{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

assgt6 - S is contained in S(Note Proving that the...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
MATH 407-10a, Assignment 6 Guidelines Answer the questions in the space provided; you may write on both sides of the paper. Put the names of all group members in the top right corner. You may append additional sheets as needed. Please staple all sheets together before submission. Due: Thursday, Feb 11th Definition. Let S be a nonempty subset of the complex plane. A complex number ω is said to be a boundary point of S if, for every r > 0, the open disc D ( ω ; r ) intersects both S and S c (the complement of S ). The boundary of S , denoted by ∂S , is the set of all boundary points of S , to wit: ∂S := { ω C : ω is a boundary point of S } 1. ( 10 marks ) Suppose that S is a nonempty subset of the complex plane. Prove that the following statements are equivalent: (i) S is closed. (ii) ∂S S , that is, the boundary of
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: S is contained in S . (Note: Proving that the statements are equivalent entails doing the following: assume (i) and prove that (ii) holds; then assume (ii) and prove that (i) holds.) 2. ( 10 marks ) Let ( z n : n ∈ N ) be a sequence of complex numbers. Prove the following state-ments: (i) If lim n →∞ | z n | 1 /n = L , and 0 ≤ L < 1, then the series ∞ ∑ n =1 z n is absolutely convergent . (ii) If lim n →∞ | z n | 1 /n = L , and L > 1, or if lim n →∞ | z n | 1 /n = + ∞ , then lim n →∞ | z n | = ∞ ; in particular the series ∞ ∑ n =1 z n diverges. (Model your argument after the proof of Theorem 1.3.10 in the book.)...
View Full Document

{[ snackBarMessage ]}