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See ", ButtonBox["license.nb ", ButtonData:>{"license.nb", None}, ButtonStyle->"Hyperlink"], "for details." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Complex Numbers", "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "In ", StyleBox["Mathematica, ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`I = \@\(-1\)\)]], " is the positive square root of ", Cell[BoxData[ \(TraditionalForm\`\(-1\)\)]], ". The typeset version, ", Cell[BoxData[ \(TraditionalForm\`\[ImaginaryI]\)]], ", is obtained by typing \[EscapeKey]ii\[EscapeKey]. ", StyleBox["Mathematica", FontSlant->"Italic"], " provides the following commands for working with complex numbers." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ FormBox[GridBox[{ {\(Complex[x, \ y]\), RowBox[{"complex", " ", "number", " ", RowBox[{ FormBox[\(x\ + \ I\ y\), "TraditionalForm"], "."}]}]}, {\(Re[z]\), RowBox[{ "real", " ", "part", " ", "of", " ", "the", " ", "complex", " ", "number", " ", RowBox[{ FormBox["z", "TraditionalForm"], "."}]}]}, {\(Im[z]\), RowBox[{ "imaginary", " ", "part", " ", "of", " ", "the", " ", "complex", " ", "number", " ", RowBox[{ FormBox["z", "TraditionalForm"], "."}]}]}, {\(Conjugate[z]\), RowBox[{"complex", " ", "congugate", " ", "of", " ", RowBox[{ FormBox["z", "TraditionalForm"], ".", " ", FormBox[\(Conjugate[x\ + \ I\ y]\ \), "TraditionalForm"]}], RowBox[{ FormBox[\( = \ x\ - \ I\ y\), "TraditionalForm"], "."}]}]}, {\(Abs[z]\), RowBox[{"absolute", " ", "value", " ", "of", " ", RowBox[{ FormBox["z", "TraditionalForm"], ".", " ", FormBox[ RowBox[{\(Abs[x\ + \ I\ y]\), " ", FormBox[\( = \@\(x\^2 + y\^2\)\), "TraditionalForm"]}], "TraditionalForm"], "."}]}]}, {\(Arg[z]\), RowBox[{ RowBox[{"principal", " ", "argument", " ", "of", " ", RowBox[{ FormBox["z", "TraditionalForm"], ".", " ", \(-\[Pi]\)}]}], "<", \(Arg[z]\), "\[LessEqual]", "\[Pi]"}]} }, ColumnAlignments->{Left}], TraditionalForm]], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell["Complex[1,2]", "Input", AspectRatioFixed->True], Cell["Re[1+2I]", "Input", AspectRatioFixed->True], Cell[TextData["Im[1+2\[ImaginaryI]]"], "Input", AspectRatioFixed->True], Cell["Conjugate[1+2I]", "Input", AspectRatioFixed->True], Cell[TextData["Abs[1+2\[ImaginaryI]]"], "Input", AspectRatioFixed->True], Cell[TextData[{ Cell[BoxData[ FormBox[ StyleBox[\(Arg[z]\), "Input"], TraditionalForm]]], " returns an argument of ", Cell[BoxData[ \(TraditionalForm\`z\)]], " that is in the range ", Cell[BoxData[ \(TraditionalForm\`\(\((\(-\[Pi]\), \ \[Pi]\)]\)\)]], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell["Arg[1+2I]", "Input", AspectRatioFixed->True], Cell["Arg[-1]", "Input", AspectRatioFixed->True], Cell["Arg[-I]", "Input", AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell["Multi-valued Functions", "Section"], Cell[TextData[{ "Recall that the functions ", Cell[BoxData[ \(TraditionalForm\`log\ z\)]], " and ", Cell[BoxData[ \(TraditionalForm\`z\^\[Alpha]\)]], " are multivalued." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(TraditionalForm \`\(log\ z = \(log | z | \(+\[ImaginaryI]\)\ arg\ z = log | z | \(+\[ImaginaryI]\)\ Arg\ z + \[ImaginaryI]\ 2\ \[Pi]\ n\), \)\)], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell[BoxData[ \(TraditionalForm \`z\^\[Alpha] = \(\[ExponentialE]\^\(\[Alpha]\ log\ z\) . \)\)], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "When you type ", Cell[BoxData[ FormBox[ StyleBox[\(Log[z]\), "Input"], TraditionalForm]]], " or ", Cell[BoxData[ \(TraditionalForm\`z\^\[Alpha]\)]], " in ", StyleBox["Mathematica", FontSlant->"Italic"], ", you get a value whose argument is in the range ", Cell[BoxData[ \(TraditionalForm\`\(\((\(-\[Pi]\), \[Pi]\)]\)\)]], ". Thus you get the principal branch of ", Cell[BoxData[ \(TraditionalForm\`log\ z\)]], " and ", Cell[BoxData[ \(TraditionalForm\`z\^\[Alpha]\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Log[\(-1\)]\)], "Input"], Cell[BoxData[ \(Log[I]\)], "Input"], Cell[BoxData[ \(Log[\(-I\)]\)], "Input"], Cell[BoxData[ \(\@\(-1\)\)], "Input"], Cell[BoxData[ \(ComplexExpand[\@\(-1\)\%3]\)], "Input"], Cell[TextData[{ "If you want all the values of ", Cell[BoxData[ \(TraditionalForm\`1\^\(1/5\)\)]], ", you can use ", StyleBox["Solve[]", "Input"], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Solve[x\^5 == 1, x]\)], "Input"], Cell[BoxData[ \(x /. %\)], "Input"], Cell[BoxData[ \(ComplexExpand[%] // TableForm\)], "Input"], Cell[TextData[{ "The same method will not work if you want to find all the values of ", Cell[BoxData[ \(TraditionalForm\`log(1)\)]], ". " }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Solve[Exp[x] == 1, x]\)], "Input"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " informed you that since it had to use the logarithm, the inverse of the \ exponential, to solve the equation, it did not find all the solutions." }], "Text", TextAlignment->Left, TextJustification->1], Cell[CellGroupData[{ Cell["Exercise 1", "Subsubsection"], Cell[TextData[{ "Why does ", StyleBox["Mathematica", FontSlant->"Italic"], " give different results for the following two inputs? Are the inputs \ algebraically equivalent or not?" }], "Text", TextJustification->1], Cell[BoxData[{ \(\((\@x)\)\^2\), \(\@x\^2\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "(Hint: Evaluate the above expressions for ", Cell[BoxData[ \(TraditionalForm\`x = \(-1\)\)]], ".) Are the following two expressions equivalent?" }], "Text", TextJustification->1], Cell[BoxData[{ \(Exp[Log[x]]\), \(Log[Exp[x]]\)}], "Input"], Cell[TextData[{ "(Hint: Evaluate the above expressions for ", Cell[BoxData[ \(TraditionalForm\`x = 2 \[Pi]\ I\)]], ".)" }], "Text", TextJustification->1] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Contour Integrals", "Section"], Cell[CellGroupData[{ Cell["Parameterizing Integrals", "Subsection"], Cell[TextData[{ "If you specify complex-valued endpoint in a definite integral, ", StyleBox["Mathematica", FontSlant->"Italic"], " will evaluate the integral along the straight line joining the two \ endpoints." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Integrate[z\^2, {z, 0, I}]\)], "Input"], Cell["\<\ If there is a singularity on this line, the integral may not \ converge.