Chapter_6

Chapter_6 - VI 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8 1 VI.1.1 1 2 3 P L K 111 LLL Find all possible Laurent expansions

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Unformatted text preview: VI 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8 1 VI.1.1 1 2 3 P L K 111 LLL Find all possible Laurent expansions centered at of the following functions. (a) 1 z 2 & z (b) z & 1 z +1 (c) 1 ( z 2 & 1)( z 2 & 4) Solution (a) In the region < j z j < 1 , we have 1 z 2 & z = & 1 z 1 1 & z = & 1 z 1 X k =0 z k = & 1 X k =0 z k & 1 = [ k & 1 = n ] = & 1 X n = & 1 z n : In the region j z j > 1 , we have 1 z 2 & z = 1 z 2 1 1 & 1 z = 1 z 2 1 X k =0 & 1 z ¡ k = 1 z 2 1 X k =0 z & k & 2 = [ & k & 2 = n ] = & 2 X n = &1 z n : (b) In the region j z j < 1 we have z & 1 z + 1 = 1 & 2 z + 1 = 1 & 2 1 & ( & z ) = 1 & 2 1 X n =0 ( & 1) n z n = & 1 & 2 1 X n =1 ( & 1) n z n : In the region j z j > 1 we have z & 1 z + 1 = 1 & 2 z + 1 = 1 & 2 z 1 1 & ¢ & 1 z £ = 1 & 2 z 1 X k =0 ( & 1) k z k = = 1 & 2 1 X k =0 ( & 1) k z k +1 = [ k + 1 = n ] = 1 & 2 1 X n =1 ( & 1) n & 1 z n = 1 + 2 1 X n =1 ( & 1) n z n : (c) From computation, we have 2 1 ( z 2 & 1) ( z 2 & 4) = & 1 3 1 z 2 & 1 + 1 3 1 z 2 & 4 In the region j z j < 1 , we have 1 ( z 2 & 1) ( z 2 & 4) = & 1 3 1 z 2 & 1 + 1 3 1 z 2 & 4 = 1 3 1 1 & z 2 & 1 12 1 1 & z 2 4 = = 1 3 1 X n =0 z 2 n & 1 12 1 X n =0 z 2 n 4 n = 1 12 1 X n =0 & 4 & 4 & n ¡ z 2 n : In the region 1 < j z j < 2 , we have 1 ( z 2 & 1) ( z 2 & 4) = & 1 3 1 z 2 & 1 + 1 3 1 z 2 & 4 = & 1 3 1 z 2 & 1 & 1 z 2 ¡ & 1 3 1 4 & 1 & z 2 4 ¡ = = & 1 3 z 2 1 X k =0 1 z 2 n & 1 12 1 X n =0 z 2 n 4 n = & 1 3 & 1 X k = &1 z 2 k & 1 12 1 X n =0 z 2 n 4 n In the region j z j > 2 , we have 1 ( z 2 & 1) ( z 2 & 4) = & 1 3 1 z 2 & 1 + 1 3 1 z 2 & 4 = = & 1 3 1 z 2 & 1 & 1 z 2 ¡ + 1 3 1 z 2 & 1 & 4 z 2 ¡ = = & 1 3 z 2 1 X n =0 1 z 2 n + 1 3 z 2 1 X n =0 4 n z 2 n = = & 1 3 1 X n =0 1 z 2( n +1) + 1 3 1 X n =0 4 n z 2( n +1) = 1 3 1 X n =0 (4 n & 1) z & 2( n +1) = [ n + 1 = & k ] = = 1 3 & 1 X k = &1 & 4 & 1 & k & 1 ¡ z 2 k : 3 VI.1.2 1 2 3 P L K 111 LLL For each of the functions in Exercise 1, &nd the Laurent expansion centered at z = & 1 that converges at z = 1 2 . Determine the largest open set on which each series converges. Solution (a) Partial fractions give us 1 z 2 & z = & 1 z + 1 z & 1 Laurent series for & 1 =z & 1 z = & 1 z + 1 & 1 = & 1 z + 1 1 1 & 1 z +1 = & 1 z + 1 1 X k =0 1 ( z + 1) k = = & 1 X k =0 1 ( z + 1) k +1 = [ k + 1 = & n ] = & & 1 X n = &1 ( z + 1) n Laurent series for 1 =z & 1 1 z & 1 = 1 z + 1 & 2 = & 1 2 1 1 & z +1 2 = & 1 2 1 X n =0 ( z + 1) n 2 n = & 1 2 n +1 1 X n =0 ( z + 1) n Thus we have 1 z 2 & z = 1 X k = &1 a n ( z + 1) n ; where a n = & & 1 for n ¡ & 1 & 1 2 n +1 for n ¢ ¡ ¡ ¡ ¡ The function 1 = ( z 2 & z ) converges on the set 1 < j z + 1 j < 2 . (b) We have z & 1 z + 1 = 1 & 2 z + 1 The function ( z & 1) = ( z + 1) converges on the set < j z + 1 j < 1 ....
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This note was uploaded on 11/13/2010 for the course MATH 132 taught by Professor Grossman during the Spring '08 term at UCLA.

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Chapter_6 - VI 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8 1 VI.1.1 1 2 3 P L K 111 LLL Find all possible Laurent expansions

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