\alpha}{2}$$ for some $0 < \alpha < 2$ , then we have $$\frac{z_0p'(z_0)}{p(z_0)}=is\alpha,$$ where $$s \ge \frac{1}{2} \left( a + \frac{1}{a} \right)$$ when $\arg \left\{ p(z_0) \right\} = \frac{\pi \alpha}{2}$ and $$s \le -\frac{1}{2}\left(a + \frac{1}{a}\right)$$ when $\arg\left\{p(z_0)\right\} = -\frac{\pi\alpha}{2}$ , where $$\{p(z_0)\}^{1/\alpha} = \pm ia$$ , and $a > 0$ .

Lemma . [[\]](#page-8-8) Let p be a positive integer. If f (z) = z^p + ∞ ⁿ=p+ anzⁿ is analytic in ^D and if it satisfies $$\mathfrak{Re}\left\{\frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right\} > 0, \quad z \in \mathbb{D},\tag{2.2}$$ then f (z) is p-valently starlike in D and $$\mathfrak{Re}\left\{\frac{zf^{(k)}(z)}{f^{(k-1)}(z)}\right\} > 0, \quad z \in \mathbb{D}$$ $$\tag{2.3}$$ for k = , , . . . , (p – ).

Lemma . ([\[\]](#page-8-8), p.) Let f ∈ Ap. If there exists a (p – k + )-valent starlike function g(z) = ∞ ⁿ=p–k+ bnzⁿ (bp–k+ = ) that satisfies $$\mathfrak{Re}\left\{\frac{zf^{(k)}(z)}{g(z)}\right\} > 0, \quad |z| < 1, \tag{2.4}$$ then f (z) is p-valent in |z| <.

Theorem . Let f ∈ A^p and suppose that <span id="page-2-0"></span> $$|f^{(p+1)}(z)| < \alpha(\beta_0)(p!), \quad |z| < 1,$$ (3.1) where β = . ··· is the positive root of the equation $$2\beta + \frac{2}{\pi} \tan^{-1} \beta = 1, \quad 0 < \beta < 1$$ (3.2) and $$\alpha(\beta_0) = \sin\left\{\frac{\pi}{2} \left(\beta_0 + (2/\pi) \tan^{-1} \beta_0\right)\right\}$$ $$= \sin\left\{\frac{(1-\beta_0)\pi}{2}\right\}$$ $$= 0.82 \cdots.$$ Then we have $$\Re\left\{\frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right\} > 0, \quad |z| < 1, \tag{3.3}$$ and, therefore, we have $$\Re\left\{\frac{zf'(z)}{f(z)}\right\} > 0, \quad |z| < 1, \tag{3.4}$$ or f(z) is p-valently starlike in |z| < 1. Proof From the hypothesis (3.1), we have <span id="page-3-1"></span> $$\begin{aligned} \left| f^{(p)}(z) - p! \right| &= \left| \int_0^z f^{(p+1)}(t) \, \mathrm{d}t \right| \\ &\leq \int_0^{|z|} \left| f^{(p+1)}(t) \right| |\mathrm{d}t| < p! \int_0^{|z|} \alpha(\beta_0) |\mathrm{d}t| \\ &= p! \alpha(\beta_0) |z| < p! \alpha(\beta_0). \end{aligned}$$ This shows that $$\left| \arg \left\{ f^{(p)}(z) \right\} \right| < \sin^{-1} \alpha(\beta_0), \quad |z| < 1.$$ (3.5) Let us put $$q(z) = \frac{f^{(p-1)}(z)}{p!z}, \qquad q(0) = 1.$$ (3.6) Then it follows that $$q(z) + zq'(z) = \frac{f^{(p)}(z)}{p!}.