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Integrate[1\/\(z - I\), {z, 0, 2 I}]\)], "Input"], Cell[TextData[{ "To evaluate the integral ", Cell[BoxData[ \(TraditionalForm \`\[Integral]\_C\ \(1\/\(sin(z)\)\) \[DifferentialD]z\)]], " where ", Cell[BoxData[ \(TraditionalForm\`C\)]], " is the unit circle centered at ", Cell[BoxData[ \(TraditionalForm\`z = 0\)]], ", we can parametrize the circle with ", Cell[BoxData[ \(TraditionalForm \`z = \[ExponentialE]\^\(\[ImaginaryI]\ \[Theta]\)\)]], ", ", Cell[BoxData[ \(TraditionalForm \`\[DifferentialD]z = \[ImaginaryI]\ \[ExponentialE]\^\(\[ImaginaryI]\ \[Theta]\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\[Theta] \[Element] \((0, 2 \[Pi])\)\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Integrate[ \(1\/Sin[Exp[I\ \[Theta]]]\) I\ Exp[I\ \[Theta]], {\[Theta], 0, 2 \[Pi]}]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Residues", "Subsection"], Cell[TextData[{ "If ", Cell[BoxData[ \(TraditionalForm\`f(z)\)]], " has the Laurent expansion," }], "Text"], Cell[BoxData[ \(TraditionalForm \`\(f(z) = \[Sum]\+\(n = \(-\[Infinity]\)\)\%\[Infinity]\ a\_n\ \((z - z\_0)\)\^n, \)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "then the residue of ", Cell[BoxData[ \(TraditionalForm\`f(z)\)]], " at ", Cell[BoxData[ \(TraditionalForm\`z = z\_0\)]], " is" }], "Text"], Cell[BoxData[ \(TraditionalForm \`Res(f(z), z\_0) = lim\+\(z \[Rule] z\_0\)\(\(( 1\/\(\((n - 1)\)!\)\ \[DifferentialD]\^\(n - 1\)\/\[DifferentialD]\ z\^\(n - 1\)\ \((\((z - z\_0)\)\^n\ \(f(z)\))\))\) . \)\)], "DisplayFormula", TextAlignment->Center], Cell["\<\ To refresh your memory we present two important results in contour \ integration.\ \>", "Text", TextJustification->1], Cell[TextData[{ StyleBox["Residue Theorem.", FontWeight->"Bold"], " Let ", Cell[BoxData[ \(TraditionalForm\`C\)]], " be a positively oriented, simple, closed contour. If ", Cell[BoxData[ \(TraditionalForm\`f(z)\)]], " is analytic in and on ", Cell[BoxData[ \(TraditionalForm\`C\)]], " except for isolated singularities at ", Cell[BoxData[ \(TraditionalForm\`z\_1, \ z\_2, \ \[Ellipsis], \ z\_m\)]], " inside ", Cell[BoxData[ \(TraditionalForm\`C\)]], " then" }], "Text", TextJustification->1], Cell[BoxData[ \(TraditionalForm \`\[Integral]\_C\( f(z)\) \[DifferentialD]z = 2 \[Pi]\ \[ImaginaryI]\ \(\[Sum]\+\(k = 1\)\%m\( Res(f(z), z\_k) . \)\)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "If in addition ", Cell[BoxData[ \(TraditionalForm\`f(z)\)]], " is analytic outside ", Cell[BoxData[ \(TraditionalForm\`C\)]], " in the finite complex plane then" }], "Text", TextJustification->1], Cell[BoxData[ \(TraditionalForm \`\[Integral]\_C\( f(z)\) \[DifferentialD]z = 2 \[Pi]\ \[ImaginaryI]\ \(Res(1\/z\^2\ \(f(1\/z)\), 0) . \)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "If ", Cell[BoxData[ \(TraditionalForm\`f(z)\)]], " has first order poles at ", Cell[BoxData[ \(TraditionalForm\`\[Zeta]\_1, \ \[Ellipsis], \ \[Zeta]\_n\)]], " on the contour and the contour does not have corners at these points then \ the principal value of the integral is" }], "Text", TextJustification->1], Cell[BoxData[ \(TraditionalForm \`\[Integral]\_C\( f(z)\) \[DifferentialD]z = 2 \[Pi]\ \[ImaginaryI]\ \(\[Sum]\+\(k = 1\)\%m Res(f(z), z\_k)\) + \[Pi]\ \[ImaginaryI]\ \(\[Sum]\+\(k = 1\)\%n\( Res(f(z), \[Zeta]\_k) . \)\)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ StyleBox["Fourier Integrals.", FontWeight->"Bold"], " Let ", Cell[BoxData[ \(TraditionalForm\`f(z)\)]], " be analytic except for isolated singularities, with only first order \ poles on the real axis. Suppose that ", Cell[BoxData[ \(TraditionalForm\`f(z)\)]], " vanishes as ", Cell[BoxData[ \(TraditionalForm\`\( | z | \) \[Rule] \[Infinity]\)]], ". If ", Cell[BoxData[ \(TraditionalForm\`\[Omega]\)]], " is a positive real number then" }], "Text", TextJustification->1], Cell[BoxData[ \(TraditionalForm \`\[Integral]\_\(-\[Infinity]\)\%\[Infinity]\( f(z)\)\ \[ExponentialE]\^\(\[ImaginaryI]\ \[Omega]\ x\)\ \[DifferentialD]x = 2 \[Pi]\ \[ImaginaryI]\ \(\[Sum]\+\(k = 1\)\%m Res( \(f(z)\)\ \[ExponentialE]\^\(\[ImaginaryI]\ \[Omega]\ z\), z\_k) \) + \[Pi]\ \[ImaginaryI]\ \(\[Sum]\+\(k = 1\)\%n \( Res(\(f(z)\)\ \[ExponentialE]\^\(\[ImaginaryI]\ \[Omega]\ z\), x\_k) . \)\)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "where ", Cell[BoxData[ \(TraditionalForm\`z\_1, \ \[Ellipsis], \ z\_m\)]], " are the singularities of ", Cell[BoxData[ \(TraditionalForm\`f(z)\)]], " in the upper half plane and ", Cell[BoxData[ \(TraditionalForm\`x\_1, \ \[Ellipsis], \ x\_n\)]], " are the first order poles on the real axis. If ", Cell[BoxData[ \(TraditionalForm\`\[Omega]\)]], " is a negative real number then" }], "Text", TextJustification->1], Cell[BoxData[ \(TraditionalForm \`\[Integral]\_\(-\[Infinity]\)\%\[Infinity]\( f(z)\)\ \[ExponentialE]\^\(\[ImaginaryI]\ \[Omega]\ x\)\ \[DifferentialD]x = \(-2\) \[Pi]\ \[ImaginaryI]\ \(\[Sum]\+\(k = 1\)\%m Res( \(f(z)\)\ \[ExponentialE]\^\(\[ImaginaryI]\ \[Omega]\ z\), z\_k) \) - \[Pi]\ \[ImaginaryI]\ \(\[Sum]\+\(k = 1\)\%n \( Res(\(f(z)\)\ \[ExponentialE]\^\(\[ImaginaryI]\ \[Omega]\ z\), x\_k) . \)\)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "where ", Cell[BoxData[ \(TraditionalForm\`z\_1, \ \[Ellipsis], \ z\_m\)]], " are the singularities of ", Cell[BoxData[ \(TraditionalForm\`f(z)\)]], " in the lower half plane and ", Cell[BoxData[ \(TraditionalForm\`x\_1, \ \[Ellipsis], \ x\_n\)]], " are the first order poles on the real axis." }], "Text", TextJustification->1], Cell[TextData[{ "Again consider the integral ", Cell[BoxData[ \(TraditionalForm \`\[Integral]\_C\ \(1\/\(sin(z)\)\) \[DifferentialD]z\)]], " where ", Cell[BoxData[ \(TraditionalForm\`C\)]], " is the unit circle centered at ", Cell[BoxData[ \(TraditionalForm\`z = 0\)]], ". We can evaluate this integral with the Residue theorem. The only \ singularity of the integrand occurs at ", Cell[BoxData[ \(TraditionalForm\`z = 0\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(2 \[Pi]\ I\ Residue[1\/Sin[z], {z, 0}]\)], "Input"], Cell[BoxData[ \(\(?Residue\)\)], "Input"], Cell[TextData[{ "The ", StyleBox["Residue[]", "Input"], " function will work for a pole of any finite order, but will choke if you \ give it an essential singularity." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Residue[Cos[x]/x^7, {x, 0}]\)], "Input"], Cell[BoxData[ \(Residue[Exp[1/x], {x, 0}]\)], "Input"], Cell[CellGroupData[{ Cell["Exercise 2", "Subsubsection"], Cell["Evaluate", "Text"], Cell[BoxData[ \(TraditionalForm \`1\/\(2 \[Pi]\ \[ImaginaryI]\)\ \(\[Integral]\_C\( sin \((z)\)\)\/\(z(z - 1)\)\ \[DifferentialD]z\)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "where ", Cell[BoxData[ \(TraditionalForm\`C\)]], " is the positively oriented circle of radius 2 centered at the origin." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Exercise 3", "Subsubsection"], Cell["Evaluate the integral,", "Text"], Cell[BoxData[ \(TraditionalForm \`\(\[Integral]\_C\ \(\(cot(z)\)\ \(coth(z)\)\)\/z\^3\ \[DifferentialD]z, \)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "where ", Cell[BoxData[ \(TraditionalForm\`C\)]], " is the positively oriented unit circle. Evaluate the reside with three \ methods:\na) Find the series expansion of the integrand.\nb) Use the \ residue formula, i.e. differentiate the integral and substitute ", Cell[BoxData[ \(TraditionalForm\`z = 0\)]], ".\nc) Use the ", StyleBox["Residue[]", FontWeight->"Bold"], " function." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Exercise 4", "Subsubsection"], Cell[TextData[{ "Evaluate the Fourier transform of ", Cell[BoxData[ \(TraditionalForm\`1/\((x\^2 + 1)\)\)]], " with the Residue Theorem. That is, evaluate" }], "Text", TextJustification->1], Cell[BoxData[ \(TraditionalForm \`1\/\(2 \[Pi]\)\ \(\[Integral]\_\(-\[Infinity]\)\%\[Infinity]\ \[ExponentialE]\^\(\(-\[ImaginaryI]\)\ \[Omega]\ x\)\/\(x\^2 + 1\)\ \(\[DifferentialD]x . \)\)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "Corroborate your result by evaluating the integral with the ", StyleBox["Integrate[]", FontWeight->"Bold"], " function." }], "Text", TextJustification->1] }, Open ]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Solutions", "Section", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Solution 1", "Subsubsection"], Cell[BoxData[ \(Clear[x]\)], "Input"], Cell["Mathematically, the two expressions,", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\((\@x)\)\^2\), \(\@x\^2\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "are not algebraically equivalent. Recall that the square root has two \ values. Both 1 and -1 are square roots of 1. The first expression has only \ one value. The second expression has two possible values, namely ", Cell[BoxData[ \(TraditionalForm\`x\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\(-x\)\)]], ". \n", StyleBox["Mathematica", FontSlant->"Italic"], " has a little bit different interpretation of the expressions. Instead of \ dealing with multiple-valued functions like the square root and the logarithm \ ", StyleBox["Mathematica", FontSlant->"Italic"], " always works with the principal branch of the function. So when ", Cell[BoxData[ \(TraditionalForm\`\@x\)]], " is input or output in ", StyleBox["Mathematica", FontSlant->"Italic"], ", the expression is the square root whose argument satisfies ", Cell[BoxData[ \(TraditionalForm\`\(-\[Pi]\)/2 < arg(\@x) \[LessEqual] \[Pi]/2\)]], ". Thus the second input cannot be simplified. Evaluating the expressions \ for ", Cell[BoxData[ \(TraditionalForm\`x = \(-1\)\)]], " illustrates why this is so.