Convex meromorphic mappings

Theorem 1. f is a member of C(p) if and only if for $z \in \Delta$ , (2.1) $$\operatorname{Re}\left[1 + p^2 - 2pz + \frac{(z-p)(1-pz)f''(z)}{f'(z)}\right] < 0.$$ Proof. If f is a member of S(p) let h(z) = f((z+p)/(1+pz)); then h has a simple pole at z=0 and $\overline{\mathbb{C}} \setminus h[\Delta] = \overline{\mathbb{C}} \setminus f[\Delta]$ . Thus, f is a member of C(p) if and only if h is convex with a simple pole at z=0. This is the case if and only if [10] $$\operatorname{Re}\left(1 + \frac{zh''(z)}{h'(z)}\right) < 0$$ for $z \in \Delta$ . A straightforward computation gives $$\operatorname{Re}\left(1 + \frac{zh''(z)}{h'(z)}\right) = \operatorname{Re}Q(z)$$ where $$Q(z) = \frac{1 - pz}{1 + pz} + \frac{\frac{(1 - p^2)z}{(1 + pz)^2} f''\left(\frac{z + p}{1 + pz}\right)}{f'\left(\frac{z + p}{1 + pz}\right)}.$$ But $\operatorname{Re} Q(z) < 0$ for $z \in \Delta$ if and only if $\operatorname{Re} Q((z-p)/(1-pz)) < 0$ for $z \in \Delta$ . However, $$Q\left(\frac{z-p}{1-pz}\right) = \frac{1-p^2-2pz}{(1-p^2)} + \frac{(z-p)(1-pz)f''(z)}{(1-p^2)f'(z)},$$ which gives (2.1). Remark. If f is a member of C(p) and $$P(z) = 2pz - 1 - p^{2} - \frac{(z-p)(1-pz)f''(z)}{f'(z)}$$ then $\operatorname{Re} P(z) > 0$ , $z \in \Delta$ , $P(p) = 1 - p^2$ and P'(p) = 0.

Lemma 1. Let P(z) satisfy $\operatorname{Re} P(z) > 0$ , $z \in \Delta$ , and P(0) = 1. If $0 , then for <math>z \in \Delta$ , $$\operatorname{Re}\left[\frac{(z-p)(1-pz)P(z)+p}{z}-pz\right]>0.$$ Proof. Let 0 < r < 1 and $P_r(z) = P(rz)$ . Then $$Q_r(z) = \frac{(z-p)(1-pz)P_r(z) + p}{z} - pz$$ is analytic for $|z| \leq 1$ . If |z| = 1, then $$Q_r(z) = \frac{(z-p)(1-pz)P_r(z)}{z} - p\left(z - \frac{1}{z}\right)$$ and $$\operatorname{Re} Q_r(z) = |1 - pz|^2 \operatorname{Re} P_r(z) > 0.$$ Since Qr(z) is analytic for |z| ≤ 1, Re Qr(z) > 0 for z ∈ ∆. Letting r → 1, we obtain for z ∈ ∆, $$\operatorname{Re}\left[\frac{(z-p)(1-pz)P(z)+p}{z}-pz\right] \geq 0.$$ But equality cannot occur in the last inequality since the quantity on the left side equals 1 − p ² when z = p.

Lemma 2. If Re P(z) > 0 for z ∈ ∆ and P(p) = 1 − p 2 , then for z ∈ ∆, $$\operatorname{Re}\left[\frac{zP(z)-p+pz^2}{(z-p)(1-pz)}\right]>0\,.$$ P r o o f. Let p < r < 1 and α = (r − 1)p/(r − p 2 ) and Lr(z) = r(z − α)/(1 − αz). It is easily verified that Lr[∆] = {z : |z| < r} and Lr(p) = p. Let $$Q_r(z) = \frac{zP(L_r(z)) - p + pz^2}{(z - p)(1 - pz)}.$$ Qr(z) is analytic for |z| ≤ 1 and Re P(Lr(z)) > 0 for |z| ≤ 1. If |z| = 1 then $$\begin{aligned} \operatorname{Re} Q_r(z) &= \operatorname{Re} \left[ \frac{z P(L_r(z))}{(z-p)(1-pz)} + \frac{pz(z-1/z)}{(z-p)(1-pz)} \right] \\ &= \frac{1}{|1-pz|^2} \operatorname{Re} P(L_r(z)) > 0 \,. \end{aligned}$$ Since Q^r is analytic for |z| ≤ 1, it follows that Re Qr(z) > 0 for z ∈ ∆. Letting r → 1, we obtain for z ∈ ∆, $$\operatorname{Re}\left[\frac{zP(z)-p+pz^2}{(z-p)(1-pz)}\right] \geq 0.$$ But equality cannot occur in the last inequality since the expression on the left equals 1 when z = 0.

