On the univalence of some analytic functions

Lemma 2. Under the hypothesis of Lemma 1 we have for |z| < 1 $$|zp'(z)/p(z)| \leq 2nz^n(1-\alpha)/\{(1-|z|^n)[1+(1-2\alpha)|z|^n]\}.$$

Lemma 3. Under the hypothesis of Lemma 1 we have for |z| < 1 Re $(p(z)) \ge [1 + (2\alpha - 1) |z|^n]/(1 + |z|^n)$ .

Lemma 4. ([6]). If $h(z) = 1 + c_n z^n + c_{n+1} z^{n+1} + \cdots$ is analytic and Re (h(z)) > 0 for |z| < 1, then $$[1 - \lambda \mid h(z) \mid]^{-1} \leq (1 - |z|^n) / [(1 - |z|^n) - \lambda (1 + |z|^n)]$$ for $|z| < [(1 - \lambda)/(1 + \lambda)]^{1/n}$ , where $0 \le \lambda < 1$ . 3. THEOREM 1. Suppose that $f(z)=z+a_{n+1}z^{n+1}+a_{n+2}z^{n+2}+\cdots$ , and $g(z)=z+b_{n+1}z^{n+1}+b_{n+2}z^{n+2}+\cdots$ are analytic and $\operatorname{Re}(g(z)/z)>0$ for |z|<1. If $\operatorname{Re}(f(z)/[\lambda f(z)+(1-\lambda)g(z)])>0$ , $0\leq \lambda<1$ , for |z|<1, then f(z) is univalent and starlike for $|z|< R^{1/n}$ , where $R=\{[(2n+\lambda-n\lambda)^2+(1-\lambda^2)]^{1/2}-(2n+\lambda-n\lambda)\}/(1+\lambda)$ . Proof. Let $$h(z) = f(z)/[\lambda f(z) + (1-\lambda)g(z)] = 1 + c_n z^n + c_{n+1} z^{n+1} + \cdots,$$ then h(z) is analytic and Re (h(z)) > 0 for |z| < 1. Now (10) $$f(z) [1 - \lambda h(z)] = (1 - \lambda)h(z)zp(z) ,$$ where $p(z) = g(z)/z = 1 + b_{n+1}z^n + b_{n+2}z^{n+1} + \cdots$ . Multiplying the logarithmic derivative of both sides of equation (10) by z we have (11) $$zf'(z)/f(z) = 1 + zp'(z)/p(z) + zh'(z)/\{h(z)[1 - \lambda h(z)]\}$$ . Equation (11) is valid for those z for which $1 - \lambda h(z) \neq 0$ and |z| < 1. Since $|h(z)| \leq (1 + |z|^n)/(1 - |z|^n)$ , $1 - \lambda h(z) \neq 0$ in particular if $|z| < [(1 - \lambda)/(1 + \lambda)]^{1/n}$ . Now from equation (11), we have $$|zf'(z)/f(z) - 1| \le |zp'(z)/p(z)| + |zh'(z)/h(z)| |1 - \lambda h(z)|^{-1}$$ and by using Lemma 2 with $\alpha = 0$ and Lemma 4, this gives $$|zf'(z)/f(z) - 1| \leq \frac{2n |z|^n}{1 - |z|^{2n}} + \frac{2n |z|^n}{(1 - |z|^{2n}) - \lambda(1 + |z|^n)^2},$$ $$= \frac{2n |z|^n [(1 - |z|^n) - \lambda(1 + |z|^n) + (1 - |z|^n)]}{(1 - |z|^{2n}) [(1 - |z|^n) - \lambda(1 + |z|^n)]}$$ provided that $|z| < [(1-\lambda)/(1+\lambda)]^{1/n}$ . The fact that |zf'(z)/f(z) - 1| < 1 implies that Re(zf'(z)/f(z)) > 0, it follows from the inequality (12) that Re(zf'(z)/f(z) > 0 if $$|z| < [(1 - \lambda)/(1 + \lambda)]^{1/n}$$ and if (13) $$G(|z|^n) \equiv (1+\lambda) |z|^{3n} + (4n+2n\lambda+\lambda-1) |z|^{2n} + (2n\lambda-4n-\lambda-1) |z|^n + (1-\lambda) > 0.