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AI Summary & Key Takeaways
AI Insight⚡ TL;DR(Too Long; Didn't Read)
We find conditions on the complex-valued function A : U → C definedin the unit disc U such that the differential inequalityRe [A(z)p2(z) − α(zp0(z) − 1)2 + 2β(zp0(z))2 + γ] > 0implies Re p(z) > 0, where p ∈ H[1, n], α, β ∈ C and n is a positiveinteger.
📌 3 Key Takeaways
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Objective & Hypothesis:We find conditions on the complex-valued function A : U → C definedin the unit disc U such that the differential inequalityRe [A(z)p2(z) − α(zp0(z) − 1)2 + 2β(zp0(z))2 + γ] > 0implies Re p(z) > 0, where p ∈ H[1, n], α, β ∈ C and n is a positiveinteger.
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Key Findings & Data:We find conditions on the complex-valued function A : U → C definedin the unit disc U such that the differential inequalityRe [A(z)p2(z) − α(zp0(z) − 1)2 + 2β(zp0(z))2 + γ] > 0implies Re p(z) > 0, where p ∈ H[1, n], α, β ∈ C and n is a positiveinteger.
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Conclusion & Impact:We find conditions on the complex-valued function A : U → C definedin the unit disc U such that the differential inequalityRe [A(z)p2(z) − α(zp0(z) − 1)2 + 2β(zp0(z))2 + γ] > 0implies Re p(z) > 0, where p ∈ H[1, n], α, β ∈ C and n is a positiveinteger.
🔬Design:Research Article
📍Scope:Comprehensive Sample
⚙️Variables:Differential subordination · Dominant
Abstract
We find conditions on the complex-valued function A : U → C definedin the unit disc U such that the differential inequalityRe [A(z)p2(z) − α(zp0(z) − 1)2 + 2β(zp0(z))2 + γ] > 0implies Re p(z) > 0, where p ∈ H[1, n], α, β ∈ C and n is a positiveinteger.
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Cite This Article
FRASIN, B. (2022). A NEW DIFFERENTIAL INEQUALITY. Hacettepe Journal of Mathematics and Statistics, Cilt: 36 - Sayı: 1, 7–10. https://ajindex.com/articles/530737