Theorem 2. C(p) = Σ(p) for 0 < p < 1. P r o o f. Let f be a member of Σ(p) and $$P(z) = -1 - \frac{zf''(z)}{f'(z)} + \frac{1+pz}{1-pz} - \frac{z+p}{z-p}.$$ Then Re P(z)>0, z∈∆, and P(0) = 1. Straightforward computations give $$2pz - 1 - p^2 - \frac{(z-p)(1-pz)f''(z)}{f'(z)} = \frac{(z-p)(1-pz)P(z) + p}{z} - pz.$$ Therefore, by Lemma 1, Re $$\left[2pz - 1 - p^2 - \frac{(z-p)(1-pz)f''(z)}{f'(z)}\right] > 0$$ for z ∈ ∆, and thus by Theorem 1, f is a member of C(p). Conversely, suppose f is a member of C(p) and let $$P(z) = 2pz - 1 - p^2 - \frac{(z-p)(1-pz)f''(z)}{f'(z)}.$$ Then by Theorem 1, $\operatorname{Re} P(z) > 0$ , $z \in \Delta$ , and $P(p) = 1 - p^2$ . Straightforward computations give $$-1 - \frac{zf''(z)}{f'(z)} - \frac{z+p}{z-p} + \frac{1+pz}{1-pz} = \frac{zP(z)-p+pz^2}{(z-p)(1-pz)}.$$ Thus, by Lemma 2. $$\operatorname{Re}\left[-1 - \frac{zf''(z)}{f'(z)} - \frac{z+p}{z-p} + \frac{1+pz}{1-pz}\right] > 0$$ for $z \in \Delta$ . Therefore f is a member of $\Sigma(p)$ 3. The coefficients $a_n$ . In this section we use Theorem 1 to study the coefficients $a_1$ and $a_2$ in (1.2), if f is a member of C(p). We will make use of the following lemma.

Lemma 3. Let P(z) be analytic in $\Delta$ and satisfy Re P(z) > 0, $z \in \Delta$ , $P(p) = 1 - p^2$ and P'(p) = 0, $0 . If <math>P(z) = (1 - p^2) + d_2(z - p)^2 + d_3(z - p)^3 + \dots$ for |z - p| < 1 - p, then $$|d_2| \le \frac{2}{1 - p^2} \,,$$ (3.2) $$\left| \frac{p}{1 - p^2} d_2 + d_3 \right| \le \frac{6p}{(1 - p^2)^2}, \quad 2/3 \le p < 1,$$ (3.3) $$\left| \frac{p}{1 - p^2} d_2 + d_3 \right| \le \frac{2(1 + \frac{9}{4}p^2)}{1 - p^2}, \quad 0$$ All the inequalities are sharp. Proof. Let $$w(z) = \frac{P(z) - (1 - p^2)}{P(z) + 1 - p^2}.$$ Then w(p) = 0 and $|w(z)| \le 1, z \in \Delta$ . Also $$w'(z) = \frac{2(1-p^2)P'(z)}{[P(z) + (1-p)^2]^2}$$ and hence w'(p) = 0. Comparing coefficients in the expansions of both sides of $$[P(z) + (1 - p^2)]w(z) = P(z) - (1 - p^2),$$ we obtain $$(3.4) d_2 = (1 - p^2)w''(p)$$ and (3.5) $$\frac{p}{1-p^2}d_2 + d_3 = pw''(p) + (1-p)\frac{w'''(p)}{3}.$$ We can write $$w(z) = \phi\left(\frac{z-p}{1-pz}\right)$$ where $\phi$ is analytic for |z| < 1, $\phi(0) = \phi'(0) = 0$ and $|\phi(z)| \le 1$ , $z \in \Delta$ . In particular, we obtain $$w''(p) = \frac{\phi''(0)}{(1 - p^2)^2}.$$ Since $|\phi''(0)/2| \leq 1$ , we have $|w''(p)| \leq 2/(1-p^2)^2$ . Thus from (3.4) we obtain $$|d_2| = (1 - p^2)|w''(p)| \le \frac{2}{1 - p^2},$$ which is (3.1). Next from (3.5) we obtain $$\frac{p}{1-p^2}d_2 + d_3 = \frac{1}{(1-p^2)^2} \left[ \frac{\phi'''(0)}{3} + 3p\phi''(0) \right].$$ If $\phi(z) = c_2 z^2 + c_3 z^3 + \dots, z \in \Delta$ , then $$\frac{p}{1-p^2}d_2+d_3=\frac{2}{(1-p^2)^2}[c_3+3pc_2].