$$ Let $|z|^n=t$ and consider the cubic polynomial G(t) for $0 \le t \le 1$ . G(t) has at most two positive zeros. Since $G(0)=(1-\lambda)>0$ , $G[(1-\lambda)/(1+\lambda)]=-4\lambda n(1-\lambda)/(1+\lambda)^2<0$ and $G(1)=4\lambda n>0$ , it follows that $G(t_1)=0$ for some $t_1$ such that $0< t_1<(1-\lambda)/(1+\lambda)$ and G(t)>0 for $0 \le t < t_1$ and G(t)<0 for $t_1< t<(1-\lambda)/(1+\lambda)$ . Hence $\operatorname{Re}(zf'(z)/f(z))>0$ for those z for which only the inequality (13) is true. Now the inequality (13) holds if, in particular $$\begin{aligned} (1+\lambda) \mid z \mid^{\mathfrak{z}_n} + (4n-2n\lambda+\lambda-1) \mid z \mid^{\mathfrak{z}_n} \ &+ (2n\lambda-4n-\lambda-1) \mid z \mid^{\mathfrak{z}} + (1-\lambda) > 0 \end{aligned}$$ or, or, $$(|z|^n-1)[(1+\lambda)|z|^{2n}+(4n-2n\lambda+2\lambda)|z|^n+(\lambda-1)]>0$$ $$(1 + \lambda) |z|^{2n} + (4n - 2n\lambda + 2\lambda) |z|^n + (\lambda - 1) < 0$$ . The last inequality holds if (14) $$|z|^n < \{[(2n + \lambda - n\lambda)^2 + (1 - \lambda^2)]^{1/2} - (2n + \lambda - n\lambda)\}/(1 + \lambda)$$ . Since f(z) is univalent and starlike for those z for which $$\operatorname{Re}\left(zf'(z)/f(z)\right)>0$$ , we have that f(z) is univalent and starlike for $|z| < R^{1/n}$ , where R is the right side of (14). If we put $\lambda = 0$ in Theorem 1 we obtain the following result which, when n = 1, reduces to a result of Ratti [5, Theorem 1].

Corollary 1. Suppose that $f(z) = z + a_{n+1}z^{n+1} + a_{n+2}z^{n+2} \cdots$ , and $g(z) = z + b_{n+1}z^{n+1} + b_{n+2}z^{n+2} + \cdots$ are analytic and Re (g(z)/z) > 0 for |z| < 1. If Re (f(z)/g(z)) > 0 for |z| < 1 then f(z) is univalent and starlike for $|z| < [(4n^2 + 1)^{1/2} - 2n]^{1/n}$ . The functions $f(z) = z(1-z^n)^2/(1+z^n)^2$ and $g(z) = z(1-z^n)/(1+z^n)$ satisfy the hypothesis of Corollary 1 and it is easy to see that the derivative of f(z) vanishes at $z = [(4n^2+1)^{1/2}-2n]^{1/n}$ and hence $[(4n^2+1)^{1/2}-2n]^{1/n}$ is in fact the radius of univalence for such functions f(z). This shows that Corollary 1 is sharp and hence Theorem 1 is sharp at least for $\lambda=0$ .

Theorem 2. Suppose $f(z) = z + a_2 z^2 + \cdots$ , and $$g(z) = z + b_0 z^2 + \cdots$$ are analytic for |z| < 1 and Re (g(z)/z) > 1/2 for |z| < 1. If Re $$(f(z)/[\lambda f(z) + (1 - \lambda)g(z)]) > 0$$ for $|z| < 1$ then f(z) is univalent and starlike for $|z| < (1 - \lambda)/(3 + \lambda)$ .

