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{SECT 0 {EXCHG {PARA 258 "" 0 "" {TEXT 257 41 "COMPLEX ANALYSIS: Maple
 Worksheets,  2001" }{TEXT 286 144 "\n(c) John H. Mathews          Rus
sell W. Howell\nmathews@fullerton.edu     howell@westmont.edu\n\nCompl
imentary software to accompany the textbook:" }}{PARA 258 "" 0 "" 
{TEXT 257 82 "COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed,
 2001, ISBN: 0-7637-1425-9" }{TEXT 285 197 "\nJones and Bartlett Publi
shers, Inc.,      40  Tall  Pine  Drive,      Sudbury,  MA  01776\nTel
e.  (800) 832-0034;      FAX:  (508)  443-8000,      E-mail:  mkt@jbpu
b.com,      http://www.jbpub.com/" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
284 1 "\n" }{TEXT 256 26 "CHAPTER 8   RESIDUE THEORY" }{TEXT 279 2 "\n
\n" }{TEXT 256 32 "Section 8.1  The Residue Theorem" }{TEXT 280 7 "\n
\n     " }{TEXT -1 116 "The Cauchy-integral formulas in Section 6.5 ar
e useful in evaluating contour integrals over a simple closed contour \+
" }{XPPEDIT 18 0 "C;" "6#%\"CG" }{TEXT -1 35 " where the integrand has
 the form  " }{XPPEDIT 18 0 "f(z)/((z-z[0])^k);" "6#*&-%\"fG6#%\"zG\"
\"\"),&F'F(&F'6#\"\"!!\"\"%\"kGF." }{TEXT -1 7 "  and  " }{XPPEDIT 18 
0 "f;" "6#%\"fG" }{TEXT -1 103 "  is an analytic function.  In this ca
se, the singularity of the integrand is at worst a pole of order " }
{XPPEDIT 18 0 "k" "6#%\"kG" }{TEXT -1 5 " at  " }{XPPEDIT 18 0 "z[0]" 
"6#&%\"zG6#\"\"!" }{TEXT -1 133 ".  In this section we extend this res
ult to integrals that have a finite number of isolated singularities a
nd lie inside the contour " }{XPPEDIT 18 0 "C" "6#%\"CG" }{TEXT -1 93 
".  This new method can be used in cases where the integrand has an es
sential singularity at  " }{XPPEDIT 18 0 "z[0]" "6#&%\"zG6#\"\"!" }
{TEXT -1 57 "  and is an important extension of the previous method.  \+
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 
24 "Definition 8.1:  Residue" }{TEXT 299 2 "  " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 300 5 "Let  " }{XPPEDIT 18 0 "f(
z)" "6#-%\"fG6#%\"zG" }{TEXT 289 57 "  have a nonremovable isolated si
ngularity at the point  " }{XPPEDIT 18 0 "z[0]" "6#&%\"zG6#\"\"!" }
{TEXT 290 10 " .  Then  " }{XPPEDIT 18 0 "f(z)" "6#-%\"fG6#%\"zG" }
{TEXT 291 42 "  has \nthe Laurent series representation  " }{XPPEDIT 
18 0 "f(z) = sum(a[n]*(z-z0)^n, n=-infinity..infinity)" "6#/-%\"fG6#%
\"zG-%$sumG6$*&&%\"aG6#%\"nG\"\"\"),&F'F0%#z0G!\"\"F/F0/F/;,$%)infinit
yGF4F8" }{TEXT 288 22 "  .  The coefficient  " }{XPPEDIT 18 0 "a[-1]" 
"6#&%\"aG6#,$\"\"\"!\"\"" }{TEXT 292 6 "  of  " }{XPPEDIT 18 0 "1/(z-z
[0])" "6#*&\"\"\"F$,&%\"zGF$&F&6#\"\"!!\"\"F*" }{TEXT 293 17 "  is \nc
alled the " }{TEXT 257 7 "residue" }{TEXT 298 5 " of  " }{XPPEDIT 18 
0 "f(z)" "6#-%\"fG6#%\"zG" }{TEXT 294 6 "  at  " }{XPPEDIT 18 0 "z[0]
" "6#&%\"zG6#\"\"!" }{TEXT 295 35 "  and we use the notation  Res[f , \+
" }{XPPEDIT 18 0 "z[0]" "6#&%\"zG6#\"\"!" }{TEXT 296 4 "] = " }
{XPPEDIT 18 0 "a[-1]" "6#&%\"aG6#,$\"\"\"!\"\"" }{TEXT 297 4 " .  " }
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT 287 93 "Load Maple's  \"residue\" procedure.\nMake sur
e this is done only ONCE during a Maple  session.\n" }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 17 "readlib(residue):" }{TEXT -1 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT 256 22 "Example 8.1, Page 307." }{TEXT 281 46 "   Use \+
Laurent series to find the residue at  " }{XPPEDIT 18 0 "z = 0" "6#/%
\"zG\"\"!" }{TEXT 264 20 "  for the function  " }{XPPEDIT 18 0 "f(z) =
 exp(2/z)" "6#/-%\"fG6#%\"zG-%$expG6#*&\"\"#\"\"\"F'!\"\"" }{TEXT 265 
3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "f:='f': s:='s': z:='z':
\nf := z -> exp(2/z):\ns := series(f(z), z=infinity, 7):\n`f(z) ` = f(
z);\n`f(z) ` = s;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT 260 20 "The coefficient of  " }{XPPEDIT 18 0 "1/
z" "6#*&\"\"\"F$%\"zG!\"\"" }{TEXT 266 6 "  is  " }{XPPEDIT 18 0 "2" "
6#\"\"#" }{TEXT 267 21 "  so the residue is  " }{XPPEDIT 18 0 "`Res[f,
0]` = 2" "6#/%)Res[f,0]G\"\"#" }{TEXT 268 2 " ." }}}{EXCHG {PARA 0 "" 
0 "" {TEXT 278 1 "\n" }{TEXT 256 22 "Example 8.2, Page 308." }{TEXT 
282 23 "  Find the residue at  " }{XPPEDIT 18 0 "z = 0" "6#/%\"zG\"\"!