$$ Using known inequalities for bounded functions, we obtain $$|c_3 + 3pc_2| \le |c_3| + 3p|c_2| \le 1 - |c_2|^2 + 3p|c_2|$$ . Therefore (3.6) $$\left| \frac{p}{1 - p^2} d_2 + d_3 \right| \le \frac{2}{(1 - p^2)^2} [1 + 3p|c_2| - |c_2|^2].$$ Let $x = |c_2|$ and $h(x) = 1 + 3px - x^2$ , $0 \le x \le 1$ . Then h'(x) = 3p - 2x. If $p \ge 2/3$ , then $h'(x) \ge 0$ for $0 \le x \le 1$ and hence $$(3.7) h(x) \le h(1) = 3p, 2/3 \le p < 1.$$ If 0 , then <math>h(x) achieves its maximum at x = 3p/2. Hence (3.8) $$h(x) \le 1 + \frac{9}{4}p^2, \quad 0$$ Combining (3.6), (3.7) and (3.8) gives (3.2) and (3.3). Equality is attained in (3.1) by the function $$P(z) = \frac{1 + p^2 - 4pz + (1 + p^2)z^2}{1 - z^2},$$ which is obtained by taking $w(z) = [(z-p)/(1-pz)]^2$ . The same function gives equality in (3.2). If 0 , let $$\phi(z) = \frac{z^2(z + \frac{3}{2}p)}{1 + \frac{3}{2}pz}$$ and $w(z) = \phi((z-p)/(1-pz))$ . The resulting function $P(z) = (1-p^2)(1+w(z))/(1-w(z))$ gives equality in (3.3).

Theorem 3. Let f be a member of C(p) and have the expansion (1.2). Then $$(3.9) |a_1| \le \frac{p^2}{(1-p^2)^3},$$ $$(3.10) |a_2| \le \frac{(4+9p^2)|a_{-1}|}{12(1-p^2)^3}, 0$$ (3.11) $$|a_2| \le \frac{p}{(1-p^2)^3} |a_{-1}| \le \frac{p^3}{(1-p^2)^4}, \quad 2/3 \le p \le 1.$$ All the inequalities are sharp. Remark. Making use of the area theorem, Komatu [5] proved inequality (3.9) for the larger class S(p). Proof of Theorem 3. Let $$P(z) = 2pz - 1 - p^2 - \frac{(z-p)(1-pz)f''(z)}{f'(z)}.$$ Then P(z) satisfies the hypotheses of Lemma 3. Comparing coefficients on both sides of the equation $$[2p(z-p)-(1-p^2)]f'(z)-(z-p)[(1-p^2)-p(z-p)]f''(z)=P(z)f'(z)$$ we obtain $$(3.12) 2a_1(1-p^2) = a_{-1}d_2$$ and (3.13) $$6(1-p^2)a_2 = 2pa_1 + a_{-1}d_3.$$ Combining (3.1) and (3.12) gives $$|a_1| \le \frac{|a_{-1}|}{(1-p^2)^2} \, .$$ However, $|a_{-1}| \le p^2/(1-p^2)$ (cf. [4], [5]), giving (3.9). Combining (3.12) and (3.13) gives (3.14) $$a_2 = \frac{1}{6(1-p^2)} \left[ \frac{p}{1-p^2} d_2 + d_3 \right] a_{-1}.$$ If $0 , then (3.3) and (3.14) gives (3.10). If <math>2/3 \le p < 1$ , then (3.2) combined with (3.14) gives (3.11). Equality is attained in (3.9) and (3.11) by f(z) = -pz/((z-p)(1-pz)). If 0 , equality is attained in (3.10) by the function <math>f which satisfies $$2pz - 1 - p^2 - \frac{(z-p)(1-pz)f''(z)}{f'(z)} = P(z)$$ where P(z) is the function satisfying the hypotheses of Lemma 3 and giving equality in (3.3). Since $\operatorname{Re} P(z) = 0$ on |z| = 1 with finitely many exceptions and since $$1 + \frac{zf''(z)}{f'(z)} = \frac{z}{(z-p)(1-pz)} \left[ p\left(z - \frac{1}{z}\right) - P(z) \right],$$ it follows that on |z| = 1. $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) = \frac{1}{|1 - pz|^2}\operatorname{Re}P(z) = 0$$ with finitely many exceptions. Laborious computations give $$P(z) = \frac{(1-p^2)(2+2p-p^2)}{2-2p-p^2} \cdot \frac{(1+z)(z-e^{i\gamma})(z-e^{-i\gamma})}{(1-z)(z-e^{i\beta})(z-e^{-i\beta})}$$ where $e^{i\gamma} \neq e^{i\beta}$ , and $e^{i\beta}$ is not real for 0 . Thus Re(1 + <math>zf''(z)/f'(z)) = 0 on |z| = 1 with 3 exceptional points. It follows that for the extremal function in the case $0 , <math>\overline{\mathbb{C}} \setminus f[\Delta]$ is the interior of a triangle. Remark. In the case $0 of Theorem 3, using the inequality <math>|a_{-1}| \le p^2/(1-p^2)$ in (3.10) does not result in a sharp inequality.

Theorem 4. If f is a member of C(p) with expansion (1.2), then $$\left| p + \frac{a_0(1-p^2)}{a_{-1}} \right| \le \frac{1+p^2}{p}$$ , and the inequality is sharp. Proof. Let $$h(z) = \frac{-a_{-1}}{(1 - p^2)f\left(\frac{p - z}{1 - pz}\right)},$$ then h is a member of S(p) and for |z-p| < 1-p, $$h(z) = z + \left(p + \frac{(1-p^2)a_0}{a_{-1}}\right)z^2 + \dots$$ Using (1.3) when n=2, we get $$\left| p + \frac{(1-p^2)a_0}{a_1} \right| \le \frac{1+p^2}{p}$$ . Equality is attained by f(z) = -pz/((z-p)(1-pz)). 4. The coefficients $b_n$ . Let f be a member of C(p) and have the expansion (1.1) for |z| < p. As remarked in the introduction, sharp upper bounds on $|b_n|$ are known for all n and a sharp lower bound on $Re(b_2)$ follows from results in [9]. In this section we will obtain a sharp lower bound on $Re(b_3)$ which suggests a conjecture concerning $Re(b_n)$ for all n.

Theorem 5. Let f be a member of C(p) with expansion (1.1). Then (4.1) $$\operatorname{Re} b_2 \ge \frac{1+p^4}{p(1+p^2)} > 1$$ and (4.2) $$\operatorname{Re} b_3 \ge \frac{1 - p^2 + p^4}{p^2} = \frac{1 + p^6}{p^2(1 + p^2)} > 1.$$ Both inequalities are sharp, each being attained by the function $$f(z) = \frac{p(1+p^2)z - 2p^2z^2}{(1-p^2)(p-z)(1-pz)}.$$ Proof. Let $$P(z) = 2pz - 1 - p^2 - \frac{(-p + (1+p^2)z - pz^2)f''(z)}{f'(z)},$$ then $\operatorname{Re} P(z) > 0$ , $z \in \Delta$ , $P(p) = 1 - p^2$ and P'(p) = 0. Let $P(z) = c_0 + c_1 z + c_2 z^2 + \ldots$ Comparing coefficients on both sides of the equation $$P(z)f'(z) = [2pz - (1+p^2)]f'(z) - [-p + (1+p^2)z - pz^2]f''(z),$$ we obtain $$(4.3) c_0 = 2pb_2 - (1+p^2)$$ and $$(4.4) 2c_0b_2 + c_1 = 2p - 4(1+p^2)b_2 + 6pb_3.$$ Using (4.3) and (4.4) we obtain $$(4.5) b_2 = \frac{c_0 + (1+p^2)}{2p}$$ and (4.6) $$6p^2b_3 = c_0^2 + 3(1+p^2)c_0 + pc_1 + 2(1+p^2+p^4).$$ Let $w(z)=[P(z)-(1-p^2)]/[P(z)+(1-p^2)],$ then |w(z)|<1 for $z\in\Delta$ and w(p)=w'(p)=0. Thus we can write $$w(z) = \left(\frac{z-p}{1-pz}\right)^2 \phi(z)$$ where $|\phi(z)| < 1$ for $z \in \Delta$ . We have $$P(z) = \frac{(1 - p^2)(1 + w(z))}{1 - w(z)}.$$ Thus (4.7) $$c_0 = P(0) = \frac{(1-p^2)(1+w(0))}{1-w(0)} = \frac{(1-p^2)(1+p^2\phi(0))}{1-p^2\phi(0)}.$$ It follows that $$\operatorname{Re} c_0 \ge (1 - p^2) \frac{1 - p^2 |\phi(0)|}{1 + p^2 |\phi(0)|} \ge \frac{(1 - p^2)^2}{1 + p^2}.