Theorem 3. Let $f(z)=z+\sum_{k=n+1}^{\infty}b_kz^k$ and $g(z)=z+\sum_{k=n+1}^{\infty}b_kz^k$ be analytic for |z|<1 and g(z) be starlike of order $\alpha$ , $0\leq\alpha<1$ , for |z|<1. If $\operatorname{Re}\left(f(z)/[\lambda f(z)+(1-\lambda)g(z)]\right)>0$ for |z|<1, then f(z) is univalent and starlike for (i) $$|z| < [(1-\lambda)/(1+\lambda+2n)]^{1/n}$$ if $\alpha = 1/2$ ; and (ii) $$|z| < R^{1/n}$$ , $if \; \alpha \neq 1/2$ , where $$R=\{[A^2+4(1-\lambda^2)(2\alpha-1)]^{1/2}-A\}/[2(1+\lambda)(2\alpha-1)]$$ with $A=2n+\lambda+1-(2\alpha-1)(1-\lambda)$ . Proof. Proceeding as in the proof of Theorem 1 we get $$\operatorname{Re} (zf'(z)/f(z)) \ge \operatorname{Re} (zg'(z)/g(z)) - |zh'(z)/h(z)| |1 - \lambda h(z)|^{-1}$$ . Applying Lemma 3 (to zg'(z)/g(z)) and Lemmas 2 and 4 we get, (20) $$\operatorname{Re}\left(\frac{zf'(z)}{f(z)}\right) \ge \frac{1 + (2\alpha - 1) |z|^n}{1 + |z|^n} - \frac{2n |z|^n}{(1 - |z|^{2n}) - \lambda(1 + |z|^n)^2}$$ provided that $|z| < [(1-\lambda)/(1+\lambda)]^{1/n}$ . Hence Re (zf'(z)/f(z)) > 0 for those z for which $|z| < [1-\lambda)/(1+\lambda)]^{1/n}$ and the right side of inequality (20) is greater than zero. The latter holds if (21) $$G(\mid z\mid^n) \equiv (1+\lambda)(2\alpha-1)\mid z\mid^{2n} \\ + \left[2n+\lambda+1-(2\alpha-1)(1-\lambda)\right]\mid z\mid^n-(1-\lambda)<0 \ .$$ Let $|z|^n=t$ and consider the quadratic G(t) for $0 \le t \le 1$ . Since $G(0)=\lambda-1<0$ , $G[(1-\lambda)/(1+\lambda)]=2n(1-\lambda)/(1+\lambda)>0$ , it follows that $G(t_1)=0$ for some $t_1$ such that $0< t_1<(1-\lambda)/(1+\lambda)$ and G(t)<0 for $0\le t< t_1$ and G(t)>0 for $t_1< t<(1-\lambda)/(1+\lambda)$ . Hence f(z) is univalent and starlike for those z for which only the inequality (21) holds. Now the inequality (21) holds if $$|z| < [(1-\lambda)/(1+\lambda+2n)]^{1/n}$$ when $\alpha = 1/2$ and $$|z|<\{[A^2+4(1-\lambda^2)(2\alpha-1)]^{1/2}-A\}^{1/n}/[2(1+\lambda)(2\alpha-1)]^{1/n}$$ when $\alpha \neq 1/2$ , where $A = 2n + \lambda + 1 - (2\alpha - 1)(1 - \lambda)$ and this completes the proof. If we put $\lambda=0$ , n=1 and $\alpha=0$ in the above result then we see that $f(z)=z+\sum_{k=2}^\infty a_k z^k$ under the modified hypothesis is univalent and starlike for $|z|<2-\sqrt{3}$ , a result obtained by MacGregor [2, Theorem 3]. On the other hand if $\lambda=0$ and n=1, Theorem 3 reduces to a result of Ratti [5, Theorem 3]. The functions $$f(z) = z(1-z^n)/(1+z^n)^{\frac{2-2\alpha}{n}+1}$$ and $g(z) = z/(1+z^n)^{\frac{2-2\alpha}{n}}$ show that Theorem 3 is sharp at least for $\lambda = 0$ and arbitrary n, since the derivative of f(z) vanishes at $$z = \{[(n+1-\alpha) - ((n+1-\alpha)^2 - (1-2\alpha))^{1/2}]/(1-2\alpha)\}^{1/n}$$ for $\alpha \neq 1/2$ and at z = -1/(2n + 1) when $\alpha = 1/2$ . 4. Let S(R) denote the functions $f(z) = z + \sum_{k=2}^{\infty} a_k z^k$ which are analytic and satisfy |zf'(z)/f(z) - 1| < 1 for |z| < R. Obviously every member of S(R) is univalent and starlike for |z| < R. We now prove the following result.

