ps6sol - ACM 95a/100a Problem Set 6 Solutions Prepared by:...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ACM 95a/100a Problem Set 6 Solutions Prepared by: Zhiyi Li 11/13/2007 Total: 126 points Include grading section: 2 points Problem 1 (3 10 points) Expand f ( z ) = 1 z (1- z ) in Laurent series that converge in the following domains: a) 0 < | z | < 1 , b) | z | > 1 , c) | z + 1 | > 2 . Solution 1 We will use the geometric series sum formula: 1 1- = X n =0 n , | | < 1 (1) a) Here we let z , since | z | < 1. Thus, we can write f ( z ) = 1 z 1 1- z = 1 z X n =0 z n = X n =0 z n- 1 , < | z | < 1 (2) b) Here we let 1 z , since | z | > 1 fl fl 1 z fl fl < 1. We manipulate f ( z ) so that it is written in terms of 1 z : f ( z ) = 1 z 1 1- z = 1 z 1 z 1 z- 1 =- 1 z 2 1 1- 1 z =- 1 z 2 X n =0 1 z n =- X n =0 1 z n +2 , | z | > 1 (3) c) | z + 1 | > 2 fl fl fl 1 z +1 fl fl fl < 1 2 and fl fl fl 2 z +1 fl fl fl < 1. Therefore, 1 f ( z ) = 1 z 1 1- z = 1 z + 1 1- z = 1 ( z + 1)- 1 + 1 2- ( z + 1) = 1 z + 1 1 1- 1 z +1- 1 z + 1 1 1- 2 z +1 = 1 z + 1 X n =0 1 z + 1 n- X n =0 2 z + 1 n ! = X n =1 1- 2 n ( z + 1) n +1 , | z + 1 | > 2 (4) Alternatively, we note that | z + 1 | > 2 is contained in | z | > 1. Thus, our series from part (b). converges in | z + 1 | > 2, f ( z ) =- X n =0 1 z n +2 , | z + 1 | > 2 (5) Problem 2 (10 points) Without determining the series, specify the region of convergence for a Laurent series representing f ( z ) = 1 / ( z 4 + 4) in powers of z- 1 that converges at z = i . Solution 2 We are considering the Laurent series about z = 1 that converges at z = i . The poles of f ( z ) occur where z 4 + 4 = 0, or at z = 4 1 4 e i ( 4 + k 2 ) , = 1 i . Thus, there is a Laurent series for each of the three domains depicted in Figure 1 (each domain boundary intersects two poles). The Laurent series that converges at z = i is associated with the domain 1 < | z- 1 | < 5 . Problem 3 (5 6 points) Classify all the singularities (removable, poles, isolated essential, branch points, non-isolated essential) of the following functions in the extended complex plane a) z z 2 + 1 , b) 1 sin z , c) log(1 + z 2 ) , d) z sin 1 z , e) exp 1 z- i log 1+ z 1- z cos ( z ) ....
View Full Document

This note was uploaded on 02/22/2010 for the course ACM 95A taught by Professor Nilesa.pierce during the Fall '06 term at Caltech.

Page1 / 9

ps6sol - ACM 95a/100a Problem Set 6 Solutions Prepared by:...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online