$$ Using this inequality in conjunction with (4.5) gives (4.1), which has also been proven by Miller [9]. Next, we have $$c_1 = P'(0) = \frac{2(1 - p^2)w'(0)}{(1 - w(0))^2}$$ $$= \frac{2(1 - p^2)[-2p(1 - p^2)\phi(0) + p^2\phi'(0)]}{(1 - p^2\phi(0))^2}.$$ Combining (4.6), (4.7) and (4.8), we eventually obtain $$(4.9) 6p^2b_3 = (1-p^2)\left[ (1-p^2) + \frac{2p^3\phi'(0)}{(1-p^2\phi(0))^2} + 3(1+p^2)\frac{1+p^2\phi(0)}{1-p^2\phi(0)} \right]$$ $$+ 2(1+p^2+p^4).$$ Now let $$Q(z) = \frac{1 + p^2 \phi(z)}{1 - p^2 \phi(z)}.$$ Then $\operatorname{Re} Q(z) \geq (1-p^2)/(1+p^2) > 0$ for $z \in \Delta$ and (4.9) can be written as $$(4.10) 6p2b3 = (1 - p2)[(1 - p2) + pQ'(0) + 3(1 + p2)Q(0)] + 2(1 + p2 + p4).$$ Let $T(z)=Q(z)-(1-p^2)/(1+p^2)$ . Since $\operatorname{Re} T(z)>0$ for $z\in \Delta,$ it is known that $$|T'(0)| < 2 \operatorname{Re} T(0)$$ . Thus $$|Q'(0)| \le 2 \operatorname{Re} \left[ Q(0) - \frac{1 - p^2}{1 + p^2} \right].$$ Hence $$2\operatorname{Re} Q(0) \ge |Q'(0)| + 2\frac{1-p^2}{1+p^2}.$$ Using the last inequality with (4.10) we obtain $$6p^{2} \operatorname{Re} b_{3} \geq (1 - p^{2}) \left[ (1 - p^{2}) - p|Q'(0)| + \frac{3}{2} (1 + p^{2})|Q'(0)| + 3(1 - p^{2}) \right]$$ $$+ 2(1 + p^{2} + p^{4})$$ $$= (1 - p^{2}) \left[ (1 - p^{2}) + \frac{3 + 3p^{2} - 2p}{2} |Q'(0)| + 3(1 - p^{2}) \right]$$ $$+ 2(1 + p^{2} + p^{4})$$ $$\geq (1 - p^{2})[4(1 - p^{2})] + 2(1 + p^{2} + p^{4})$$ $$= 6(1 - p^{2} + p^{4}),$$ which gives (4.2). An examination of the proof indicates that equality holds in (4.1) and (4.2) if and only if $\phi(z) \equiv -1$ . This leads to the extremal function stated in the theorem. Remark. It seems reasonable to expect that the extremal function for Theorem 5 is extremal for all n. That is, we expect that if f is a member of C(p), then $\text{Re}(b_n) \geq (1+p^{2n})/(p^{n-1}(1+p^2))$ for all n. 5. Starlike functions. Miller [7]–[9] considered functions f of S(p) for which there exists $\varrho$ , $0 < \varrho < 1$ , so that $\text{Re}[zf'(z)/(f(z) - w_0)] < 0$ for $\varrho < |z| < 1$ and a fixed $w_0 \in \mathbb{C}, w_0 \neq 0$ . These functions map $\Delta$ onto the complement of a set which is starlike with respect to $w_0$ . This class of functions is a subclass of the class $\Sigma^*(p, w_0)$ defined as the class of functions f in S(p) such that for $z \in \Delta$ , Re $$\left[\frac{pz}{1-pz} - \frac{p}{z-p} - \frac{zf'(z)}{(f(z)-w_0)}\right] > 0$$ . Actually, the two classes are the same if $0 (cf. [9]). But for <math>p \ge \sqrt{3 - 2\sqrt{2}}$ and proper choice of $w_0$ the first class is a proper subset of the second. We will prove that $\Sigma^*(p, w_0)$ is the class of all functions f in S(p) such that $\overline{\mathbb{C}} \setminus f[\Delta]$ is starlike with respect to $w_0$ , which we denote by $\Sigma^{\mathrm{s}}(p, w_0)$ .