$$ If there exists a point $z_0$ , $|z_0| < 1$ , such that $$\left|\arg\left\{q(z)\right\}\right| < \frac{\pi\beta_0}{2} \quad \text{for } |z| < |z_0|$$ and $$\left|\arg\left\{q(z_0)\right\}\right| = \frac{\pi\beta_0}{2}$$ , then by Lemma 2.2 we have <span id="page-3-2"></span><span id="page-3-0"></span> $$\frac{z_0q'(z_0)}{q(z_0)}=i\beta_0k,$$ where $$k \ge 1$$ when $\arg\{q(z_0)\} = \frac{\pi \beta_0}{2}$ (3.7) and $$k \le -1$$ when $\arg\{q(z_0)\} = -\frac{\pi\beta_0}{2}$ . (3.8) For the case (3.7), we have $$\arg \left\{ f^{(p)}(z) \right\} = \arg \left\{ \frac{f^{(p)}(z)}{p!} \right\}$$ $$= \arg \left\{ q(z_0) + z_0 q'(z_0) \right\}$$ $$= \arg \left\{ q(z_0) \left( 1 + \frac{z_0 q'(z_0)}{q(z_0)} \right) \right\}$$ $$= \frac{\pi \beta_0}{2} + \arg \{ 1 + i\beta_0 k \}$$ $$\geq \frac{\pi \beta_0}{2} + \tan^{-1} \beta_0$$ $$= \frac{\pi}{2} \left\{ \beta_0 + (2/\pi) \tan^{-1}(\beta_0) \right\}$$ $$= \sin^{-1} \alpha(\beta_0).$$ This contradicts (3.5), and for the case (3.8), applying the same method as above, we have <span id="page-4-0"></span> $$\arg\{f^{(p)}(z)\} \le -\sin^{-1}\alpha(\beta_0).$$ This also contradicts (3.5) and, therefore, it shows that $$\left| \arg \left\{ \frac{f^{(p-1)}(z)}{p!z} \right\} \right| < \frac{\pi \beta_0}{2}, \quad |z| < 1.$$ (3.9) Applying (3.5) and (3.9), we have $$\left| \arg \left\{ \frac{zf^{(p)}(z)}{f^{(p-1)}(z)} \right\} \right| = \left| \arg \left\{ f^{(p)}(z) \right\} + \arg \left\{ \frac{z}{f^{(p-1)}(z)} \right\} \right|$$ $$\leq \left| \arg \left\{ f^{(p)}(z) \right\} \right| + \left| \arg \left\{ \frac{z}{f^{(p-1)}(z)} \right\} \right|$$ $$< \sin^{-1} \alpha(\beta_0) + \frac{\pi \beta_0}{2}$$ $$= \frac{\pi}{2} \left( 2\beta_0 + \frac{2}{\pi} \tan^{-1} \beta_0 \right)$$ $$= \frac{\pi}{2}.$$ This shows that $$\Re e \left\{ \frac{z f^{(p)}(z)}{f^{(p-1)}(z)} \right\} > 0, \quad |z| < 1,$$ and by Lemma 2.3 we have $$\Re \left\{ \frac{zf'(z)}{f(z)} \right\} > 0, \quad |z| < 1,$$ or f(z) is p-valently starlike in |z| < 1. Remark Note that if m(x)=x + ^π tan– x, then $$m(0.383) = 0.9988537761 \cdots$$ , $m(0.384) = 1.001408771 \cdots$ Hence, if $$2\beta_0 + \frac{2}{\pi} \tan^{-1} \beta_0 = 1,$$ then β = . ··· . Moreover, $$0.9673808495 \cdots < \sin^{-1}\alpha(\beta_0) = \frac{\pi}{2} \left(\beta_0 + \frac{2}{\pi} \tan^{-1}\beta_0\right) = \frac{(1-\beta_0)\pi}{2} < 0.96982343 \cdots$$ and $$\alpha(\beta_0) = \sin\left\{\frac{\pi}{2}\left(\beta_0 + \frac{2}{\pi}\tan^{-1}\beta_0\right)\right\} = 0.824669\cdots$$ For p = , Theorem [.](#page-2-2) becomes the following corollary which extends the result contained in Lemma [.](#page-1-1).

