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-1 "Monaco" 1 9 255 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "R3 Font 2" -1 257 1 {CSTYLE " " -1 -1 "Symbol" 1 10 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "R3 Font 0" -1 258 1 {CSTYLE "" -1 -1 "Geneva" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 258 "" 0 "" {TEXT 257 41 "COMPLEX ANALYSIS: Maple Worksheets, 2001" }{TEXT 318 144 "\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 317 197 "\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 315 1 "\n" }{TEXT 256 26 "CHAPTER 8 RESIDUE THEORY" }{TEXT 307 2 "\n \n" }{TEXT 256 36 "Section 8.2 Calculation of Residues" }{TEXT 308 5 "\n\n " }{TEXT -1 102 "The calculation of a Laurent series expansion is tedious in most circumstances. Since the residue at " }{XPPEDIT 18 0 "z[0]" "6#&%\"zG6#\"\"!" }{TEXT -1 32 " involves only the coeffi cient " }{XPPEDIT 18 0 "a[-1];" "6#&%\"aG6#,$\"\"\"!\"\"" }{TEXT -1 134 " in the Laurent expansion, we seek a method to calculate the resi due from special information about the nature of the singularity at \+ " }{XPPEDIT 18 0 "z[0]" "6#&%\"zG6#\"\"!" }{TEXT -1 12 ".\n \011\n I f " }{XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT -1 34 " has a removable sing ularity at " }{XPPEDIT 18 0 "z[0]" "6#&%\"zG6#\"\"!" }{TEXT -1 8 ", t hen " }{XPPEDIT 18 0 "a[-1] = 0;" "6#/&%\"aG6#,$\"\"\"!\"\"\"\"!" } {TEXT -1 7 " for " }{XPPEDIT 18 0 "n = 1,2,`...`;" "6%/%\"nG\"\"\"\" \"#%$...G" }{TEXT -1 16 ". Therefore if " }{XPPEDIT 18 0 "z[0]" "6#&% \"zG6#\"\"!" }{TEXT -1 36 " is a removable singularity, then " } {XPPEDIT 18 0 "`Res[`*f*`,`*z[0]*`]` = 0;" "6#/*,%%Res[G\"\"\"%\"fGF&% \",GF&&%\"zG6#\"\"!F&%\"]GF&F," }{TEXT -1 71 ". The following theorem \+ gives methods for evaluating residues at poles." }{TEXT 316 5 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 32 "Theorem 8.2 (Residues at Poles)" }{TEXT -1 4 " " }}{PARA 0 "" 0 "" {TEXT -1 5 " " }}{PARA 0 "" 0 "" {TEXT 256 3 "(i)" }{TEXT 327 2 " " }{TEXT -1 4 "If " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 23 " has a simple pole at " }{XPPEDIT 18 0 "z[0]" "6#&%\"zG6#\"\"!" } {TEXT -1 8 ", then " }{TEXT 323 7 "Res[f, " }{XPPEDIT 18 0 "z[0]" "6# &%\"zG6#\"\"!" }{TEXT 329 4 "] = " }{XPPEDIT 18 0 "Limit((z-z[0])*f(z) ,z = z[0]);" "6#-%&LimitG6$*&,&%\"zG\"\"\"&F(6#\"\"!!\"\"F)-%\"fG6#F(F )/F(&F(6#F," }{TEXT 324 5 " . " }{TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT 256 4 "(ii)" }{TEXT -1 6 " If " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 22 " has a pole of order " }{XPPEDIT 18 0 "2" "6#\"\"#" } {TEXT -1 4 " at " }{XPPEDIT 18 0 "z[0]" "6#&%\"zG6#\"\"!" }{TEXT -1 7 ", then " }{TEXT 321 8 " Res[f, " }{XPPEDIT 18 0 "z[0]" "6#&%\"zG6#\" \"!" }{TEXT 330 4 "] = " }{XPPEDIT 18 0 "Limit(Diff((z-z[0])^2*f(z),z) ,z = z[0]);" "6#-%&LimitG6$-%%DiffG6$*&,&%\"zG\"\"\"&F+6#\"\"!!\"\"\" \"#-%\"fG6#F+F,F+/F+&F+6#F/" }{TEXT 322 6 " . \n" }}{PARA 0 "" 0 "" {TEXT 256 5 "(iii)" }{TEXT 328 2 " " }{TEXT -1 4 "If " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 22 " has a pole of order " }{XPPEDIT 18 0 "3" "6#\"\"$" }{TEXT -1 4 " at " }{XPPEDIT 18 0 "z[0]" "6#&%\"zG6#\"\" !" }{TEXT -1 7 ", then " }{TEXT 325 8 " Res[f, " }{XPPEDIT 18 0 "z[0] " "6#&%\"zG6#\"\"!" }{TEXT 331 4 "] = " }{XPPEDIT 18 0 "1/`2!`;" "6#*& \"\"\"F$%#2!G!\"\"" }{XPPEDIT 18 0 "Limit(Diff((z-z[0])^3*f(z),z,z),z \+ = z[0]);" "6#-%&LimitG6$-%%DiffG6%*&,&%\"zG\"\"\"&F+6#\"\"!!