Theorem 6. f is a member of $\Sigma^{s}(p, w_0)$ if and only if, for $z \in \Delta$ , $$\operatorname{Re}\left[\frac{(z-p)(1-pz)f'(z)}{f(z)-w_0}\right]<0.$$ Proof. Suppose f is a member of S(p) and let $$g(z) = f\left(\frac{z+p}{1+pz}\right).$$ f is a member of Σ^s (p, w0) if and only if C \ g(∆) is starlike with respect to w0. This is the case if and only if F(z) = g(z) − w⁰ maps ∆ onto the complement of a set which is starlike with respect to the origin. Since F has its pole at the origin, C \ F[∆] is starlike with respect to the origin if and only if Re[zF⁰ (z)/F(z)] < 0 for z ∈ ∆. The last inequality is true if and only if $$\operatorname{Re}\left[\left(\frac{z-p}{1-pz}\right)F'\left(\frac{z-p}{1-pz}\right)\middle/F\left(\frac{z-p}{1-pz}\right)\right]<0$$ for z ∈ ∆. A straightforward computation gives $$\left(\frac{z-p}{1-pz}\right)F'\left(\frac{z-p}{1-pz}\right) / F\left(\frac{z-p}{1-pz}\right) = \frac{(z-p)(1-pz)f'(z)}{(1-p^2)(f(z)-w_0)}$$ and the theorem follows.

Theorem 7. Σ^s (p, w0) = Σ^∗ (p, w0) for all p, 0 < p < 1, and all w⁰ 6= 0. P r o o f. Let f be a member of Σ^∗ (p, w0) and $$P(z) = \frac{pz}{1 - pz} - \frac{p}{z - p} - \frac{zf'(z)}{f(z) - w_0},$$ then Re P(z) > 0 for z ∈ ∆ and P(0) = 1. From this we obtain (5.1) $$\frac{(z-p)(1-pz)f'(z)}{f(z)-w_0} = -\frac{(z-p)(1-pz)P(z)+p(1-z^2)}{z}.$$ Let 0 < r < 1 and $$Q_r(z) = \frac{(z-p)(1-pz)P(rz) + p(1-z^2)}{z},$$ then Qr(z) is analytic for |z| ≤ 1, and $$\operatorname{Re} Q_r(z) = |1 - pz|^2 \operatorname{Re} P(rz) > 0$$ for |z| = 1. Thus Re Qr(z) > 0 for z ∈ ∆. If we let r → 1, we obtain (5.2) $$\operatorname{Re} \left[ \frac{(z-p)(1-pz)P(z) + p(1-z^2)}{z} \right] \ge 0.$$ However, the expression on the left side of (5.2) is strictly positive for z = p. Thus equality cannot occur in (5.2). Hence from (5.1), $$\operatorname{Re}\left[\frac{(z-p)(1-pz)f'(z)}{f(z)-w_0}\right] < 0$$ for z ∈ ∆. Thus by Theorem 6, f is a member of Σ^s (p, w0). Conversely, suppose f is a member of Σ^s (p, w0) and let 0 , $$P(z) = -\frac{(z-p)(1-pz)f'(z)}{f(z) - w_0}$$ then Re P(z) > 0 for $z \in \Delta$ and $P(p) = 1 - p^2$ . We obtain (5.3) $$\frac{pz}{1-pz} - \frac{p}{(z-p)} - \frac{zf'(z)}{f(z)-w_0} = \frac{zP(z)-p(1-z^2)}{(z-p)(1-pz)}.$$ By Lemma 2 the real part of the expression on the right side of (5.3) is strictly positive for z in $\Delta$ . Thus f is a member of $\Sigma^*(p, w_0)$ . Miller [9] has given some estimates of coefficients in the expansion (1.1) if f is a member of $\Sigma^*(p, w_0) = \Sigma^{\mathrm{s}}(p, w_0)$ . We will next give sharp bounds on a few coefficients in the expansion (1.2).

