New Sum-Product Estimates for Real and Complex Numbers

A variation on the sum-product problem seeks to show that a set which is defined by additive and multiplicative operations will always be large. In this paper, we prove new results of this type. In particular, we show that for any finite set A\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A$$\end{document} of positive real numbers, it is true that |{a+bc+d:a,b,c,d∈A}|≥2|A|2-1.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \Big |\Big \{\frac{a+b}{c+d}:a,b,c,d\in {A}\Big \}\Big |\ge {2|A|^2-1}. \end{aligned}$$\end{document}As a consequence of this result, it is also established that |4k-1A(k)|:=|A…A⏟ktimes+⋯+A…A⏟4k-1times|≥|A|k.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} |4^{k-1}A^{(k)}|:=|\underbrace{\underbrace{A\ldots {A}}_{k\,\,\text {times}}+\cdots {+A\ldots {A}}}_{4^{k-1} \mathrm{times}}|\ge {|A|^k}. \end{aligned}$$\end{document}Later on, it is shown that both of these bounds hold in the case when A\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A$$\end{document} is a finite set of complex numbers, although with smaller multiplicative constants.