Assignment-daixieTM为您提供格拉斯哥大学University of Glasgow METHODS IN COMPLEX ANALYSIS MATHS4076复分析代写代考和辅导服务！
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It deals with functions that are differentiable in the complex sense, meaning that they satisfy the Cauchy-Riemann equations. These functions are also called holomorphic functions.
The theory of analytic functions has many important applications in various fields, including physics, engineering, and mathematics. Some of the key topics that you might encounter in this course include power series expansions, Laurent series, singularities, residues, conformal mappings, and the Cauchy integral formula.
Throughout the course, you will learn about the properties and behavior of analytic functions, and you will develop techniques for analyzing and manipulating these functions. You will also study the connections between analytic functions and other areas of mathematics, such as complex geometry and number theory.
Overall, the study of analytic functions is a fascinating and important topic in mathematics, and it is essential for anyone interested in pursuing a career in pure or applied mathematics.
Let $\sum_{n=1}^{\infty} z_n$ be a convergent series of complex numbers. (1) Let $\alpha<\pi / 2$. Show that if for all $n,\left|\operatorname{Arg} z_n\right|<\alpha$, then the series $\sum_{n=1}^{\infty} z_n$ converges absolutely.
(2) Does the conclusion of the previous part hold if $\alpha=\pi / 2$ ?
(2) The conclusion of part (1) does not hold if $\alpha = \frac{\pi}{2}$. For example, consider the series $\sum_{n=1}^\infty \frac{i^n}{n}$, where $i = \sqrt{-1}$. For all $n$, we have $|\operatorname{Arg} (i^n/n)| = \frac{\pi}{2}$. However, the series converges conditionally by the alternating series test.
(3) Assume that for all $n, \Re z_n \geq 0$ and that the series $\sum_{n=1}^{\infty} z_n^2$ also converges. Show that the series $\sum_{n=1}^{\infty} z_n^2$ converges absolutely.
(3) Since $\Re z_n \geq 0$ for all $n$, we have $|z_n| = \sqrt{z_n \overline{z_n}} = \sqrt{z_n^2} = z_n$. Therefore, we have $$|z_n^2| = |z_n|^2 = z_n^2$$ Since the series $\sum_{n=1}^\infty z_n^2$ converges, we have $$\sum_{n=1}^\infty |z_n^2| = \sum_{n=1}^\infty z_n^2$$ converges, and hence the series $\sum_{n=1}^\infty z_n$ converges absolutely.