\"\"\"\"$- %\"fG6#F+F,F+F+/F+&F+6#F/" }{TEXT 326 6 " . " }{TEXT -1 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 320 93 "Load Maple's \"residue\" procedure.\nMake sure this is done o nly ONCE during a Maple session.\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "readlib(residue):" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 260 1 "\n" }{TEXT 256 22 "Example 8.4, Page 311." }{TEXT 309 24 " Fi nd the residue of " }{XPPEDIT 18 0 "f(z)= pi*cot(pi*z)/z^2" "6#/-%\"f G6#%\"zG*(%#piG\"\"\"-%$cotG6#*&F)F*F'F*F**$F'\"\"#!\"\"" }{TEXT 261 6 " at " }{XPPEDIT 18 0 "z[0] = 0" "6#/&%\"zG6#\"\"!F'" }{TEXT 262 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "f:='f': z:='z':\nf := z \+ -> Pi*cot(Pi*z)/z^2:\n`f(z) ` = f(z);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 263 14 "The function " }{XPPEDIT 18 0 "f(z)" "6#-%\"fG6#%\"zG" }{TEXT 286 23 " has a pole of order " }{XPPEDIT 18 0 "3" "6#\"\"$" }{TEXT 285 6 " at " } {XPPEDIT 18 0 "z[0] = 0" "6#/&%\"zG6#\"\"!F'" }{TEXT 264 28 " . \nNex t, the residue at " }{XPPEDIT 18 0 "z[0] = 0" "6#/&%\"zG6#\"\"!F'" } {TEXT 265 35 " is determined using derivatives:\n" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 210 "`z^3 f(z) ` = z^3 * f(z);\nD1 := diff(z^3 * f(z), \+ z):\nD2 := factor(diff(z^3 * f(z), z$2)):\nL := 1/2! * limit(D2, z=0) :\nDiff(z^3*`f(z)`,z) = D1;\nDiff(z^3*`f(z)`,z$2) = D2;\nLimit(Diff(z^ 3*`f(z)`,z$2),z=0)/2! = L;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} }{EXCHG {PARA 0 "" 0 "" {TEXT 266 71 "We compare this with Maple's res idue procedure for computing residues.\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "`f(z) ` = f(z);\n`Res[f,0] ` = residue(f(z), z=0);" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 267 29 "Which is the coefficient of " }{XPPEDIT 18 0 "1/z" "6#* &\"\"\"F$%\"zG!\"\"" }{TEXT 287 39 " in the Laurent series expansion \+ for " }{XPPEDIT 18 0 "f(z)" "6#-%\"fG6#%\"zG" }{TEXT 288 3 " .\n" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "`f(z) ` = f(z);\n`f(z) ` = series(f (z), z=0, 7);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 256 22 "Example 8.5, Page 312." }{TEXT 310 31 " \+ Use residues to integrate " }{XPPEDIT 18 0 "int(1/(z^4 + z^3 - 2*z ^2), z=C..` `)" "6#-%$intG6$*&\"\"\"F',(*$%\"zG\"\"%F'*$F*\"\"$F'*& \"\"#F'*$F*F/F'!\"\"F1/F*;%\"CG%$~~~G" }{TEXT 268 11 " around " } {XPPEDIT 18 0 "C" "6#%\"CG" }{TEXT 290 2 ": " }{XPPEDIT 18 0 "abs(z) = 3" "6#/-%$absG6#%\"zG\"\"$" }{TEXT 289 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "f:='f': F:='F': z:='z':\nf := z -> 1/(z^4 + z^3 - 2*z ^2):\n`f(z) ` = f(z);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 269 27 "Find th e singularities of " }{XPPEDIT 18 0 "f(z) = 1/(z^4 + z^3 - 2*z^2)" "6 #/-%\"fG6#%\"zG*&\"\"\"F),(*$F'\"\"%F)*$F'\"\"$F)*&\"\"#F)*$F'F0F)!\" \"F2" }{TEXT 270 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 221 "Zn := \+ sort([solve(denom(f(z))=0, z)]):\n`For f(z) ` = f(z);\n`The singulari ties are:`;\nz1 := subs(z=Zn[1],z): z[1] = z1;\nz2 := subs(z=Zn[2],z): z[2] = z2;\nz3 := subs(z=Zn[3],z): z[3] = z3;\nz4 := subs(z=Zn[4],z): z[4] = z4;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 271 50 "Find out which singularities lie within a circ le " }{XPPEDIT 18 0 "abs(z) < 3" "6#2-%$absG6#%\"zG\"\"$" }{TEXT 291 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 255 "print(abs(z[1]) ,`< 3 \+ `, abs(z1)<3, evalb(evalf(abs(z1))<3));\nprint(abs(z[2]) ,`< 3 `, a bs(z2)<3, evalb(evalf(abs(z2))<3));\nprint(abs(z[3]) ,`< 3 `, abs(z3 )<3, evalb(evalf(abs(z3))<3));\nprint(abs(z[4]) ,`< 3 `, abs(z4)<3, \+ evalb(evalf(abs(z4))<3));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 272 75 "Rem ark. Sometimes Maple will form the list of values in a different orde r." }}{PARA 0 "" 0 "" {TEXT 295 79 "It is always necessary to visually inspect the above results before proceeding." }}{PARA 0 "" 0 "" {TEXT 296 26 "\nCompute the residues at " }{XPPEDIT 18 0 "z[1]" "6#&% \"zG6#\"\"\"" }{TEXT 273 4 " , " }{XPPEDIT 18 0 "z[2]" "6#&%\"zG6#\" \"#" }{TEXT 274 7 " and " }{XPPEDIT 18 0 "z[4]" "6#&%\"zG6#\"\"%" } {TEXT 275 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 146 "r1 := residue (f(z), z=z1): `Res[f`,z1,`] ` = r1;\nr2 := residue(f(z), z=z2): `Res[f `,z2,`] ` = r2;\nr4 := residue(f(z), z=z4): `Res[f`,z4,`] ` = r4;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 276 69 "The value of the integral is comp uted by using the residue calculus:\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "`f(z)` = f(z);\nval := 2*Pi*I*(r1 + r2 + r4):\nInt(f(z),z=C..``) = val;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 256 22 "Example 8.6, Page 312." }{TEXT 311 31 " Use resid ues to integrate " }{XPPEDIT 18 0 "int(1/(z^4 + 4), z=C..` `)" "6# -%$intG6$*&\"\"\"F',&*$%\"zG\"\"%F'F+F'!\"\"/F*;%\"CG%$~~~G" }{TEXT 277 11 " around " }{XPPEDIT 18 0 "C" "6#%\"CG" }{TEXT 292 2 ": " } {XPPEDIT 18 0 "abs(z-1) = 2" "6#/-%$absG6#,&%\"zG\"\"\"F)!\"\"\"\"#" } {TEXT 293 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "f:='f': F:='F' : z:='z':\nf := z -> 1/(z^4 + 4):\n`f(z) ` = f(z);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 278 27 "Find the singularities of " }{XPPEDIT 18 0 "f(z) = 1/(z^4 + 4)" "6#/-%\"fG6#% \"zG*&\"\"\"F),&*$F'\"\"%F)F,F)!\"\"" }{TEXT 279 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 221 "Zn := sort([solve(denom(f(z))=0, z)]):\n`For f(z) ` = f(z);\n`The singularities are:`;\nz1 := subs(z=Zn[1],z): z[ 1] = z1;\nz2 := subs(z=Zn[2],z): z[2] = z2;\nz3 := subs(z=Zn[3],z): z[ 3] = z3;\nz4 := subs(z=Zn[4],z): z[4] = z4;" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 280 50 "Find out which singularities lie within a circle " }{XPPEDIT 18 0 "abs(z-1) < 2" "6 #2-%$absG6#,&%\"zG\"\"\"F)!\"\"\"\"#" }{TEXT 294 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 271 "print(abs(z[1]-1),`< 2 `,abs(z1-1)<2, eval b(evalf(abs(z1-1))<2));\nprint(abs(z[2]-1),`< 2 `,abs(z2-1)<2, evalb (evalf(abs(z2-1))<2));\nprint(abs(z[3]-1),`< 2 `,abs(z3-1)<2, evalb( evalf(abs(z3-1))<2));\nprint(abs(z[4]-1),`< 2 `,abs(z4-1)<2, evalb(e valf(abs(z4-1))<2));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 297 75 "Remark. Sometimes Maple will for m the list of values in a different order." }}{PARA 0 "" 0 "" {TEXT 298 79 "It is always necessary to visually inspect the above results b efore proceeding." }}{PARA 0 "" 0 "" {TEXT 281 25 "\nCompute the resid ue at " }{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\"\"" }{TEXT 282 7 " and \+ " }{XPPEDIT 18 0 "z[2]" "6#&%\"zG6#\"\"#" }{TEXT 283 3 " .\n" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "r1 := residue(f(z), z=z1): `Res[f`, z1,`] ` = r1;\nr2 := residue(f(z), z=z2): `Res[f`,z2,`] ` = r2;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 284 69 "The value of the integral is comp uted by using the residue calculus:\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "val := 2*Pi*(r1 + r2):\nInt(f(z),z=C..