Theorem 8. If f(z) is a member of $\Sigma^*(p, w_0)$ and has expansion (1.2) for |z-p|<1-p then $$|a_0 - w_0| \le \frac{2+p}{1-p^2} |a_{-1}|$$ and $$|a_1| \le \frac{|a_{-1}|}{(1-p^2)^2}$$ Both inequalities are sharp. Proof. We first prove inequality (5.5). Let $$P(z) = \frac{-(z-p)(1-pz)f'(z)}{f(z) - w_0},$$ then Re P(z) > 0 for $z \in \Delta$ and $P(p) = 1 - p^2$ . Let $$P(z) = (1 - p^{2}) + \sum_{n=1}^{\infty} c_{n}(z - p)^{n}$$ for |z-p| < 1-p. Comparing coefficients on both sides of the equation $$(f(z) - w_0)P(z) = -(z - p)(1 - pz)f'(z),$$ we obtain $$(5.6) a_{-1}c_1 + (1-p^2)(a_0 - w_0) = -pa_{-1},$$ $$(5.7) a_{-1}c_2 + (a_0 - w_0)c_1 + (1 - p^2)a_1 = -a_1(1 - p^2).$$ Combining (5.6) and (5.7), we eventually obtain $$(5.8) -2(1-p^2)^2 a_1 = a_{-1}[(1-p^2)c_2 - pc_1 - c_1^2].$$ We now claim that $|(1-p^2)c_2-pc_1-c_1^2| \leq 2$ . To prove this, let $$Q(z) = \frac{1}{1 - p^2} P\left(\frac{z + p}{1 + pz}\right),$$ then $\operatorname{Re} P(z)>0$ for $z\in \Delta$ and Q(0)=1. Thus [11] there exists m(t) increasing on $[0,2\pi]$ with $\int_0^{2\pi}\,dm(t)=1$ , such that $$\frac{1}{1-p^2} P\bigg(\frac{z+p}{1+pz}\bigg) = \int\limits_0^{2\pi} \frac{1+e^{it}z}{1-e^{it}z} \, dm(t) \, .$$ Thus $$P(z) = (1 - p^2) \int_0^{2\pi} \frac{(1 - pz) + e^{it}(z - p)}{(1 - pz) - e^{it}(z - p)} dm(t).$$ Expanding the integrand in powers of z - p and integrating we obtain $$c_1 = 2\int\limits_0^{2\pi} e^{it} \, dm(t)$$ and $$c_2 = \frac{2}{1 - p^2} \int_0^{2\pi} \left( e^{2it} + pe^{it} \right) dm(t) = \frac{2}{1 - p^2} \int_0^{2\pi} e^{2it} dm(t) + \frac{pc_1}{1 - p^2}.$$ Thus (5.9) $$(1-p^2)c_2 - pc_1 - c_1^2 = 2 \int_0^{2\pi} e^{2it} dm(t) - c_1^2.$$ Now let $$T(z) = \int_{0}^{2\pi} \frac{1 + e^{it}z}{1 - e^{it}z} dm(t).$$ Then $\operatorname{Re} T(z) > 0$ for $z \in \Delta$ and T(0) = 1. If $$T(z) = 1 + p_1 z + p_2 z^2 + \dots, \quad z \in \Delta,$$ then $$p_1 = 2 \int_{0}^{2\pi} e^{it} dm(t) = c_1$$ and $p_2 = 2 \int_{0}^{2\pi} e^{2it} dm(t)$ . Thus from (5.9), $$(1 - p^2)c_2 - pc_1 - c_1^2 = p_2 - p_1^2.$$ But it is known [6] that $|p_2 - p_1^2| \le 2$ . Thus $$|(1-p^2)c_2 - pc_1 - c_1^2| \le 2.$$ Therefore from (5.8) we obtain $$2(1-p^2)^2|a_1| < 2|a_{-1}|$$ . which is (5.5). Next, from (5.6), $$|a_0 - w_0| = \frac{|a_{-1}||c_1 + p|}{1 - p^2} = \frac{|a_{-1}||p_1 + p|}{1 - p^2} \le \frac{|a_{-1}|(2 + p)}{1 - p^2}.