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"\tThe point ", Cell[BoxData[ \(\[Alpha]\)]], " is called a ", ButtonBox["Singular Point", ButtonData:>{ URL[ "http://mathworld.wolfram.com/SingularPointFunction.html"], None}, ButtonStyle->"Hyperlink"], ", or ", StyleBox["singularity", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " of the complex function f(z) if f is ", StyleBox["not", FontColor->RGBColor[1, 0, 1]], " analytic at ", Cell[BoxData[ \(z = \[Alpha]\)]], ", but every neighborhood ", Cell[BoxData[ \(\(D\_R\) \((\[Alpha])\)\ \ of\ \ \[Alpha]\)]], " contains at least one point at which f(z) is analytic. For example, \ the function ", Cell[BoxData[ \(f \((z)\) = 1\/\(1\ - \ z\)\)]], " is not analytic at ", Cell[BoxData[ \(z = 1\)]], ", but is analytic for all other values of z. Thus the point ", Cell[BoxData[ \(z = 1\)]], " is a singular point of f(z). As another example, consider ", Cell[BoxData[ \(g \((z)\)\ = \ Log \((z)\)\)]], ". We saw in ", ButtonBox["Section 5.2", ButtonData:>{"ca0502.nb", None}, ButtonStyle->"Hyperlink"], " that g is analytic for all z except at the origin and at all points \ on the negative real-axis. Thus, the origin and each point on the negative \ real axis is a singularity of ", Cell[BoxData[ \(g \((z)\)\ = \ Log \((z)\)\)]], ".\n\n\tThe point ", Cell[BoxData[ \(\[Alpha]\)]], " is called a ", ButtonBox["Isolated Singularity", ButtonData:>{ URL[ "http://mathworld.wolfram.com/IsolatedSingularity.html"], None}, ButtonStyle->"Hyperlink"], " of the complex function f(z) if f is ", StyleBox["not", FontColor->RGBColor[1, 0, 1]], " analytic at ", Cell[BoxData[ \(z = \[Alpha]\)]], ", but there exists a real number ", Cell[BoxData[ \(R > 0\)]], " such that f is analytic everywhere in the punctured disk ", Cell[BoxData[ \(\(D\_R\%*\) \((\[Alpha])\)\)]], ". Our function ", Cell[BoxData[ \(f \((z)\) = 1\/\(1\ - \ z\)\)]], " has an isolated singularity at ", Cell[BoxData[ \(z = 1\)]], ". However, for the function ", Cell[BoxData[ \(g \((z)\)\ = \ Log \((z)\)\)]], ", the singularity at ", Cell[BoxData[ \(z = 0\)]], " (or in general for any point ", Cell[BoxData[ \(z = \(-a\)\)]], " on the negative real-axis) is a non-isolated singularity for ", Cell[BoxData[ \(g \((z)\)\ = \ Log \((z)\)\)]], ". Functions with isolated singularities have a Laurent series, since the \ punctured disk ", Cell[BoxData[ \(\(D\_R\%*\) \((\[Alpha])\)\)]], " is the same as the annulus ", Cell[BoxData[ \(A \((r, R, \[Alpha])\)\)]], ". The logarithm function ", Cell[BoxData[ \(g \((z)\)\ = \ Log \((z)\)\)]], " does ", StyleBox["not", FontColor->RGBColor[1, 0, 1]], " have a ", ButtonBox["Laurent", ButtonData:>{ URL[ "http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Laurent_\ Pierre.html"], None}, ButtonStyle->"Hyperlink"], " series at any point ", Cell[BoxData[ \(z = \(-a\)\)]], " on the negative real-axis. We now look at this special case of \ Laurent's theorem in order to classify three types of isolated singularities. \ " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ StyleBox["\nDefinition 7.5 (", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ButtonBox["Removable Singularity", ButtonData:>{ URL[ "http://mathworld.wolfram.com/RemovableSingularity.html"], None}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ButtonBox["Pole of order k", ButtonData:>{ URL[ "http://mathworld.wolfram.com/Pole.html"], None}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ButtonBox["Essential Singularity", ButtonData:>{ URL[ "http://mathworld.wolfram.com/EssentialSingularity.html"], None}, ButtonStyle->"Hyperlink"], StyleBox["), Page 290.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Let f have an isolated singularity at ", Cell[BoxData[ \(\[Alpha]\)]], " with Laurent series expansion \n\n\t", Cell[BoxData[ \(f \((z)\) = \[Sum]\+\(n = \(-\[Infinity]\)\)\%\[Infinity]\( c\_n\) \((z - \[Alpha])\)\^n\)]], " valid for ", Cell[BoxData[ \(z \[Element] A \((0, R, \[Alpha])\)\)]], ". \n\nThen we distinguish the following types of singularities at ", Cell[BoxData[ \(\[Alpha]\)]], ". \n\n", StyleBox["(i)", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " If ", Cell[BoxData[ \(c\_n = \(0\ \ for\ \ n = \(-1\)\), \(-2\), \(-3\), ... \)]], ", then we say that f has a ", StyleBox["removable singularity", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " at ", Cell[BoxData[ \(\[Alpha]\)]], ". \n\n", StyleBox["(ii)", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " If k is a positive integer such that ", Cell[BoxData[ \(c\_\(-k\) \[NotEqual] 0\)]], " and ", Cell[BoxData[ \(c\_n = \(0\ \ for\ \ n = \(-k\) - 1\), \(-k\) - 2, \(-k\) - 3, ... \)]], ", then we say that f has a ", StyleBox["pole of order k", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " at ", Cell[BoxData[ \(\[Alpha]\)]], ". \n\n", StyleBox["(iii)", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " If ", Cell[BoxData[ \(c\_n \[NotEqual] 0\)]], " for infinitely many negative integers n, then we say that f has an \ ", StyleBox["essential singularity", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " at ", Cell[BoxData[ \(\[Alpha]\)]], ". " }], "Text"], Cell[TextData[{ "\n", StyleBox["Example of a Pole, Page 291.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " The following example will help this concept. Consider the function f(z) \ = cot(z). The leading term in the Laurent series expansion S(z) is ", Cell[BoxData[ \(1\/z\)]], " and S(z) goes to ", Cell[BoxData[ \(\[Infinity]\ \ as\ \ z \[Rule] 0\)]], " in the same manner as cot(z). " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution for a Pole, Page 291.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[z];\)\ \), "\[IndentingNewLine]", \(\(Clear[f, lim, S, z];\)\ \), "\n", \(\(f[z_]\ = \ Cot[z];\)\ \), "\n", \(\(S[z_]\ = \ Normal[Series[f[z], {z, 0, 13}]];\)\ \), "\[IndentingNewLine]", \(\(lim = Limit[f[z], z \[Rule] 0];\)\ \), "\n", \(\(Print["\", f[z]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", S[z]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", S[z], "\< - ...\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(lim\+\(z \[Rule] 0\)\) \>\"", f[z], "\< = \>", lim];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nThe Laurent series for f(z) involves only ", Cell[BoxData[ \(1\/z\)]], " for negative powers of z. 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"\n", StyleBox["Example of a Removable Singularity (i), Page ", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], StyleBox["291", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], StyleBox[".", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], StyleBox[" ", FontColor->RGBColor[1, 0, 0]], " The function ", Cell[BoxData[ \(f \((z)\) = \(sin \((z)\)\)\/z\)], AspectRatioFixed->True], " has a removable singularity at z = 0. 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55, ColorFunction \[Rule] Hue, DisplayFunction \[Rule] Identity];\)\ \), "\[IndentingNewLine]", \(\(Show[g1, g2, PlotRange \[Rule] {{\(-0.52\), 0.52}, {\(-0.52\), 0.52}}, AspectRatio \[Rule] 1, DisplayFunction \[Rule] $DisplayFunction];\)\ \), \ "\[IndentingNewLine]", \(\(Print["\", Im[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", v[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"] }, Closed]], Cell[TextData[{ StyleBox["\nDefinition 7.6 (Zero of order k), Page 292.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " A function f(z) analytic in ", Cell[BoxData[ \(\(D\_r\) \((\[Alpha])\)\)]], " is said to have ", StyleBox["a zero of order k", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " at the point ", Cell[BoxData[ \(z = \[Alpha]\)]], " if and only if ", Cell[BoxData[ \(\(f\^\((n)\)\) \((\[Alpha])\) = \(0\ \ for\ \ n = 0\), 1, 2, ... , k - 1\)]], ", and ", Cell[BoxData[ \(\(f\^\((k)\)\) \((\[Alpha])\) \[NotEqual] 0\)]], ".\n\nA zero of order one is sometimes called a ", StyleBox["simple zero", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ". We also speak about the ", ButtonBox["Roots", ButtonData:>{ URL[ "http://mathworld.wolfram.com/Root.html"], None}, ButtonStyle->"Hyperlink"], " of an equation." }], "Text"], Cell[TextData[{ StyleBox["\nTheorem 7.10, Page 292.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " A function ", Cell[BoxData[ \(f \((z)\)\)]], " analytic in ", Cell[BoxData[ \(\(D\_R\) \((\[Alpha])\)\)]], " has a zero of order k at the point ", Cell[BoxData[ \(\(\(\ \)\(z = \[Alpha]\)\)\)]], " if and only if its Taylor series given by ", Cell[BoxData[ \(f \((z)\) = \[Sum]\+\(n = 0\)\%\[Infinity]\( c\_n\) \((z - \[Alpha])\)\^n\)]], " has ", Cell[BoxData[ \(c\_0 = \(c\_1 = \ \(\(...\)\(=\)\(c\_\(k - 1\) = 0\ \ \ and\ \ \ c\_k \[NotEqual] 0\)\)\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[StyleBox["Proof of Theorem 7.10, see text Page 292.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ "\n", StyleBox["Example 7.10, Page 292.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Show that ", Cell[BoxData[ \(f \((z)\) = z\ sin \((z\^2)\)\)], AspectRatioFixed->True], " has a zero of order k = 3 at z = 0. " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 7.10.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ "Enter the function ", Cell[BoxData[ \(f[z] = z\ sin \((z\^2)\)\)]], " and find the first few terms of the Maclaurin series. " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[z];\)\ \), "\[IndentingNewLine]", \(\(Clear[f, i, L, m, p, S];\)\ \), "\n", \(\(f[z_]\ = \ z\ Sin[z\^2];\)\ \), "\n", \(\(S[z_]\ = \ Normal[Series[f[z], {z, 0, 23}]];\)\ \), "\n", \(\(L\ = \ CoefficientList[S[z], z];\)\ \), "\n", \(i = 1; \ \ m\ = \ Length[L];\ \), "\n", \(\(While[ L[\([i]\)] == 0\ && \ i < m, \n\ \ term\ = \ z\^i; \ \ i = i + 1];\)\ \ \), "\n", \(\(fact\ = \ Simplify[S[z]\/term];\)\ \), "\n", \(\(p[z_]\ = \ term\ fact;\)\ \), "\n", \(\(Print["\", f[z]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", S[z], "\< + ...\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", term\ , "\<(\>", fact, "\< + ...)\>"];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nThus, ", Cell[BoxData[ \(f \((z)\) = z\ sin \((z\^2)\)\)], AspectRatioFixed->True], " has a zero of order k = 3 at z = 0." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ Investigate the graph for real variables.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\n\(Plot[f[x], {x, \(-2\), 2}, PlotRange \[Rule] {{\(-2\), 2}, {\(-2\), 2}}, Ticks \[Rule] {Range[\(-2\), 2, 1], Range[\(-4\), 4, 1]}, PlotStyle \[Rule] Magenta];\)\ \ \), "\[IndentingNewLine]", \(\(Print["\", f[x]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nWe can use ", StyleBox["Mathematica", FontSlant->"Italic"], " to investigate the real and imaginary parts and the absolute value of ", Cell[BoxData[ \(f[z] = z\ sin \((z\^2)\)\)]], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(u[x_, y_] = FullSimplify[ ComplexExpand[Re[f[x + \[ImaginaryI]\ y]], TargetFunctions \[Rule] {Im, Re}]];\)\ \), "\n", \(\(v[x_, y_] = FullSimplify[ ComplexExpand[Im[f[x + \[ImaginaryI]\ y]], TargetFunctions \[Rule] {Im, Re}]];\)\ \), "\[IndentingNewLine]", \(\(w[x_, y_] = FullSimplify[ ComplexExpand[Abs[f[x + \[ImaginaryI]\ y]], TargetFunctions \[Rule] {Im, Re}]];\)\ \), "\[IndentingNewLine]", \(\(Plot3D[u[x, y], {x, \(-1\), 1}, {y, \(-1\), 1}, PlotPoints \[Rule] {21, 21}, \[IndentingNewLine]PlotRange \[Rule] {{\(-1\), 1}, {\(-1\), 1}, {\(-4\), 4}}, \[IndentingNewLine]AxesLabel \[Rule] {"\", "\", \ "\<\>"}, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Re[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", u[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(Plot3D[v[x, y], {x, \(-1\), 1}, {y, \(-1\), 1}, PlotPoints \[Rule] {21, 21}, \[IndentingNewLine]PlotRange \[Rule] {{\(-1\), 1}, {\(-1\), 1}, {\(-4\), 4}}, \[IndentingNewLine]AxesLabel \[Rule] {"\", "\", \ "\<\>"}, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Im[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", v[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(Plot3D[w[x, y], {x, \(-1\), 1}, {y, \(-1\), 1}, PlotPoints \[Rule] {21, 21}, \[IndentingNewLine]PlotRange \[Rule] {{\(-1\), 1}, {\(-1\), 1}, {0, 5}}, \[IndentingNewLine]AxesLabel \[Rule] {"\", \ "\", "\<\>"}, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Abs[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", w[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(ContourPlot[ Re[f[x\ + \ \[ImaginaryI]\ y]], {x, \(-1\), 1}, {y, \(-1\), 1}, \[IndentingNewLine]PlotPoints \[Rule] 35, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Re[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", u[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(ContourPlot[ Im[f[x\ + \ \[ImaginaryI]\ y]], {x, \(-1\), 1}, {y, \(-1\), 1}, \[IndentingNewLine]PlotPoints \[Rule] 35, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Im[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", v[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(ContourPlot[ Abs[f[x\ + \ \[ImaginaryI]\ y]], {x, \(-1\), 1}, {y, \(-1\), 1}, \[IndentingNewLine]PlotPoints \[Rule] 35, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Abs[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", w[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"] }, Closed]], Cell[TextData[{ StyleBox["\nTheorem 7.11, Page 292.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Suppose f(z) is analytic in ", Cell[BoxData[ \(\(D\_R\) \((\[Alpha])\)\)]], ". Then f(z) has a zero of order k at the point ", Cell[BoxData[ \(z = \[Alpha]\)]], " if and only if it can be expressed in the form ", Cell[BoxData[ \(f \((z)\) = \(\((z - \[Alpha])\)\^k\) g \((z)\)\)]], ", where g(z) is analytic at ", Cell[BoxData[ \(z = \[Alpha]\ \ and\ \ g \((\[Alpha])\) \[NotEqual] 0\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[StyleBox["Proof of Theorem 7.11, see text Page 292.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ StyleBox["\nCorollary 7.4, Page 293.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " If f(z) and g(z) are analytic at ", Cell[BoxData[ \(z = \[Alpha]\)]], ", and have zeros of orders m and n, respectively at ", Cell[BoxData[ \(z = \[Alpha]\)]], ", then their product ", Cell[BoxData[ \(h \((z)\) = f \((z)\) g \((z)\)\)]], " has a zero of order ", Cell[BoxData[ \(m + n\ \ at\ \ z = \[Alpha]\)]], ". \n" }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ "\n", StyleBox["Example 7.11, Page 293.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Show that ", Cell[BoxData[ \(f \((z)\) = z\^3\ sin \((z)\)\)], AspectRatioFixed->True], " has a zero of order k = 4 at z = 0. " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 7.11.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ "Enter the function ", Cell[BoxData[ \(f[z] = z\^3\ Sin[z]\)]], " and find the first few terms of the Maclaurin series. " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[z];\)\ \), "\[IndentingNewLine]", \(\(Clear[f, i, L, m, p, S];\)\ \), "\n", \(\(f[z_]\ = \ z\^3\ Sin[z];\)\ \), "\n", \(\(S[z_]\ = \ Normal[Series[f[z], {z, 0, 16}]];\)\ \), "\n", \(\(L\ = \ CoefficientList[S[z], z];\)\ \), "\n", \(i = 1; \ \ m\ = \ Length[L];\ \), "\n", \(\(While[ L\[LeftDoubleBracket]i\[RightDoubleBracket] == 0\ && \ i < m, \n\ \ \ term\ = \ z\^i; \ \ i = i + 1];\)\ \ \), "\n", \(\(fact\ = \ Simplify[S[z]\/term\ ];\)\ \), "\n", \(\(p[z_]\ = \ term\ fact;\)\ \), "\n", \(\(Print["\", f[z]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", S[z], "\< + ...\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", term\ , "\<(\>", fact, "\< + ...)\>"];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nThus, ", Cell[BoxData[ \(f \((z)\) = z\^3\ sin \((z)\)\)], AspectRatioFixed->True], " has a zero of order k = 4 at z = 0.", "\n", "\nInvestigate the graph for real variables." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\n\(Plot[f[x], {x, \(-2\), 2}, PlotRange \[Rule] {{\(-2\), 2}, {0, 4}}, Ticks \[Rule] {Range[\(-2\), 2, 1], Range[0, 4, 1]}, PlotStyle \[Rule] Magenta];\)\ \ \), "\[IndentingNewLine]", \(\(Print["\", f[x]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nWe can use ", StyleBox["Mathematica", FontSlant->"Italic"], " to investigate the real and imaginary parts and the absolute value of ", Cell[BoxData[ \(f \((z)\) = z\^3\ sin \((z)\)\)]], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(u[x_, y_] = FullSimplify[ ComplexExpand[Re[f[x + \[ImaginaryI]\ y]], TargetFunctions \[Rule] {Im, Re}]];\)\ \), "\n", \(\(v[x_, y_] = FullSimplify[ ComplexExpand[Im[f[x + \[ImaginaryI]\ y]], TargetFunctions \[Rule] {Im, Re}]];\)\ \), "\[IndentingNewLine]", \(\(w[x_, y_] = FullSimplify[ ComplexExpand[Abs[f[x + \[ImaginaryI]\ y]], TargetFunctions \[Rule] {Im, Re}]];\)\ \), "\[IndentingNewLine]", \(\(Plot3D[ Re[f[x\ + \ \[ImaginaryI]\ y]], {x, \(-1\), 1}, {y, \(-1\), 1}, PlotPoints \[Rule] {21, 21}, \[IndentingNewLine]PlotRange \[Rule] {{\(-1\), 1}, {\(-1\), 1}, {\(-4\), 1}}, \[IndentingNewLine]AxesLabel \[Rule] {"\", "\", \ "\<\>"}, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Re[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", u[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(Plot3D[ Im[f[x\ + \ \[ImaginaryI]\ y]], {x, \(-1\), 1}, {y, \(-1\), 1}, PlotPoints \[Rule] {21, 21}, \[IndentingNewLine]PlotRange \[Rule] {{\(-1\), 1}, {\(-1\), 1}, {\(-2\), 2}}, \[IndentingNewLine]AxesLabel \[Rule] {"\", "\", \ "\<\>"}, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Im[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", v[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(Plot3D[ Abs[f[x\ + \ \[ImaginaryI]\ y]], {x, \(-1\), 1}, {y, \(-1\), 1}, PlotPoints \[Rule] {21, 21}, \[IndentingNewLine]PlotRange \[Rule] {{\(-1\), 1}, {\(-1\), 1}, {0, 5}}, \[IndentingNewLine]AxesLabel \[Rule] {"\", \ "\", "\<\>"}, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Abs[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", w[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(ContourPlot[ Re[f[x\ + \ \[ImaginaryI]\ y]], {x, \(-1\), 1}, {y, \(-1\), 1}, \[IndentingNewLine]PlotPoints \[Rule] 35, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Re[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", u[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(ContourPlot[ Im[f[x\ + \ \[ImaginaryI]\ y]], {x, \(-1\), 1}, {y, \(-1\), 1}, \[IndentingNewLine]PlotPoints \[Rule] 35, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Im[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", v[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(ContourPlot[ Abs[f[x\ + \ \[ImaginaryI]\ y]], {x, \(-1\), 1}, {y, \(-1\), 1}, \[IndentingNewLine]PlotPoints \[Rule] 35, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Abs[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", w[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"] }, Closed]], Cell[TextData[{ StyleBox["\nTheorem 7.12, Page 294.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " A function f(z) analytic in the punctured disk ", Cell[BoxData[ \(\(D\_R\^*\) \((\[Alpha])\)\)]], " has a pole of order k at ", Cell[BoxData[ \(z = \[Alpha]\)]], " if and only if it can be expressed in the form \n\n\t\t\t", Cell[BoxData[ \(f \((z)\) = \(h \((z)\)\)\/\((z\ - \ \[Alpha])\)\^k\)]], ", \n\nwhere h(z) is analytic at ", Cell[BoxData[ \(z = \[Alpha]\ \ and\ \ h \((\[Alpha])\) \[NotEqual] 0\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[StyleBox["Proof of Theorem 7.12, see text Page 294.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell["\<\ \tCorollaries 7.5-7.8 are useful in determining the order of a zero or a \ pole. The proofs follow easily from Theorems 7.10 and 7.12 and are left as \ exercises.\ \>", "Text"], Cell[TextData[{ StyleBox["\nCorollary 7.5, Page 295.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontWeight->"Bold"], "If f(z) is analytic and has a zero of order k at the point ", Cell[BoxData[ \(z = \[Alpha]\)]], ", then ", Cell[BoxData[ \(g \((z)\) = 1\/\(f \((z)\)\)\)]], " has a pole of order k at ", Cell[BoxData[ \(z = \[Alpha]\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ StyleBox["\nCorollary 7.6, Page 295. ", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], "If f(z) has a pole of order k at the point ", Cell[BoxData[ \(\(\(\ \)\(z = \[Alpha]\)\)\)]], ", then ", Cell[BoxData[ \(g \((z)\) = 1\/\(f \((z)\)\)\)]], " has a removable singularity at ", Cell[BoxData[ \(z = \[Alpha]\)]], ". If we define ", Cell[BoxData[ \(g \((\[Alpha])\) = 0\)]], ", then g(z) has a zero of order k at ", Cell[BoxData[ \(z = \[Alpha]\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ StyleBox["\nCorollary 7.7, Page 295.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " If f(z) and g(z) have poles of orders m and n, respectively at \ the point ", Cell[BoxData[ \(z = \[Alpha]\)]], ", then their product ", Cell[BoxData[ \(h \((z)\) = f \((z)\) g \((z)\)\)]], " has a pole of order ", Cell[BoxData[ \(m + n\ \ at\ \ z = \[Alpha]\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ StyleBox["\nCorollary 7.8, Page 295.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Let f(z) and g(z) be analytic with zeros of orders m and n, \ respectively at ", Cell[BoxData[ \(z = \[Alpha]\)]], ". Then their quotient ", Cell[BoxData[ \(h \((z)\) = \(f \((z)\)\)\/\(g \((z)\)\)\)]], " has the following behavior:\n\n", StyleBox["(i)", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " If ", Cell[BoxData[ \(m > n\)]], ", then h(z) has a removable singularity at ", Cell[BoxData[ \(z = \[Alpha]\)]], ". If we define ", Cell[BoxData[ \(h \((\[Alpha])\) = 0\)]], ", then h(z) has a zero of order ", Cell[BoxData[ \(m - n\ \ at\ \ z = \[Alpha]\)]], ". \n\n", StyleBox["(ii)", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " If ", Cell[BoxData[ \(m < n\)]], ", then h(z) has a pole of order ", Cell[BoxData[ \(n - m\ \ at\ \ z = \[Alpha]\)]], ". \n\n", StyleBox["(iii)", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " If ", Cell[BoxData[ \(m = n\)]], ", then h(z) has a removable singularity at ", Cell[BoxData[ \(z = \[Alpha]\)]], ", and can be defined so that h(z) is analytic at ", Cell[BoxData[ \(z = \[Alpha]\)]], ", by ", Cell[BoxData[ \(h \((\[Alpha])\) = lim\+\(z\ \[Rule] \ \[Alpha]\)\ \ h \((z)\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ "\n", StyleBox["Example 7.12, Page 295.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Locate the zeros and poles of ", Cell[BoxData[ \(h \((z)\) = Tan[z]\/z\)], AspectRatioFixed->True], ", and determine their order. " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 7.12.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ "Enter the function ", Cell[BoxData[ \(h[z] = Tan[z]\/z\)]], " and find the first few terms of the Maclaurin series. " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[z];\)\ \), "\[IndentingNewLine]", \(\(Clear[h, i, S];\)\ \), "\n", \(\(h[z_]\ = \ Tan[z]\/z;\)\ \), "\n", \(\(S[z_]\ = \ Normal[Series[h[z], {z, 0, 13}]];\)\ \), "\n", \(\(Print["\", h[z]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", S[z]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", S[z], "\< + ...\>"];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nThus, ", Cell[BoxData[ \(h \((z)\) = Tan[z]\/z\)], AspectRatioFixed->True], " has a removable singularity at z = 0.\n\nNow look for the zeros of \ h[x]." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Print["\< h[z] = \>", h[z]];\)\ \), "\n", \(\(For[k = \(-5\), k \[LessEqual] \(-1\), \(k++\), \n\ \ \ Print["\", k, "\< \[Pi]] = \>", h[k\ \[Pi]]]\ ];\)\ \), "\[IndentingNewLine]", \(\(For[k = 1, k \[LessEqual] 5, \(k++\), \n\ \ \ Print["\", k, "\< \[Pi]] = \>", h[k\ \[Pi]]]\ ];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell["Or we could consider the following method of investigation.", "Text"], Cell[BoxData[{ \(\(Print["\", h[x]\ ];\)\ \), "\[IndentingNewLine]", \(\(Print["\", h[n\ \[Pi]]\ ];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Simplify[h[n\ \[Pi]], Element[n, Integers]]\ ];\)\ \)}], "Input"], Cell[TextData[{ "\nThus, ", Cell[BoxData[ \(h \((z)\) = Tan[z]\/z\)], AspectRatioFixed->True], " has simple zeros at ", Cell[BoxData[ \(z = \(n\ \[Pi]\ \ for\ \ n = \(\[PlusMinus]1\)\), \(\[PlusMinus]2\), \ ... \)]], ". \n\nNow consider the other singular points." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Print["\< h[z] = \>", h[z]];\)\ \), "\n", \(\(For[i = \(-4\), i <= 5, \(i++\), \[IndentingNewLine]Print["\", \(2 i - 1\)\/2, \ "\< \[Pi]] = \>", \[IndentingNewLine]Limit[h[z], z \[Rule] \(\((2 i - 1)\) \[Pi]\)\/2]\ ]\ ];\)\ \), "\ \[IndentingNewLine]", \(\)}], "Input"], Cell["Or we could consider the following method of investigation.", "Text"], Cell[BoxData[{ \(\(Print["\", h[x]\ ];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"", h[\((2 n + 1)\)\ \[Pi]\/2]\ ];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"", Simplify[h[\((2 n + 1)\)\ \[Pi]\/2], Element[n, Integers]]\ ];\)\ \)}], "Input"], Cell[TextData[{ "\nHence, ", Cell[BoxData[ \(h \((z)\) = Tan[z]\/z\)], AspectRatioFixed->True], " has simple poles at ", Cell[BoxData[ \(z = \(\((2 n - 1)\) \[Pi]\)\/2\)]], " where n is an integer." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ Investigate the graph for real variables.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\n\(Plot[h[x], {x, \(-3\) \[Pi], 3 \[Pi]}, PlotRange \[Rule] {{\(-3\) \[Pi], 3 \[Pi]}, {\(-10\), 10}}, Ticks \[Rule] {Range[\(\(-7\) \[Pi]\)\/2, \(7 \[Pi]\)\/2, \[Pi]], Range[\(-10\), 10, 5]}, PlotStyle \[Rule] Magenta];\)\ \ \), "\[IndentingNewLine]", \(\(Print["\", h[x]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nWe can use ", StyleBox["Mathematica", FontSlant->"Italic"], " to investigate the real part, imaginary part and absolute value of ", Cell[BoxData[ \(h \((z)\) = Tan[z]\/z\)]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(u[x_, y_] = FullSimplify[ ComplexExpand[Re[h[x + \[ImaginaryI]\ y]], TargetFunctions \[Rule] {Im, Re}]];\)\ \), "\[IndentingNewLine]", \(\(v[x_, y_] = FullSimplify[ ComplexExpand[Im[h[x + \[ImaginaryI]\ y]], TargetFunctions \[Rule] {Im, Re}]];\)\ \), "\[IndentingNewLine]", \(\(w[x_, y_] = FullSimplify[ ComplexExpand[Abs[h[x + \[ImaginaryI]\ y]], TargetFunctions \[Rule] {Im, Re}]];\)\ \), "\[IndentingNewLine]", \(\(Plot3D[ u[x, y], {x, \(-6\), 6}, {y, \(-6\), 6}, \[IndentingNewLine]PlotPoints \[Rule] {41, 41}, PlotRange \[Rule] {{\(-6\), 6}, {\(-6\), 6}, {0, 0.15}}, \[IndentingNewLine]ViewPoint \[Rule] {4, 4, 1}, ClipFill \[Rule] None, \[IndentingNewLine]AxesLabel \[Rule] {"\", "\", \ "\<\>"}, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", h[z], "\<] = \>", u[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(Plot3D[ u[x, y], {x, \(-6\), 6}, {y, \(-6\), 6}, \[IndentingNewLine]PlotPoints \[Rule] {41, 41}, PlotRange \[Rule] {{\(-6\), 6}, {\(-6\), 6}, {0, 2}}, \[IndentingNewLine]ViewPoint \[Rule] {4, 4, 1}, ClipFill \[Rule] None, \[IndentingNewLine]AxesLabel \[Rule] {"\", "\", \ "\<\>"}, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", h[z], "\<] = \>", u[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(Plot3D[ v[x, y], {x, \(-6\), 6}, {y, \(-6\), 6}, \[IndentingNewLine]PlotPoints \[Rule] {41, 41}, PlotRange \[Rule] {{\(-6\), 6}, {\(-6\), 6}, {\(-0.15\), 0.15}}, \[IndentingNewLine]ViewPoint \[Rule] {4, 2, 2}, ClipFill \[Rule] None, \[IndentingNewLine]AxesLabel \[Rule] {"\", "\", \ "\<\>"}, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", h[z], "\<] = \>", v[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(Plot3D[ v[x, y], {x, \(-6\), 6}, {y, \(-6\), 6}, \[IndentingNewLine]PlotPoints \[Rule] {41, 41}, PlotRange \[Rule] {{\(-6\), 6}, {\(-6\), 6}, {\(-1\), 1}}, \[IndentingNewLine]ViewPoint \[Rule] {4, 2, 2}, ClipFill \[Rule] None, \[IndentingNewLine]AxesLabel \[Rule] {"\", "\", \ "\<\>"}, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", h[z], "\<] = \>", v[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(Plot3D[ w[x, y], {x, \(-6\), 6}, {y, \(-6\), 6}, \[IndentingNewLine]PlotPoints \[Rule] {41, 41}, PlotRange \[Rule] {{\(-6\), 6}, {\(-6\), 6}, {0, 0.2}}, \[IndentingNewLine]ViewPoint \[Rule] {4, 4, 1}, ClipFill \[Rule] None, \[IndentingNewLine]AxesLabel \[Rule] {"\", "\", \ "\<\>"}, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", h[z], "\<] = \>", w[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(Plot3D[ w[x, y], {x, \(-6\), 6}, {y, \(-6\), 6}, \[IndentingNewLine]PlotPoints \[Rule] {41, 41}, PlotRange \[Rule] {{\(-6\), 6}, {\(-6\), 6}, {0, 2}}, \[IndentingNewLine]ViewPoint \[Rule] {4, 4, 1}, ClipFill \[Rule] None, \[IndentingNewLine]AxesLabel \[Rule] {"\", "\", \ "\<\>"}, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", h[z], "\<] = \>", w[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"] }, Closed]], Cell[TextData[{ "\n", StyleBox["Example 7.13, Page 296.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Locate the poles of ", Cell[BoxData[ \(g \((z)\) = 1\/\(5 z\^4\ + \ 26 z\^2\ + \ 5\)\)], AspectRatioFixed->True], ", and specify their order. " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 7.13.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ "Enter the function ", Cell[BoxData[ \(g[z] = 1\/\(5 z\^4\ + \ 26 z\^2\ + \ 5\)\)]], " and find roots of the polynomial in the denominator. " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[z, Z];\)\ \), "\[IndentingNewLine]", \(\(Clear[g, solset];\)\ \), "\n", \(\(g[z_]\ = \ 1\/\(5 z\^4\ + \ 26 z\^2\ + \ 5\);\)\ \), "\n", \(\(solset\ = \ Solve[Denominator[g[z]] == 0, z];\)\ \), "\n", \(\(Print["\", g[z]];\)\ \), "\n", \(\(Print["\"];\)\ \), "\n", \(\(Print[solset];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell["\<\ Evaluate g[z] at the poles. \ \>", "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[{ \(\[IndentingNewLine]\(Z\_1\ = \ solset\_\(\(\[LeftDoubleBracket]\)\(1, 1, \ 2\)\(\[RightDoubleBracket]\)\);\)\ \), "\n", \(\(Z\_2\ = \ solset\_\(\(\[LeftDoubleBracket]\)\(2, 1, \ 2\)\(\[RightDoubleBracket]\)\);\)\ \), "\n", \(\(Z\_3\ = \ solset\_\(\(\[LeftDoubleBracket]\)\(3, 1, \ 2\)\(\[RightDoubleBracket]\)\);\)\ \), "\n", \(\(Z\_4\ = \ solset\_\(\(\[LeftDoubleBracket]\)\(4, 1, \ 2\)\(\[RightDoubleBracket]\)\);\)\ \), "\n", \(\(Print["\", g[z]];\)\ \), "\n", \(\(Print["\", Z\_1, "\<] = \>", Limit[g[z], z \[Rule] Z\_1]];\)\ \), "\n", \(\(Print["\", Z\_2, "\<] = \>", Limit[g[z], z \[Rule] Z\_2]];\)\ \), "\n", \(\(Print["\", Z\_3, "\<] = \>", Limit[g[z], z \[Rule] Z\_3]];\)\ \), "\n", \(\(Print["\", Z\_4, "\<] = \>", Limit[g[z], z \[Rule] Z\_4]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nThus ", Cell[BoxData[ \(g \((z)\) = 1\/\(5 z\^4\ + \ 26 z\^2\ + \ 5\)\)], AspectRatioFixed->True], " has four simple poles.", "\n", "\nWe can use ", StyleBox["Mathematica", FontSlant->"Italic"], " to investigate the absolute value of ", Cell[BoxData[ \(g \((z)\) = 1\/\(5 z\^4\ + \ 26 z\^2\ + \ 5\)\)]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(u[x_, y_] = FullSimplify[ ComplexExpand[Re[g[x + \[ImaginaryI]\ y]], TargetFunctions \[Rule] {Im, Re}]];\)\ \), "\[IndentingNewLine]", \(\(v[x_, y_] = FullSimplify[ ComplexExpand[Im[g[x + \[ImaginaryI]\ y]], TargetFunctions \[Rule] {Im, Re}]];\)\ \), "\[IndentingNewLine]", \(\(w[x_, y_] = FullSimplify[ ComplexExpand[Abs[g[x + \[ImaginaryI]\ y]], TargetFunctions \[Rule] {Im, Re}]];\)\ \), "\[IndentingNewLine]", \(\(Plot3D[ u[x, y], {x, \(-3\), 3}, {y, \(-3\), 3}, \[IndentingNewLine]PlotPoints \[Rule] {41, 41}, PlotRange \[Rule] {{\(-3\), 3}, {\(-3\), 3}, {\(-0.05\), 0.4}}, \[IndentingNewLine]ClipFill \[Rule] None, AxesLabel \[Rule] {"\", "\", "\<\>"}, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Re[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", u[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(ContourPlot[ u[x, y], {x, \(-3\), 3}, {y, \(-3\), 3}, \[IndentingNewLine]PlotPoints \[Rule] 75, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Re[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", u[x, y]];\)\ \), 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\[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", v[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(ContourPlot[ v[x, y], {x, \(-3\), 3}, {y, \(-3\), 3}, \[IndentingNewLine]PlotPoints \[Rule] 75, ColorFunction \[Rule] Hue];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Im[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", v[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(Plot3D[v[x, y], {x, \(-3\), 3}, {y, \(-3\), 3}, PlotPoints \[Rule] 55, \[IndentingNewLine]PlotRange \[Rule] {{\(-3\), 3}, {\(-3\), 3}, {0, 0.4}}, BoxRatios \[Rule] {1, 1, 1}, Mesh \[Rule] False, AxesLabel \[Rule] {"\", "\", "\<\>"}, ViewPoint \[Rule] {\(-3\), 3, 2}, Boxed \[Rule] False, LightSources \[Rule] {{{4, 7, 20}, Red}, {{7, 10, 10}, Green}, {{\(-13\), \(-8\), 15}, Blue}}];\)\ \), "\n", \(\(Print["\", Im[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", 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PlotPoints \[Rule] 55, \[IndentingNewLine]PlotRange \[Rule] {{\(-3\), 3}, {\(-3\), 3}, {0, 0.4}}, BoxRatios \[Rule] {1, 1, 1}, Mesh \[Rule] False, AxesLabel \[Rule] {"\", "\", "\<\>"}, ViewPoint \[Rule] {\(-3\), 3, 2}, Boxed \[Rule] False, LightSources \[Rule] {{{4, 7, 20}, Red}, {{7, 10, 10}, Green}, {{\(-13\), \(-8\), 15}, Blue}}];\)\ \), "\n", \(\(Print["\", Abs[f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", w[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"] }, Closed]], Cell[TextData[{ "\n", StyleBox["Example 7.14, Page 296.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Locate the zeros and poles of ", Cell[BoxData[ \(g \((z)\) = \(\[Pi]\ cot \((\[Pi]\ z)\)\)\/z\^2\)], AspectRatioFixed->True], ", and determine their order." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 7.14.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ "Enter the function ", Cell[BoxData[ \(g[z] = \(\[Pi]\ Cot[\[Pi]\ z]\)\/z\^2\)]], " and find roots of the polynomial in the denominator. " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[z];\)\ \), "\[IndentingNewLine]", \(\(Clear[g, S];\)\ \), "\n", \(\(g[z_]\ = \ \(\[Pi]\ Cot[\[Pi]\ z]\)\/z\^2;\)\ \), "\n", \(\(S[z_]\ = \ Normal[Series[g[z], {z, 0, 9}]];\)\ \), "\n", \(\(Print["\", g[z]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\", S[z]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\", S[z], "\< + ...\>"];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nThus ", Cell[BoxData[ \(g \((z)\) = \(\[Pi]\ cot \((\[Pi]\ z)\)\)\/z\^2\)], AspectRatioFixed->True], ", has a pole of order n = 3 at z = 0.\n\nNow consider the other \ singularities." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Print["\", g[z]];\)\ \), "\[IndentingNewLine]", \(\(For[k = \(-5\), k \[LessEqual] \(-1\), \(k++\), \[IndentingNewLine]\(Print["\", k, "\<] = \>", Limit[g[z], z \[Rule] k]];\)\ \ ];\)\ \), "\[IndentingNewLine]", \(\(For[k = 1, k \[LessEqual] 5, \(k++\), \[IndentingNewLine]\(Print["\", k, "\<] = \>", Limit[g[z], z \[Rule] k]];\)\ \ ];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell["Or we could consider the following method of investigation.", "Text"], Cell[BoxData[{ \(\(Print["\", g[x]\ ];\)\ \), "\[IndentingNewLine]", \(\(Print["\", g[n]\ ];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Simplify[g[n], Element[n, Integers]]\ ];\)\ \)}], "Input"], Cell[TextData[{ "\nThus ", Cell[BoxData[ \(g \((z)\) = \(\[Pi]\ cot \((\[Pi]\ z)\)\)\/z\^2\)], AspectRatioFixed->True], " has simple poles at ", Cell[BoxData[ \(z = \(\[PlusMinus]1\), \(\[PlusMinus]2\), \ ... \)]], ". ", "\n", "\nInvestigate the graph for real variables." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\n\(Plot[g[x], {x, \(-3\), 3}, PlotRange \[Rule] {{\(-3\), 3}, {\(-10\), 10}}, Ticks \[Rule] {Range[\(-3\), 3, 1], Range[\(-10\), 10, 5]}, PlotStyle \[Rule] Magenta];\)\ \ \), "\[IndentingNewLine]", \(\(Print["\", g[x]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nWe can use ", StyleBox["Mathematica", FontSlant->"Italic"], " to investigate the real part, imaginary part, and absolute value of ", Cell[BoxData[ \(g \((z)\) = \(\[Pi]\ cot \((\[Pi]\ z)\)\)\/z\^2\)]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(u[x_, y_] = FullSimplify[ ComplexExpand[Re[g[x + \[ImaginaryI]\ y]], TargetFunctions \[Rule] {Im, Re}]];\)\ \), "\[IndentingNewLine]", \(\(v[x_, y_] = FullSimplify[ ComplexExpand[Im[g[x + \[ImaginaryI]\ y]], TargetFunctions \[Rule] {Im, Re}]];\)\ \), 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