On the multivalence of a class of meromorphic functions

Theorem 1. The function g(z) defined by (2) is univalent in the disk $$|z|<\left(\frac{a}{1+a}\right)^{1/2}$$ . This estimate is sharp for a = 1 and for all a the best estimate of the radius of the disk about the origin for which all functions of type g(z) are univalent does not exceed the number $$\min\left[\frac{(2a)^{1/2}}{1+a^{1/2}},\,a^{1/2}\right]$$ .

Theorem 2. Let g(z) be a function as defined by (2). Then g(z) is completely p-valent in the disk $|z| < r_p$ where $r_p$ is the unique positive root (<1) of the equation $$x^{(p+3)/2} = ap!(1-x^{2/p})^p$$ .

Theorem 3. Let f(z) and g(z) be as defined by (3) and (2) respectively and let conditions (6) be satisfied. Then (a) The function f(z) is univalent in the region $|z| r_i(l)$ where $r_i(l)$ is the largest positive root of the equation $$x^{2l+2} - r^2(l)x^{2l} - 1 = 0$$ and where r(l) $(1 < r(l) < r_1(l))$ is the largest positive root of the equation $$y^{2l} - ly^{l-1} - 1 = 0.$$ (b) The function g(z) is univalent in the disk $|z| < r_0(l)$ , where $r_0(l)$ is the positive root of the equation $$lz^{l+1} = a(1-z^{2l})$$ .

Theorem 4. Let f(z) be a function defined by (1). Then if $R > (27/11)^{1/2}$ then the image of |z| = R is starlike with respect to the origin. Proof. By (1) and (3) $$g(z) = \frac{A}{f(1/z)} = \frac{1}{z} - \alpha \left(\frac{1}{z}\right) = \frac{1}{z} + \beta(z)$$ defined for 0 < |z| < 1 with $\beta(z)$ analytic and $|\beta(z)| \le 1$ in the unit disk. Let $$h(z) = \frac{1}{g(z)}.$$ h(z) is analytic in the unit disk and h(0) = 0. We investigate the starlikeness of h(z). To this effect we notice that the number $$\operatorname{Re}\left\{z\frac{h'(z)}{h(z)}\right\} = \operatorname{Re}\left\{\frac{1-z^2\beta'(z)}{1+z\beta(z)}\right\}$$ will be positive if and only if (8) $$\operatorname{Re}\left[1+\overline{z\beta}-z^2\beta'-z\,|z|^2\,\overline{\beta}\beta'\right]>0.$$ Let |z|=r, $|\beta(z)|=a$ . Then $|\beta'(z)| \leq (1-a^2)/1-r^2$ and (8) reduces to $$ar + r^2 \frac{1-a^2}{1-r^2} + ar^3 \frac{1-a^2}{1-r^2} < 1$$ or $$\phi(a) = ar + r^2(2 - a^2) - a^3r^3 < 1$$ . Now $\phi(a)$ is positive and increasing for $0 \le a \le (3r)^{-1}$ so that $\phi(a) < \phi(1) < 1$ if $0 \le a < 1$ and $0 \le r \le 1/3$ if, however, r > 1/3 then $$\phi(a) \leq \phi\Bigl(\frac{1}{3r}\Bigr) = 2r^2 + \frac{5}{27} < 1$$ for $0 \le a < 1$ and $r^2 < 11/27$ . The theorem is established. It is perhaps worthwhile to mention that the function f(z) of the type (1) possesses the property $$\text{Re}\left[\frac{1}{f(z)}\right]'>0 \quad \text{for} \quad |z|>\sqrt{2}$$ . It is known (Noshiro-Warschawski-Wolff [2] p. 17) that an analytic function $\psi(z)$ in a convex domain is univalent if $\operatorname{Re} \psi'(z) > 0$ there. However, the geometric properties of the range that characterize these functions $\psi(z)$ , even in simple domains as the unit disk, are not known. (See [9], p. 317.) 3. Multivalence of lacunary polynomials. There are two related problems concerning the valence of polynomials in a disk on the basis of the knowledge of the distribution of one of its values. PROBLEM 1. If a polynomial of degree n assumes p times $(1 \le p < n)$ a certain value (counting multiplicities) in the center of a circle and does not take this value elsewhere in this circle, find the largest concentric circle in which the polynomial is p-valent. PROBLEM 2. If a polynomial of degree n assumes p times $(1 \le p < n)$ the values it has at the center, find the largest concentric circle in which the polynomial is necessarily p-valent. For simplicity we shall assume that the circle in Problems 1 and 2 is the unit circle. In this case it is easy to show that there exist numbers depending only on p and n which are solutions to these problems. For arbitrary polynomials a sharp bound for Problem 1 was given by M. Biernacki [1], p. 627, who also gave estimates for the solution of Problem 2. (See [1], p. 632, and [3], p. 92.) We shall here generalize Problems 1 and 2 to lacunary polynomials, for which we solve Problem 1 and give bounds for the solution of Problem 2. The following results shall be applied in the proofs. THEOREM B. [10], Theorem 5. Let $$P(z) = a_p z^p + a_{p-s} z^{p-s} + \cdots + a_0$$ , $Q(z)=b_qz^q+b_{q-t}z^{q-t}+\cdots+b_0,\, a_pb_q\neq 0,\, q>p,\, s\geq 1,\, t\geq 1,\,\, have\,\, their$ zeros in the disks $|z|\leq R_1$ and $|z|\leq R_2$ respectively. Let $r={\rm Min}\,(s,\,t)$ . Then at least p zeros of the polynomial $$P(z) + \lambda Q(z)$$ lie in the disk (9) $$|z| \leq \operatorname{Max} \left\{ \left( \frac{qR_1^r + pR_1^r}{q - p} \right)^{1/r}, R_2 \right\}.$$ REMARK. In [10] the bound (9) was deduced but no discussion about its sharpness was made. However, following the analysis M. Biernacki [1], pp. 625-627, made on the location of the polynomial (10) $$(X+P)^p + a(X-Q)^q = 0$$ where q>p are integers, P and Q positive real numbers and a is such that the equation (10) has a double real root, one deduces that the polynomial: $$(X^r + R_1^r)^{\alpha} + a(X^r - R_2^r)^{\beta},$$ where $$r=s=t,\, p=\alpha r,\, q=\beta r,\, a=(-1)^{\alpha+\beta+1}\frac{\alpha^\alpha}{\beta^\beta}\!\!\left(\!\frac{\beta-\alpha}{R_1^r+R_2^r}\!\right)^{\!\beta-\alpha}$$ has exactly p zeros in the disk (9) and r double zeros on its circumference. One sees, therefore, that if s=t then the bound of Theorem B is sharp. THEOREM C. [11], Lemma 2(a). If $|\alpha_k| \leq r$ and $m_k > 0$ , k = 1, 2, ..., s, there exists an analytic function $\alpha(z)$ defined for all |z| > r satisfying $$\sum_{k=1}^{s} \frac{m_k}{z - \alpha_k} = \sum_{k=1}^{s} \frac{m_k}{z - \alpha(z)}$$ and such that $|\alpha(z)| \leq r$ for |z| > r. If in addition $\sum_{k=1}^{s} m_k \alpha_k^l = 0$ for $l = 1, 2, \dots, p$ , then $|\alpha(z)| \leq r^{p+1}/|z|^p$ for |z| > r. We are now in a position to prove