Theorem 4. Let $f(z) = z + a_{n+1}z^{n+1} + a_{n+2}z^{n+2} + \cdots$ , and $g(z) = z + b_{n+1}z^{n+1} + b_{n+2}z^{n+2} + \cdots$ be analytic and satisfy Re (g(z)/z) > 0 for |z| < 1. If $|f(z)/[\lambda f(z) + (1-\lambda)g(z)] - 1| < 1$ , $0 \le \lambda < 1$ , for |z| < 1, then $f(z) \in S(R^{1/n})$ , where R is the smallest positive root of the equation $(2n\lambda + \lambda - n - 1) R^2 - (3n + \lambda - 2n\lambda) R + (1 - \lambda) = 0$ . Proof. Let (22) $$h(z) = \frac{f(z)}{[\lambda f(z) + (1 - \lambda)g(z)]} - 1 = c_n z^n + c_{n+1} z^{n+1} + \cdots$$ By hypothesis, h(z) is analytic and |h(z)| < 1 for |z| < 1 and hence by a result of Goluzin [1] we have that for |z| < 1 $$|h'(z)| \leq n |z|^{n-1} (1 - |h(z)|^2)/(1 - |z|^{2n})$$ and by Schwarz's lemma for |z| < 1 $$|h(z)| \leq |z|^n.$$ If we let g(z) = zp(z), then we have from (22) $$f(z)[1 - \lambda - \lambda h(z)] = (1 - \lambda)zp(z)[1 + h(z)]$$ . Hence, $$\frac{zf'(z)}{f(z)}=1+\frac{zp'(z)}{p(z)}+\frac{zh'(z)}{\left[1+h(z)\right]\left[1-\lambda-\lambda h(z)\right]}$$ and this gives $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \left| \frac{zp'(z)}{p(z)} \right| + \frac{\left| zh'(z) \right|}{\left| 1 + h(z) \right| \left| 1 - \lambda - \lambda h(z) \right|}.$$ Applying Lemma 2, with $\alpha=0$ , we get, in view of (23), for |z|<1 $$\bigg|\frac{zf'(z)}{f(z)}-1\bigg| \leq \frac{2n\mid z\mid^n}{1-\mid z\mid^{2n}} + \frac{n\mid z\mid^n(1-\mid h(z)\mid^2)}{(1-\mid z\mid^{2n})\mid 1+h(z)\mid \mid 1-\lambda-\lambda h(z)\mid} \ \leq \frac{2n\mid z\mid^n}{1-\mid z\mid^{2n}} + \frac{n\mid z\mid^n(1+\mid h(z)\mid)}{(1-\mid z\mid^{2n})\mid 1-\lambda-\lambda h(z)\mid}$$ by using (24), we have $$\Big| \, \frac{z f'(z)}{f(z)} - 1 \, \Big| \, \leq \frac{2 \, n \, |z|^n}{1 - |z|^{2n}} + \frac{n \, |z|^n}{(1 - |z|^n) \, (1 - \lambda - \lambda \, |z|^n)}$$ valid for $|z| < [(1-\lambda)/\lambda]^{1/n}$ . Hence |zf'(z)/f(z) - 1| < 1 if $$|z| < [(1-\lambda)/\lambda]^{1/n}$$ and $$2n |z|^n (1-\lambda-\lambda |z|^n) + n |z|^n (1+|z|^n) < (1-|z|^{2n}) (1-\lambda-\lambda |z|^n)$$ . The last inequality holds if (25) $$G(|z|^n) \equiv \lambda |z|^{3n} + (2n\lambda + \lambda - n - 1) |z|^{2n} - (3n + \lambda - 2n\lambda) |z|^n + (1 - \lambda) > 0.$$ Let $|z|^n = t$ and consider the cubic polynomial G(t) for $0 \le t \le 1$ . G(t) has at most two positive zeros. Since $G(0)=(1-\lambda)>0$ and $G((1-\lambda)/\lambda)=-(n(1-\lambda)/\lambda^2<0$ , it follows that $G(t_1)=0$ for some $t_1$ such that $0< t_1<(1-\lambda)/\lambda$ and G(t)>0 for $0\le t< t_1$ and G(t)<0 for some values of t between $t_1$ and $(1-\lambda)/\lambda$ . Hence $$|zf'(z)/f(z)-1|<1$$ for those values of z for which only the inequality (25) holds. Now inequality (25) holds if, in particular $$(2n\lambda + \lambda - n - 1) |z|^{2n} - (3n + \lambda - 2n\lambda) |z|^n + (1 - \lambda) > 0$$ and this completes the proof. 248 If we set $\lambda = 0$ and n = 1 in the above result we have the following.

Corollary 2. Suppose $f(z) = z + a_2 z^2 + a_3 z^3 + \cdots$ and $g(z) = z + b_2 z^2 + b_3 z^3 + \cdots$ are analytic and satisfy Re (g(z)/z) > 0 for |z| < 1. If |f(z)/g(z) - 1| < 1 for |z| < 1, then |zf'(z)/f(z) - 1| < 1 for $|z| < 1/4(\sqrt{17} - 3)$ . It may be noted that Corollary 2 implies, in particular, that f(z) is univalent and starlike for |z| < 1/4 ( $\sqrt{17} - 3$ ) and hence includes a result of Ratti [5, Theorem 4]. If we take $f(z) = z(1-z^n)^2/(1+z^n)$ and $g(z) = z(1-z^n)/(1+z^n)$ , it is easy to see that these functions satisfy the hypothesis of Theorem 4 with $\lambda = 0$ . We see that f'(z) vanishes at $z_0 = [-3n + (9n^2 + 4n + 4)^{1/2}]/(2n + 2)$ and hence $$|z_{\scriptscriptstyle 0}f'(z_{\scriptscriptstyle 0})/f(z_{\scriptscriptstyle 0})-1|=1$$ . This shows that Theorem 4 is sharp for at least $\lambda=0$ and also that Corollary 2 is sharp.