\n\t", Cell[BoxData[ \(TraditionalForm \`\((\@\(-1\))\)\^2 \[Congruent] \((I)\)\^2 \[Congruent] \(-1\)\)]], "\n\t", Cell[BoxData[ \(TraditionalForm \`\@\((\(-1\))\)\^2 \[Congruent] \@1 \[Congruent] 1\)]] }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[{ \(\((\@\(-1\))\)\^2\), \(\@\((\(-1\))\)\^2\)}], "Input", AspectRatioFixed->True], Cell["Consider the following two inputs.", "Text"], Cell[BoxData[{ \(Exp[Log[x]]\), \(Log[Exp[x]]\)}], "Input"], Cell[TextData[{ "First we consider the mathematical meaning of the expressions. The \ logarithm is a multivalued function,\n\t", Cell[BoxData[ \(TraditionalForm \`log(x) = Log( | x | ) + \[ImaginaryI]\ \(Arg(x)\) + \[ImaginaryI]\ 2\ \[Pi]\ n\)]], ", ", Cell[BoxData[ \(TraditionalForm\`n \[Element] \[DoubleStruckCapitalZ]\)]], ".\nThe expression ", Cell[BoxData[ \(TraditionalForm\`\[ExponentialE]\^\(log\ x\)\)]], " has only one value since ", Cell[BoxData[ \(TraditionalForm \`\[ExponentialE]\^\(\[ImaginaryI]\ 2\ \[Pi]\ n\) = 1\)]], ". Thus ", StyleBox["Mathematica", FontSlant->"Italic"], " simplifies this expression to ", Cell[BoxData[ \(TraditionalForm\`x\)]], ". The expression ", Cell[BoxData[ \(TraditionalForm\`log(\[ExponentialE]\^x)\)]], " has an infinite number of values,\n\t", Cell[BoxData[ \(TraditionalForm \`log(\[ExponentialE]\^x) = x + \[ImaginaryI]\ 2\ \[Pi]\ n\)]], ", ", Cell[BoxData[ \(TraditionalForm\`n \[Element] \[DoubleStruckCapitalZ]\)]], ".\n", StyleBox["Mathematica", FontSlant->"Italic"], " sees things a little differently. The principal branch of the logarithm \ satisfies ", Cell[BoxData[ \(TraditionalForm\`\(-\[Pi]\) < Log(z) \[LessEqual] \[Pi]\)]], ". So if the argument of ", Cell[BoxData[ \(TraditionalForm\`z\)]], " is not in this range, then ", Cell[BoxData[ \(TraditionalForm\`Log[Exp[z]]\)]], " does not have the same value as ", Cell[BoxData[ \(TraditionalForm\`z\)]], ". Thus ", StyleBox["Mathematica", FontSlant->"Italic"], " leaves the second input unchanged. A simple example helps to illustrate \ this." }], "Text", TextJustification->1], Cell[BoxData[{ \(Exp[Log[2 \[Pi]\ I]]\), \(Log[Exp[2 \[Pi]\ I]]\)}], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Solution 2", "Subsubsection"], Cell[TextData[{ "The integrand has a removable singularity at ", Cell[BoxData[ \(TraditionalForm\`z = 0\)]], ". We can verify this by checking that the residue vanishes or by looking \ at the Laurent series expansion about ", Cell[BoxData[ \(TraditionalForm\`z = 0\)]], " and noting that there are no negative powers of ", Cell[BoxData[ \(TraditionalForm\`z\)]], " in the series." }], "Text", TextJustification->1], Cell[BoxData[ \(Residue[Sin[z]\/\(z \((z - 1)\)\), {z, 0}]\)], "Input"], Cell[BoxData[ \(Series[Sin[z]\/\(z \((z - 1)\)\), {z, 0, 3}]\)], "Input"], Cell[TextData[{ "There is a first order pole at ", Cell[BoxData[ \(TraditionalForm\`z = 1\)]], ". From the residue theorem," }], "Text"], Cell[BoxData[ \(TraditionalForm \`1\/\(2 \[Pi]\ \[ImaginaryI]\)\ \(\[Integral]\_C\( sin \((z)\)\)\/\(z(z - 1)\)\ \[DifferentialD]z\) = \(Res(\(sin \((z)\)\)\/\(z(z - 1)\), 1) . \)\)], "DisplayFormula", TextAlignment->Center], Cell[BoxData[ \(Residue[Sin[z]\/\(z \((z - 1)\)\), {z, 1}]\)], "Input"], Cell[TextData[{ "The value of the integral is ", Cell[BoxData[ \(TraditionalForm\`sin(1)\)]], "." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Solution 3", "Subsubsection"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`cot(z)\)]], " has first order poles at ", Cell[BoxData[ \(TraditionalForm\`z = n\ \[Pi]\)]], ", ", Cell[BoxData[ \(TraditionalForm\`n \[Element] \[DoubleStruckCapitalZ]\)]], ". ", Cell[BoxData[ \(TraditionalForm\`coth(z)\)]], " has first order poles at ", Cell[BoxData[ \(TraditionalForm\`z = \[ImaginaryI]\ n\ \[Pi]\)]], ", ", Cell[BoxData[ \(TraditionalForm\`n \[Element] \[DoubleStruckCapitalZ]\)]], ". The only singularity inside the contour is the fifth order pole at ", Cell[BoxData[ \(TraditionalForm\`z = 0\)]], ". By the residue theorem," }], "Text", TextJustification->1], Cell[BoxData[ \(TraditionalForm \`\[Integral]\_C\ \(\(cot(z)\)\ \(coth(z)\)\)\/z\^3\ \[DifferentialD]z = \[ImaginaryI]\ 2\ \[Pi]\ \(Res(\(\(cot(z)\)\ \(coth(z)\)\)\/z\^3, 0) . \)\)], "DisplayFormula",\ TextAlignment->Center], Cell["\<\ We can find the residue by calculating the Laurent expansion of the \ function.\ \>", "Text"], Cell[BoxData[ \(Series[\(Cot[z]\ Coth[z]\)\/z\^3, {z, 0, 0}]\)], "Input"], Cell[TextData[{ "Since the residue is ", Cell[BoxData[ \(TraditionalForm\`\(-7\)/45\)]], ", the value of the integral is " }], "Text"], Cell[BoxData[ \(TraditionalForm \`\[Integral]\_C\ \(\(cot(z)\)\ \(coth(z)\)\)\/z\^3\ \[DifferentialD]z = \(-\[ImaginaryI]\)\ \[Pi]\ \(14\/45\ . \)\)], "DisplayFormula", TextAlignment->Center], Cell["We can also evaluate the residue with the residue formula,", "Text"], Cell[BoxData[ \(TraditionalForm \`Res(f(z), z\_0) = lim\+\(z \[Rule] z\_0\)\(\(( 1\/\(\((n - 1)\)!\)\ \[DifferentialD]\^\(n - 1\)\/\[DifferentialD]\ z\^\(n - 1\)\ \((\((z - z\_0)\)\^n\ \(f(z)\))\))\) . \)\)], "DisplayFormula", TextAlignment->Center], Cell[BoxData[ \(Limit[1\/\(4!\)\ D[z\^5\ \(Cot[z]\ Coth[z]\)\/z\^3, {z, 4}], z \[Rule] 0]\)], "Input"], Cell[TextData[{ "Finally we calculate the residue with the built-in ", StyleBox["Mathematica", FontSlant->"Italic"], " function." }], "Text"], Cell[BoxData[ \(Residue[\(Cot[z]\ Coth[z]\)\/z\^3, {z, 0}]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Solution 4", "Subsubsection"], Cell[TextData[{ "For ", Cell[BoxData[ \(TraditionalForm\`\[Omega] > 0\)]], " we close the path of integration in the lower half plane." }], "Text", TextJustification->1], Cell[BoxData[ \(\(-2\) \[Pi]\ I\ Residue[1\/\(2 \[Pi]\)\ Exp[\(-I\)\ \[Omega]\ x]\/\(x\^2 + 1\), {x, \(-I\)}]\)], "Input"], Cell[TextData[{ "For ", Cell[BoxData[ \(TraditionalForm\`\[Omega] < 0\)]], " we close the path of integration in the upper half plane." }], "Text", TextJustification->1], Cell[BoxData[ \(2 \[Pi]\ I\ Residue[1\/\(2 \[Pi]\)\ Exp[\(-I\)\ \[Omega]\ x]\/\(x\^2 + 1\), {x, I}] \)], "Input"], Cell[TextData[{ "Combining this results, we see that the Fourier transform is ", Cell[BoxData[ \(TraditionalForm\`\[ExponentialE]\^\(-\( | \[Omega] | \)\)/2\)]], ". 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