$$ To see sharpness, consider $$f(z) = w_0 + pw_0 \frac{(1-z)^2}{(z-p)(1-pz)}$$ . Since $$\frac{(z-p)(1-pz)f'(z)}{f(z)-w_0} = -(1-p^2)\frac{1+z}{1-z},$$ f(z) is a member of $\Sigma^{\mathrm{s}}(p, w_0)$ . Moreover, $\overline{\mathbb{C}} \setminus f[\Delta]$ is the line segment $\xi = tw_0$ , $(1+p^2)/(1+p)^2 \le t \le (1+p^2)/(1-p)^2$ . Also, for |z-p| < 1-p, $$f(z) = \frac{pw_0(1-p)}{(1+p)(z-p)} + \left[w_0 + \frac{p(p-2+p^2)}{(1+p)(1-p^2)}w_0\right] + \frac{pw_0}{(1-p)(1+p)^3}(z-p) + \dots,$$ from which we can see that equality is attained in (5.4) and (5.5).

Theorem 9. With the notation of Theorem 8, $$|a_{-1}| \le \frac{p(1-p)}{1+p}|w_0|$$ and the inequality is sharp. Proof. With P(z) as in the proof of Theorem 8, $$\frac{d}{dz}\log(z-p)(f(z)-w_0) = \frac{(1-pz)-P(z)}{(z-p)(1-pz)}.$$ Integrating, we obtain $$f(z) - w_0 = \frac{pw_0}{z - p} \exp \int_0^z \frac{(1 - p\xi) - P(\xi)}{(\xi - p)(1 - p\xi)} d\xi.$$ Thus $$a_{-1} = \lim_{z \to p} (z - p)(f(z) - w_0) = pw_0 \exp \int_0^p \frac{(1 - p\xi) - P(\xi)}{(\xi - p)(1 - p\xi)} d\xi$$ and (5.10) $$|a_{-1}| = p|w_0| \exp \int_0^p \frac{(1-p\xi) - \operatorname{Re} P(\xi)}{(\xi-p)(1-p\xi)} d\xi.$$ We can write $$P(z) = (1 - p^2)Q\left(\frac{z - p}{1 - pz}\right)$$ where $\operatorname{Re} Q(z) > 0$ for $z \in \Delta$ and Q(0) = 1. Using the well-known inequality $\operatorname{Re} Q(z) \geq (1 - |z|)/(1 + |z|)$ , we obtain for $\xi$ real and $0 \leq \xi \leq p$ , (5.11) $$\operatorname{Re} P(\xi) \ge (1 - p^2) \frac{1 - \left| \frac{\xi - p}{1 - p\xi} \right|}{1 + \left| \frac{\xi - p}{1 - p\xi} \right|}$$ $$= (1 - p^2) \frac{1 - \frac{p - \xi}{1 - p\xi}}{1 + \frac{p - \xi}{1 - p\xi}} = (1 - p)^2 \frac{1 + \xi}{1 - \xi}.$$ Combining (5.10) and (5.11) gives $$|a_{-1}| \le p|w_0| \exp \int_0^p \frac{(1-p\xi) - (1-p)^2 \frac{1+\xi}{1-\xi}}{(\xi-p)(1-p\xi)} d\xi$$ $$= p|w_0| \exp \int_0^p \frac{p\xi + (p-2)}{(1-\xi)(1-p\xi)} d\xi$$ $$= p|w_0| \exp \int_0^p \left(\frac{-2}{1-\xi} + \frac{p}{1-p\xi}\right) d\xi = p|w_0| \left(\frac{1-p}{1+p}\right),$$ which is the inequality to be proven. Equality is attained by the function given in Theorem 8. COROLLARY. With notation of Theorem 8, $$|a_0 - w_0| \le \frac{p(p+2)}{(1+p)^2} |w_0|$$ and $$|a_1| \le \frac{p}{(1-p)(1+p)^3} |w_0|.$$ Both inequalities are sharp.