Theorem 5. Let $f(z) = z + a_{n+1}z^{n+1} + a_{n+2}z^{n+2} + \cdots$ and $g(z) = z + b_{n+1}z^{n+1} + b_{n+2}z^{n+2} + \cdots$ be analytic for |z| < 1 and g(z) be starlike of order $\alpha$ for |z| < 1, $0 \le \alpha < 1$ . If $$\mid f(z)/[\lambda f(z) + (1-\lambda)g(z)] - 1 < 1$$ , $0 \leqq \lambda < 1$ , for $\mid z \mid < 1$ , then f(z) is univalent and starlike for $|z| < R^{1/n}$ , where R is the smallest positive root of the equation (26) $$(2\alpha - 1)\lambda R^3 - (n + 2\alpha - 1 - \lambda)R^2 + (2\alpha - 2 - 2\alpha\lambda + \lambda - n)R + (1 - \lambda) = 0.$$ Proof. Proceeding as in the proof of Theorem 4 we have $$\frac{zf'(z)}{f(z)} = \frac{zg'(z)}{g(z)} + \frac{zh'(z)}{\left[1 + h(z)\right]\left[1 - \lambda - \lambda h(z)\right]}.$$ Hence, $$\operatorname{Re}\left(\frac{zf'(z)}{f(z)}\right) \geq \operatorname{Re}\left(\frac{zg'(z)}{g(z)}\right) - \frac{\mid zh'(z)\mid}{\mid 1 \, + \, h(z)\mid \mid 1 \, - \, \lambda - \, \lambda h(z)\mid} \; .$$ Since Re $(zg'(z)/g(z)) > \alpha$ and $zg'(z)/g(z) = 1 + c_n z^n + c_{n+1} z^{n+1} + \cdots$ , we have by Lemma 3 and inequalities (23) and (24) that (27) $$\operatorname{Re} (zf'(z)/f(z)) \ge [1 + (2\alpha - 1) | z|^n]/(1 + | z|^n) \\ - n | z|^n/[(1 - | z|^n) (1 - \lambda - \lambda | z|^n)]$$ valid for $|z| < [(1-\lambda)/\lambda]^{1/n}$ . Hence Re (zf'(z)/f(z)) > 0 if $|z| < [(1 - \lambda)/\lambda]^{1/n}$ and if (in view of inequality (27)) (28) $$G(|z|^n) \equiv (2\alpha - ) \lambda |z|^{3n} - (n + 2\alpha - 1 - \lambda) |z|^{2n} + (2\alpha - 2 - 2\alpha\lambda + \lambda - n) |z|^n + (1 - \lambda) > 0.$$ Let |z|=t and consider the cubic polynomial G(t) for $0 \le t \le 1$ . Since $G(0)=1-\lambda>0$ and $G((1-\lambda)/\lambda)=(-n(1-\lambda))/\lambda^2<0$ , it follows that $G(t_1)=0$ for some $t_1$ such that $0 < t_1 < (1-\lambda)/\lambda$ and G(t)>0 for $0 \le t < t_1$ and G(t)<0 for some t between $t_1$ and $(1-\lambda)/\lambda$ . Hence f(z) is starlike and univalent for $|z|< R^{1/n}$ , in view of inequality (28), where R is the smallest positive root of the equation (26). The case when $\lambda=0$ in Theorem 5 is of special interest. In this case equation (26) becomes $$(n+2\alpha-1)R^2-(2\alpha-2-n)R-1=0$$ which gives R=1/3 in case $\alpha=0$ and n=1 and $$(29) \quad R = \{(2\alpha-2-n) + [(2\alpha-2-n)^2 + 4(n+2\alpha-1)]^{1/2}\}/[2(n+2\alpha-1)]$$ if $\alpha \neq 0$ . This proves the following result, which includes a result of Ratti [5, Theorem 6].

Corollary 3. Suppose $f(z)=z+a_{n+1}z^{n+1}+a_{n+2}z^{n+2}+\cdots$ and $g(z)=z+b_{n+1}z^{n+1}+b_{n+2}z^{n+2}+\cdots$ are analytic for |z|<1 and g(z) is starlike of order $\alpha$ for |z|<1, $0\leq \alpha<1$ . If |f(z)/g(z)-1|<1 for |z|<1 then f(z) is univalent and starlike for (i) $$|z| < 1/3$$ if $\alpha = 0$ and $n = 1$ (ii) $|z| < R^{1/n}$ , where R is given by (29) if $\alpha \neq 0$ . It is easy to see that the functions $f(z)=z(1-z^n)/(1+z^n)^{(2-2\alpha)/n}$ and $g(z)=z/(1+z^n)^{(2-2\alpha)/n}$ satisfy the hypothesis of Corollary 3 and also that the derivative of f(z) vanishes at z=1/3 if $\alpha=0$ and n=1, and at $z=\{[(n+2-2\alpha)^2+4(n+2\alpha-1)]^{1/2}-(n+2-2\alpha)\}^{1/n}/[2(n+2\alpha-1)]^{1/n}$ if $\alpha\neq 0$ . This shows that Corollary 3 is sharp.