" }{TEXT 269 20 "  for the function  " }{XPPEDIT 18 0 "g(z) = 3/(2*z +
 z^2 - z^3)" "6#/-%\"gG6#%\"zG*&\"\"$\"\"\",(*&\"\"#F*F'F*F**$F'F-F**$
F'F)!\"\"F0" }{TEXT 261 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 110 
"g:='g': s:='s': z:='z':\ng := z -> 3/(2*z + z^2 - z^3):\ns := series(
g(z), z=0, 5):\n`g(z) ` = g(z);\n`g(z) ` = s;" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 262 20 "The coefficie
nt of  " }{XPPEDIT 18 0 "1/z" "6#*&\"\"\"F$%\"zG!\"\"" }{TEXT 270 6 " \+
 is  " }{XPPEDIT 18 0 "3/2" "6#*&\"\"$\"\"\"\"\"#!\"\"" }{TEXT 271 21 
"  so the residue is  " }{XPPEDIT 18 0 "`Res[g,0]` = 3/2" "6#/%)Res[g,
0]G*&\"\"$\"\"\"\"\"#!\"\"" }{TEXT 272 75 " .\nWe compare this with Ma
ple's  residue procedure for computing residues.\n" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 62 "`g(z) ` = g(z);\n`g(z) ` = s;\n`Res[g,0] ` = residu
e(g(z), z=0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT 256 22 "Example 8.3, Page 308." }{TEXT 283 30 " \+
 Use residues to integrate   " }{XPPEDIT 18 0 "int(exp(2/z), z=C..`   \+
`)" "6#-%$intG6$-%$expG6#*&\"\"#\"\"\"%\"zG!\"\"/F,;%\"CG%$~~~G" }
{TEXT 273 11 "   around  " }{XPPEDIT 18 0 "C" "6#%\"CG" }{TEXT 274 2 "
: " }{XPPEDIT 18 0 "abs(z) = 1" "6#/-%$absG6#%\"zG\"\"\"" }{TEXT 275 
21 " .\nFrom Example 8.1  " }{XPPEDIT 18 0 "`Res[f,0]` = 2" "6#/%)Res[
f,0]G\"\"#" }{TEXT 276 40 " .   Thus the value of the integral is  " }
{XPPEDIT 18 0 "2*pi*i*`Res[f,0]`" "6#**\"\"#\"\"\"%#piGF%%\"iGF%%)Res[
f,0]GF%" }{TEXT 277 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 218 "f:=
'f': F:='F': s:='s': z:='z':\nf := z -> exp(2/z):\n`F(z) ` = f(z);\ns \+
:= series(f(z), z=infinity, 5):\nres := 2:\n`Res[F,0] ` = res;  \nprin
t(int(F(z),z=C..``) = `2*Pi*I*Res[f,0])`);\nprint(int(F(z),z=C..``) = \+
2*Pi*I*res);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT 256 39 "Theorem 8.1  (Cauchy's Residue Theorem)" }
{TEXT 301 2 "  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 5 "Let  " }{XPPEDIT 18 0 "D;" "6#%\"DG" }{TEXT -1 40 "  be a \+
simply connected domain and let  " }{XPPEDIT 18 0 "C;" "6#%\"CG" }
{TEXT -1 63 "  be a simple closed positively oriented contour that lie
s in  " }{XPPEDIT 18 0 "D" "6#%\"DG" }{TEXT -1 3 ".  " }}{PARA 0 "" 0 
"" {TEXT -1 4 "If  " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 21 "  is \+
analytic inside " }{XPPEDIT 18 0 "C" "6#%\"CG" }{TEXT -1 8 " and on " 
}{XPPEDIT 18 0 "C" "6#%\"CG" }{TEXT -1 24 ", except at the points  " }
{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\"\"" }{TEXT 303 3 " , " }{XPPEDIT 
18 0 "z[2]" "6#&%\"zG6#\"\"#" }{TEXT 304 8 " , ..., " }{XPPEDIT 18 0 "
z[n];" "6#&%\"zG6#%\"nG" }{TEXT -1 18 "  that lie inside " }{XPPEDIT 
18 0 "C" "6#%\"CG" }{TEXT -1 8 ", then  " }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "       " }{XPPEDIT 18 0 "Int(f(z),z
 = C .. ``);" "6#-%$IntG6$-%\"fG6#%\"zG/F);%\"CG%!G" }{TEXT -1 5 "  = \+
 " }{XPPEDIT 18 0 "2*pi*i" "6#*(\"\"#\"\"\"%#piGF%%\"iGF%" }{XPPEDIT 
18 0 "Sum(`Res[ f , `*z[k]*`]`,k = 1 .. n);" "6#-%$SumG6$*(%*Res[~f~,~
G\"\"\"&%\"zG6#%\"kGF(%\"]GF(/F,;F(%\"nG" }{TEXT -1 10 " .        " }
{TEXT 302 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT 263 19 "End of Section 8.1." }}}}{MARK "0 0 0" 
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