Corollary . Let f ∈ A() and suppose that $$\left|f''(z)\right| < \alpha(\beta_0), \quad |z| < 1,$$ (3.10) where β = . ··· is the positive root of the equation $$2\beta + \frac{2}{\pi} \tan^{-1} \beta = 1, \quad 0 < \beta < 1,$$ (3.11) and $$\alpha(\beta_0) = \sin\left\{\frac{(1-\beta_0)\pi}{2}\right\}$$ $$= 0.82\cdots.$$ Then we have $$\Re\left\{\frac{zf'(z)}{f(z)}\right\} > 0, \quad |z| < 1, \tag{3.12}$$ or f (z) is starlike univalent in |z| < . An analytic function f (z) is said to be typically real if the inequality ImzImf (z) ≥ holds for all z ∈ D. From the definition of a typically real function it follows that z ∈ D⁺ ⇔ f (z) ∈ C⁺ and z ∈ D– ⇔ f (z) ∈ C–. The symbols D+, D–, C+, C– stand for the following sets: the upper and the lower halves of the disk D, the upper half-plane and the lower half-plane, respectively.

Theorem 3.3. Let $f(z) \in A_p$ and suppose that <span id="page-6-1"></span> $$\left| \arg \left\{ \frac{z f^{(p)}(z)}{g(z)} \right\} \right| < \frac{\pi}{2} + \tan^{-1} \frac{1 - |z|}{1 + |z|}, \quad z \in \mathbb{D},$$ (3.13) where g(z) is univalent starlike in $\mathbb{D}$ and the functions <span id="page-6-2"></span> $$\frac{zg'(z)}{g(z)}$$ and $\frac{z(\frac{zf'^{p-1}(z)}{g(z)})'}{\frac{zf'^{p-1}(z)}{g(z)}}$ (3.14) are typically real in $\mathbb{D}$ . Then we have $$\left| \arg \frac{z f^{(p-1)}(z)}{g(z)} \right| < \frac{\pi}{2}, \quad z \in \mathbb{D}.$$ (3.15) Proof Let us put <span id="page-6-0"></span> $$q(z) = \frac{zf^{(p-1)}(z)}{p!g(z)}, \qquad q(0) = 1.$$ Then it follows that $$\frac{zf^{(p)}(z)}{g(z)} = p!q(z) \left( \frac{zg'(z)}{g(z)} + \frac{zq'(z)}{q(z)} \right). \tag{3.16}$$ From the hypothesis $$\frac{zg'(z)}{g(z)} + \frac{zq'(z)}{q(z)}$$ is typically real in $\mathbb D$ . If there exists a point $z_0$ , $|z_0|<1$ , such that $$\left|\arg\{q(z)\}\right| < \frac{\pi}{2}$$ for $|z| < |z_0|$ and $$\left| \arg \left\{ q(z_0) \right\} \right| = \frac{\pi}{2}, \quad q(z_0) = \pm ia, \text{ and } a > 0,$$ then by Lemma 2.2 we have $$\frac{z_0 q'(z_0)}{q(z_0)} = is, (3.17)$$ where $$s \ge \frac{1}{2} \left( a + \frac{1}{a} \right)$$ when $\arg \left\{ q(z_0) \right\} = \frac{\pi}{2}$ and $$s \le -\frac{1}{2}\left(a + \frac{1}{a}\right)$$ when $\arg\left\{q(z_0)\right\} = -\frac{\pi}{2}$ . For the case <span id="page-7-1"></span> $$\arg\{q(z_0)\}=\frac{\pi}{2}, \quad q(z_0)=ia, a>0,$$ we have <span id="page-7-0"></span> $$\frac{z_0 q'(z_0)}{q(z_0)} = is, \quad s \ge \frac{1}{2} \left( a + \frac{1}{a} \right) > 1, \tag{3.18}$$ hence <span id="page-7-2"></span> $$\mathfrak{Im}\{z_0\} > 0 \tag{3.19}$$ because zq'(z)/q(z) is typically real. Therefore, (3.19) yields that <span id="page-7-3"></span> $$\Im \left\{ \frac{z_0 g'(z_0)}{g(z_0)} \right\} > 0, \tag{3.20}$$ because zg'(z)/g(z) is typically real. Moreover, $$\frac{1-|z_0|}{1+|z_0|} \le \Re \left\{ \frac{z_0 g'(z_0)}{g(z_0)} \right\} \le \frac{1+|z_0|}{1-|z_0|} \tag{3.21}$$ because g(z) is a univalent starlike function, see [10]. Applying (3.18), (3.20) and (3.21) in (3.16), we have $$\arg\left\{\frac{z_0 f^{(p)}(z_0)}{g(z_0)}\right\} = \arg\left\{q(z_0)\right\} + \arg\left\{\frac{z_0 g'(z_0)}{g(z_0)} + \frac{z_0 q'(z_0)}{q(z_0)}\right\} \\ = \arg\left\{q(z_0)\right\} + \arg\left\{\frac{z_0 g'(z_0)}{g(z_0)} + is\right\} \\ \ge \frac{\pi}{2} + \tan^{-1}\frac{1 - |z_0|}{1 + |z_0|}.$$ This contradicts (3.13), and for the case $$\arg\{q(z_0)\} = -\frac{\pi}{2},$$ applying the same method as above, we have $$\arg\left\{\frac{z_0 f^{(p)}(z_0)}{g(z_0)}\right\} = \arg\left\{q(z_0)\right\} + \arg\left\{\frac{z_0 g'(z_0)}{g(z_0)} + \frac{z_0 q'(z_0)}{q(z_0)}\right\}$$ $$\leq -\left\{\frac{\pi}{2} + \tan^{-1}\frac{1 - |z_0|}{1 + |z_0|}\right\}.$$ This also contradicts (3.13) and, therefore, it shows that (3.15) holds.

Corollary 3.4. Let $f(z) \in A_p$ and all the coefficients of f(z) are real and suppose that $$\left|\arg\left\{\frac{zf^{(p)}(z)}{g(z)}\right\}\right| < \frac{\pi}{2} + \tan^{-1}\frac{1-|z|}{1+|z|}, \quad z \in \mathbb{D},$$ where g(z) is univalent starlike and typically real in $\mathbb{D}$ . Then we have <span id="page-8-1"></span> $$\left|\arg \frac{zf^{(p-1)}(z)}{g(z)}\right| < \frac{\pi}{2}, \quad z \in \mathbb{D}.$$