Theorem 5. Suppose that the polynomial $$f(z) = z^p(1 + a_r z^r + \cdots + a_{n-p} z^{n-p})$$ $a_{n-p} \neq 0$ , $1 \leq p < n$ , $1 \leq r \leq n-p$ , does not vanish in the unit disk except for a p-fold zero at the origin. Then f(z) is necessarily p-valent in the disk $$|z| \leq \left(\frac{p}{n}\right)^{1/r}$$ . Furthermore the polynomials. $$z^p(1+z^r)^k$$ for $n = k_l + p$ , $k = 1, 2 \cdots$ satisfy the hypotheses of the theorem and are not p-valent in any disk about the origin of radius larger than $(p/n)^{1/r}$ .

Theorem 6. Suppose that the polynomial $$f(z) = a_n z^q + \cdots + a_n z^n ,$$ $a_q \neq 0$ , $1 \leq q < n$ , vanishes exactly p times (counting multiplicities) in the disk |z| < 1 and assume, furthermore, that the (n-p) zeros of f(z), which lie outside the unit disk, say $z_1, \dots, z_{n-p}$ possess the symmetry property (11) $$\sum_{i=1}^{n-p} z_i^{-j} = 0, j = 1, 2, \cdots, k-1,$$ where $k = 1, 2, \dots$ . When k = 1, we shall agree that no conditions of the form (11) are imposed. Then the polynomial f(z) is p-valent in the disk $$|z| < 2pA^{-1}e^{-2p}$$ , where $$A = \operatorname{Max}\left[\left(\frac{2n-p+1}{p+1}\right)^{1/k}, \frac{n-p}{k}\right]$$ . REMARK. Actually $A \le A' = \text{Max} \{(n-p)^{1/k}, (n-p)/k\}$ if $p \le n-2$ and $p \ne 1, 2$ or if $p \ne 1$ and $n \ge 5$ . The case p=1 has been solved in Problem 1 and the case p=n-1 has been solved by G. Szego [12], Theorem 12', p. 46.

Theorem 7. If the polynomial of degree n $$f(z) = a_1 z + a_2 z^2 + \cdots + a_{n-q} z^{n-q} + a_n z^n$$ has (n-1) zeros in the unit disk, then f(z) is at most (n-1)-valent in the disk |z| < x(q) where x(q) is as defined in Theorem D.