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Abstract
In this article we consider functions $f$ meromorphic in the unit disk. We give an elementary proof for a condition that is sufficient for the univalence of such functions. This condition simplifies and generalizes known conditions. We present some typical problems of geometrical function theory and give elementary solutions in the case of the above functions.
Results & Lemmas (5)
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Theorem 1.
Theorem 1. Let f be meromorphic in D and such that f(0) = f ′(0) −1 = 0. If for all z ∈D the inequality (1) is valid, then f is univalent…
Theorem 1. Let f be meromorphic in D and such that f(0) = f ′(0) −1 = 0. If for all z ∈D the inequality (1) is valid, then f is univalent in D.
Theorem 1
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Theorem 1, the function f is allowed to have a pole in the unit disk. The proof of
Theorem 1, the function f is allowed to have a pole in the unit disk. The proof of
Theorem 1
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Theorem 1 shows that such pole is necessarily a simple pole. We want to add the hint that similar meromorphic functions have been…
Theorem 1 shows that such pole is necessarily a simple pole. We want to add the hint that similar meromorphic functions have been considered in [2]. In what follows, we shall derive some analytic properties of the meromorphic functions satisfying (1). To that end we refine this condition, namely, we let λ ∈(0, 1] and consider the condition (5)
Theorem 2.
Theorem 2. (a) Let f be meromorphic in D and such that f(0) = f ′(0)−1 = 0. If for all z ∈D the inequality (1) is valid, then f is…
Theorem 2. (a) Let f be meromorphic in D and such that f(0) = f ′(0)−1 = 0. If for all z ∈D the inequality (1) is valid, then f is continuous in D with a possible exception at a zero of (3). (b) Let f be meromorphic in D and such that f(0) = f ′(0) −1 = 0. If for all z ∈D the inequality (5) is valid for some λ ∈(0, 1), then f is univalent in D.
Theorem 3.
Theorem 3. Let λ ∈(0, 1] and f be meromorphic in D and such that f(0) = f ′(0) −1 = 0. If for all z ∈D the inequality (5) is valid and f(z)…
Theorem 3. Let λ ∈(0, 1] and f be meromorphic in D and such that f(0) = f ′(0) −1 = 0. If for all z ∈D the inequality (5) is valid and f(z) = z + ∞ X n=2 anzn has no pole in the disk {z : |z| < p}, p ∈(0, 1], then the inequality (6) |a2| ≤1 + λp2 p is valid. The inequality is sharp and equality is attained if and only if (7) f(z) = z
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