``) = val;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 258 "" 0 "" {TEXT 256 22 "Example 8.8, Page 314." }{TEXT 312 41 " Find the partial fraction \+ expansion of " }{XPPEDIT 18 0 "f(z) = (3*z + 2)/(z*(z - 1)*(z - 2))" "6#/-%\"fG6#%\"zG*&,&*&\"\"$\"\"\"F'F,F,\"\"#F,F,*(F'F,,&F'F,F,!\"\"F, ,&F'F,F-F0F,F0" }{TEXT 306 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 645 "f:='f': z:='z':\nf := z -> (3*z + 2)/(z*(z - 1)*(z - 2)):\nZn := \+ sort([solve(denom(f(z))=0, z)]):\nRn := array(1..nops(Zn)):\nSn := arr ay(1..nops(Zn)):\nF := 0:\nfor i from 1 to nops(Zn) do \n if i=1 t hen p:=1 fi;\n if 10 then\n Z[p]:=Zn [i]; R[p]:=Rn[i]; S[p]:=Sn[i]; p:=p+1 fi;\nod:\n`f(z) ` = f(z);\nprint (`Singularities of f(z)`, Z);\n`f(z) ` = F;\n" }}}{EXCHG {PARA 258 "" 0 "" {TEXT 299 84 "Compare this with Maple's \"parfrac\" procedure f or the partial fraction expansion.\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "`f(z) ` = f(z);\n`f(z) ` = convert(f(z), parfrac, z);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 258 "" 0 "" {TEXT 256 22 "Example 8.9, Page 315." }{TEXT 313 42 " Find the partia l fraction expansion of " }{XPPEDIT 18 0 "f(z) = (z^2 + 3*z + 2)/(z^2 *(z - 1))" "6#/-%\"fG6#%\"zG*&,(*$F'\"\"#\"\"\"*&\"\"$F,F'F,F,F+F,F,*& F'F+,&F'F,F,!\"\"F,F1" }{TEXT 300 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 645 "f:='f': z:='z':\nf := z -> (z^2 + 3*z + 2)/(z^2 *(z \+ - 1)):\nZn := sort([solve(denom(f(z))=0, z)]):\nRn := array(1..nops(Zn )):\nSn := array(1..nops(Zn)):\nF := 0:\nfor i from 1 to nops(Zn) do \+ \n if i=1 then p:=1 fi;\n if 10 then \n Z[p]:=Zn[i]; R[p]:=Rn[i]; S[p]:=Sn[i]; p:=p+1 fi;\nod:\n`f(z) ` \+ = f(z);\nprint(`Singularities of f(z)`, Z);\n`f(z) ` = F;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 258 "" 0 "" {TEXT 301 84 "Compare this with Maple's \"parfrac\" procedure for the partial \+ fraction expansion.\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "`f(z) ` = \+ f(z);\n`f(z) ` = convert(f(z), parfrac, z);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 258 "" 0 "" {TEXT 256 27 "An Extra Exa mple, Page 314." }{TEXT 314 42 " Find the partial fraction expansion \+ of " }{XPPEDIT 18 0 "f(z) = 1/(z^4 - 1)" "6#/-%\"fG6#%\"zG*&\"\"\"F), &*$F'\"\"%F)F)!\"\"F-" }{TEXT 302 3 " .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 626 "f:='f': z:='z':\nf := z -> 1/(z^4 - 1):\nZn := sort( [solve(denom(f(z))=0, z)]):\nRn := array(1..nops(Zn)):\nSn := array(1. .nops(Zn)):\nF := 0:\nfor i from 1 to nops(Zn) do \n if i=1 then \+ p:=1 fi;\n if 10 then\n Z[p]:=Zn[i]; \+ R[p]:=Rn[i]; S[p]:=Sn[i]; p:=p+1 fi;\nod:\n`f(z) ` = f(z);\nprint(`Sin gularities of f(z)`, Z);\n`f(z) ` = F;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 258 "" 0 "" {TEXT 303 84 "Compare this w ith Maple's \"parfrac\" procedure for the partial fraction expansion .\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "`f(z) ` = f(z);\n`f(z) ` = c onvert(f(z), parfrac, z);" }}}{EXCHG {PARA 258 "" 0 "" {TEXT 304 64 "N otice that Maple does not expand the quadratic term involving " } {XPPEDIT 18 0 "z^2 + 1" "6#,&*$%\"zG\"\"#\"\"\"F'F'" }{TEXT 305 2 " . " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 319 19 "End of Section 8.2." }{TEXT -1 0 "" }}}}{MARK "0 0 0" 24 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }