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AspectRatioFixed->True, FontSize->18, CellTags->"Section 6.2"], Cell[TextData[{ "\tIn ", ButtonBox["Section 6.1", ButtonData:>{"ca0601.nb", None}, ButtonStyle->"Hyperlink"], " we learned how to evaluate integrals of the form ", Cell[BoxData[ \(\[Integral]\_a\%b\ f \((t)\) \[DifferentialD]t\)]], ", where f was complex-valued and ", Cell[BoxData[ \(\([a, b]\)\)]], " was an interval on the real axis (so that t was real, with ", Cell[BoxData[ \(t \[Element] \([a, b]\)\)]], "). In this section we shall define and evaluate integrals of the form ", Cell[BoxData[ \(\[Integral]\_C\ f \((z)\)\ \[DifferentialD]z\)]], ", where f is complex-valued and C is a contour in the plane (so that \ z is complex, with ", Cell[BoxData[ \(z \[Element] C\)]], "). Our main result is Theorem 6.1, which will show how to transform the \ latter type of integral into the kind we investigated in ", ButtonBox["Section 6.1", ButtonData:>{"ca0601.nb", None}, ButtonStyle->"Hyperlink"], ". \n\n\tWe will use concepts first introduced in ", ButtonBox["Section 1.6", ButtonData:>{"ca0106.nb", None}, ButtonStyle->"Hyperlink"], ", where we define the concept of a curve in the plane. Recall that to \ represent a curve C we used the parametric notation\n\n\t", Cell[BoxData[ \(C : \ z \((t)\) = \ x \((t)\) + \[ImaginaryI]\ y \((t)\)\)]], " for ", Cell[BoxData[ \(\(\(\ \)\(a \[LessEqual] t \[LessEqual] b\)\)\)]], ",\n\nwhere x(t) and y(t) are continuous functions. We now want to \ place a few more restrictions on the type of curve that we will be studying. \ The following discussion will lead to the concept of a contour, which is a \ type of curve that is adequate for the study of integration. \n\n\tRecall \ that C is said to be ", StyleBox["simple", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " if it does not cross itself, which is expressed by requiring that ", Cell[BoxData[ \(z \((t\_1)\) \[NotEqual] z \((t\_2)\)\)]], " whenever ", Cell[BoxData[ \(t\_1 \[NotEqual] t\_2\)]], ". A curve C with the property that ", Cell[BoxData[ \(z \((a)\) = z \((b)\)\)]], " is said to be a ", StyleBox["closed curve", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ". If ", Cell[BoxData[ \(z \((a)\) = z \((b)\)\)]], " is the only point of intersection, then we say that C is a ", StyleBox["simple closed curve", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ". As the parameter t increases from the value a to the value b, the \ point z(t) starts at the ", StyleBox["initial point", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " z(a), moves along the curve C, and ends up at the ", StyleBox["terminal point", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " z(b). If C is simple, then z(t) moves continuously from z(a) to \ z(b) as t increases, and the curve is given an ", StyleBox["orientation", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ", which we indicate by drawing arrows along the curve. \n\n\tThe \ complex-valued function z(t) is said to be ", StyleBox["differentiable", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " if both x(t) and y(t) are differentiable for ", Cell[BoxData[ \(\(\(\ \)\(a \[LessEqual] t \[LessEqual] b\)\)\)]], ". Here the one-sided derivatives of x(t) and y(t) are required to \ exist at the endpoints of the interval. The derivative ", Cell[BoxData[ \(z' \((t)\)\)]], " with respect to t is defined by the equation \n\n\t", Cell[BoxData[ \(\(z'\)[t] = \ \(x'\)[t] + \[ImaginaryI]\ \(y'\)[t]\)]], " for ", Cell[BoxData[ \(\(\(\ \)\(a \[LessEqual] t \[LessEqual] b\)\)\)]], ". \n\n\tThe curve C is said to be smooth if ", Cell[BoxData[ \(z' \((t)\)\)]], ", is continuous and nonzero on the interval. If C is a smooth curve, \ then C has a nonzero tangent vector at tach point z(t), which is given by \ the vector ", Cell[BoxData[ \(z' \((t)\)\)]], ". \n\t" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\tIf C is a smooth curve, then ds, the ", StyleBox["differential of arc length", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ", is given by ", Cell[BoxData[ \(\[DifferentialD]s = \(\@\(\([x' \((t)\)]\)\^2 + \([y' \((t)\)]\)\^2\)\ \) \[DifferentialD]t\)]], ". Since ", Cell[BoxData[ \(x' \((t)\)\)]], " and ", Cell[BoxData[ \(y' \((t)\)\)]], " are continuous functions, then so is the function ", Cell[BoxData[ \(\@\(\([x' \((t)\)]\)\^2 + \([y' \((t)\)]\)\^2\)\)]], ", and the ", ButtonBox["Length", ButtonData:>{ URL[ "http://mathworld.wolfram.com/ArcLength.html"], None}, ButtonStyle->"Hyperlink"], " L of the curve C is given by the definite integral \n\n\t", Cell[BoxData[ \(L \((C)\)\ = \(\[Integral]\_a\%b\ \(\@\(\([x' \((t)\)]\)\^2 + \([y' \ \((t)\)]\)\^2\)\) \[DifferentialD]t\ = \ \[Integral]\_a\%b\ \(\(|\)\(z' \ \((t)\)\)\(|\)\(\[DifferentialD]t\)\)\)\)]], ". \n\n\n\tNow consider C to be a curve with parameterization ", Cell[BoxData[ \(C : \ \(z\_1\) \((t)\) = \ x \((t)\) + \[ImaginaryI]\ y \((t)\)\ \ for\ \ a \[LessEqual] t \[LessEqual] b\)]], ", The ", StyleBox["opposite curve ", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " ", Cell[BoxData[ \(\(-C\)\)]], " traces out the same set of points in the plane, but in the reverse \ order, and it has the parameterization\n\n\t", Cell[BoxData[ \(\(-\(C : \ \(z\_2\) \((t)\)\)\) = \ x \((\(-t\))\) + \[ImaginaryI]\ y \((\(-t\))\)\ \ for\ \ - b \[LessEqual] t \[LessEqual] \(-a\)\)]], ".\n\nSince ", Cell[BoxData[ \(\(z\_2\) \((t)\) = \(z\_1\) \((\(-t\))\)\)]], ", it is easy to see that ", Cell[BoxData[ \(\(-C\)\)]], " is merely ", Cell[BoxData[ \(C\)]], " traversed in the opposite sense. \n\n\tA curve ", Cell[BoxData[ \(C\)]], " that is constructed by joining finitely many smooth curves end to end is \ called a ", StyleBox["contour", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ". Let ", Cell[BoxData[ \(C\_1, C\_2, ... , C\_n\)]], " denote n smooth curves such that the terminal point of ", Cell[BoxData[ \(C\_k\)]], " coincides with the initial point of ", Cell[BoxData[ \(C\_\(k + 1\)\)]], " for ", Cell[BoxData[ \(k = 1, 2, ... , n - 1\)]], ". Then the contour C is expressed by the equation\n\n\t", Cell[BoxData[ \(C = \(C\_1 + C\_2 + ... \) + C\_n\)]], ".\n\nA synonym for contour is ", StyleBox["path", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ". \n" }], "Text"], Cell[TextData[{ "\n", StyleBox["Example 6.5, Page 208.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find a parameterization of the polygonal path C from - 1 + i to 3 - \ i consisting of the three line segments:\nLine segment #1 from - 1 + i to \ - 1 ,\nLine segment #2 from - 1 to 1 + i ,\nLine segment #3 from 1 + i \ to 3 - i ." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 6.5.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell["Enter the four points and construct the three line segments.", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[t, x, y, z];\)\ \), "\[IndentingNewLine]", \(\(Clear[V1, V2, V3, Z1, Z2, Z3];\)\ \), "\n", \(\(z\_0\ = \ \(-\ 1\)\ + \ \[ImaginaryI];\)\ \), "\n", \(\(z\_1\ = \ \(-\ 1\);\)\ \), "\n", \(x\_0 = Re[z\_0]; \ y\_0 = Im[z\_0]; \ x\_1 = Re[z\_1]; \ y\_1 = Im[z\_1];\ \), "\n", \(\(Z1[t_]\ = \ x\_0\ + \ \((x\_1 - x\_0)\) t\ + \ \[ImaginaryI] \((y\_0\ + \ \((y\_1 - y\_0)\) t)\);\)\ \), "\n", \(\(z\_0\ = \ \(-\ 1\);\)\ \), "\n", \(\(z\_1\ = \ 1\ + \ \[ImaginaryI];\)\ \), "\n", \(x\_0 = Re[z\_0]; \ y\_0 = Im[z\_0]; \ x\_1 = Re[z\_1]; \ y\_1 = Im[z\_1];\ \), "\n", \(\(Z2[t_]\ = \ x\_0\ + \ \((x\_1 - x\_0)\) t\ + \ \[ImaginaryI] \((y\_0\ + \ \((y\_1 - y\_0)\) t)\);\)\ \), "\n", \(\(z\_0\ = \ 1\ + \ \[ImaginaryI];\)\ \), "\n", \(\(z\_1\ = \ 3\ - \ \[ImaginaryI];\)\ \), "\n", \(x\_0 = Re[z\_0]; \ y\_0 = Im[z\_0]; \ x\_1 = Re[z\_1]; \ y\_1 = Im[z\_1];\ \), "\n", \(\(Z3[t_]\ = \ x\_0\ + \ \((x\_1 - x\_0)\) t\ + \ \[ImaginaryI] \((y\_0\ + \ \((y\_1 - y\_0)\) t)\);\)\ \), "\n", \(\(Print[\*"\"\<\!\(Z\_1\)[t] = \>\"", Z1[t], "\<, for 0 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\n\ ", \(\(Print[\*"\"\<\!\(Z\_2\)[t] = \>\"", Z2[t], "\<, for 0 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\n\ ", \(\(Print[\*"\"\<\!\(Z\_3\)[t] = \>\"", Z3[t], "\<, for 0 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\ \[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(Z\_1\)[0] = \>\"", Z1[0]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(Z\_1\)[1] = \>\"", Z1[1]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(Z\_2\)[0] = \>\"", Z2[0]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(Z\_2\)[1] = \>\"", Z2[1]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(Z\_3\)[0] = \>\"", Z3[0]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(Z\_3\)[1] = \>\"", Z3[1]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nWe can use ", StyleBox["Mathematica", FontSlant->"Italic"], " to graph these line segments." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(V1[t_] = ComplexExpand[{Re[Z1[t]], Im[Z1[t]]}];\)\ \), "\n", \(\(V2[t_] = ComplexExpand[{Re[Z2[t]], Im[Z2[t]]}];\)\ \), "\n", \(\(V3[t_] = ComplexExpand[{Re[Z3[t]], Im[Z3[t]]}];\)\ \), "\n", \(\(ParametricPlot[Evaluate[{V1[t], V2[t], V3[t]}], {t, 0, 1}, Ticks \[Rule] {Range[\(-1\), 3, 1], Range[\(-1\), 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {Red, Blue, Green}];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"", Z1[t], "\<, for 0 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\n\ ", \(\(Print[\*"\"\< \!\(Z\_2\)[t] = \>\"", Z2[t], "\<, for 0 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\n\ ", \(\(Print[\*"\"\< \!\(Z\_3\)[t] = \>\"", Z3[t], "\<, for 0 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\ \[IndentingNewLine]", \(\)}], "Input"] }, Closed]], Cell[TextData[{ "\tWe are now ready to define the integral of a complex function along a \ contour C in the complex plane with initial point A and terminal point \ B. Our approach is to mimic what is done in calculus. We create a partition \ ", Cell[BoxData[ \(P\_n = {z\_0 = A, z\_1, z\_2, ... , z\_n = B}\)]], " of points that proceed along C from A to B and form the \ differences ", Cell[BoxData[ \(\[CapitalDelta]z\_k = \(z\_k - z\_\(k - 1\)\ \ for\ \ k = 1\), 2, ... , n\)]], ". Between each pair of partition points ", Cell[BoxData[ \(z\_\(k - 1\)\ \ and\ \ z\_k\)]], " we select a point ", Cell[BoxData[ \(c\_k\)]], " on C, where the function ", Cell[BoxData[ \(f \((c\_k)\)\)]], " is evaluated. These values are used to make a ", ButtonBox["Riemann Sum", ButtonData:>{ URL[ "http://mathworld.wolfram.com/RiemannSum.html"], None}, ButtonStyle->"Hyperlink"], " for the partition: \n\n\t", Cell[BoxData[ \(S \((P\_n)\)\ = \ \(\[Sum]\+\(k = 1\)\%n f \((c\_k)\) \((z\_k - z\_\(k - 1\))\)\ = \ \[Sum]\+\(k = 1\)\%n f \((c\_k)\) \[CapitalDelta]z\_k\)\)]], "\n\n\tAssume now that there exists a unique number L that is the limit \ of every sequence ", Cell[BoxData[ \({S \((P\_n)\)}\)]], " of ", ButtonBox["Riemann", ButtonData:>{ URL[ "http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Riemann.\ html"], None}, ButtonStyle->"Hyperlink"], " sums, where the maximum of ", Cell[BoxData[ \(\(\(|\)\(\[CapitalDelta]z\_k\)\(|\)\)\)]], " tends toward 0, for the sequence of partitions. We define the number \ L as the value of the ", StyleBox["integral of f(z) taken along the contour C", FontColor->RGBColor[0, 0, 1]], ". We thus have the following.\n" }], "Text"], Cell[TextData[{ StyleBox["Definition 6.2 (Contour Integral), Page 209.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Let C be a contour. Then \n\n\t\t\t", Cell[BoxData[ \(\[Integral]\_C\ f \((z)\)\ \[DifferentialD]z = \ lim\+\(n\ \[Rule] \ \[Infinity]\)\ \(\[Sum]\+\(k = 1\)\%n f \((c\_k)\) \[CapitalDelta]z\_k\)\)]], ", \n\nprovided that the limit exists in the sense previously discussed." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\tNotice that in definition 6.2 the value of the integral depends on the \ contour. In ", ButtonBox["Section 6.3", ButtonData:>{"ca0603.nb", None}, ButtonStyle->"Hyperlink"], " the ", ButtonBox["Cauchy", ButtonData:>{ URL[ "http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Cauchy.\ html"], None}, ButtonStyle->"Hyperlink"], StyleBox["-", FontSize->14, FontWeight->"Bold"], ButtonBox["Goursat", ButtonData:>{ URL[ "http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Goursat.\ html"], None}, ButtonStyle->"Hyperlink"], " theorem will establish the remarkable fact that if f is analytic, then ", Cell[BoxData[ \(\[Integral]\_C\ f \((z)\)\ \[DifferentialD]z\)]], " is ", StyleBox["independent", FontSlant->"Italic"], " of the contour." }], "Text"], Cell[TextData[{ "\n", StyleBox["Example 6.6, Page 209.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Use a Riemann sum to construct an approximation for the contour integral \ ", Cell[BoxData[ \(\[Integral]\_C\ \[ExponentialE]\^z\ \[DifferentialD]z\)]], " where C is a the line segment joining the point ", Cell[BoxData[ \(a\ = \(0\ \ \ to\ \ \ b = 2 + \(\[ImaginaryI]\ \[Pi]\)\/4\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 6.6.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ "Enter the function ", Cell[BoxData[ \(f[z] = \[ExponentialE]\^z\)], AspectRatioFixed->True], " and compute the approximating Riemann sum. \n", StyleBox["For illustration.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 1]], " We use n = 8 terms in the sum." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[a, b, c, z];\)\ \), "\n", \(\(Clear[dz, f, k, n, RiemannSum];\)\ \), "\n", \(\(f[z_]\ = \ \[ExponentialE]\^z;\)\ \), "\n", \(\(a\ = \ 0;\)\ \), "\n", \(\(b\ = \ 2\ + \(\(\ \)\(\[ImaginaryI]\ \[Pi]\)\)\/4;\)\ \), "\n", \(\(n\ = \ 8;\)\ \), "\n", \(\(dz\ = \ \(b - a\)\/n;\)\ \), "\n", \(\(z\_k_\ = \ Expand[a\ + \ k\ dz];\)\ \), "\n", \(\(c\_k_\ = \ Expand[\(z\_\(k - 1\)\ + \ z\_k\)\/2];\)\ \), "\n", \(\(RiemannSum\ = \[Sum]\+\(k = 1\)\%n f[c\_k] dz\ ;\)\ \), "\n", \(\(Print["\", f[z]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", n];\)\ \), "\n", \(\(Print[\*"\"\<{\!\(z\_k\)} = \>\"", Table[z\_k, {k, 1, n}]];\)\ \), "\n", \(\(Print[\*"\"\<{\!\(c\_k\)} = \>\"", Table[c\_k, {k, 1, n}]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Sum]\+\(k = 1\)\)\>\""\&n, \ \*"\"\\""];\)\ \), "\n", \(\(Print[RiemannSum];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Sum]\+\(k = 1\)\)\>\""\&n, \ \*"\"\\""];\)\ \), "\n", \(\(Print[\((\[Sum]\+\(k = 1\)\%n f[c\_k])\) dz];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Sum]\+\(k = 1\)\)\>\""\&n, \ \*"\"\\"", N[RiemannSum]];\)\ \)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ StyleBox["\nTheorem 6.1, Page 211.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Suppose f(z) is a continuous complex-valued function defined on a set \ containing the contour C. Let z(t) be any parameterization of C for ", Cell[BoxData[ \(a \[LessEqual] t \[LessEqual] b\)]], ". Then \n\n\t", Cell[BoxData[ \(\[Integral]\_C\ f \((z)\)\ \[DifferentialD]z = \[Integral]\_a\%b f \((z \((t)\))\)\ z' \((t)\) \[DifferentialD]t\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["Proof of Theorem 6.1, see text Page 211.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 1]], " We omit the proof of this theorem since it involves ideas (such as the \ theory of the Riemann-Stieltjes integral) that are beyond the scope of this \ book. A more rigorous development of the contour integral based on Riemann \ sums is found in advanced texts such as ", ButtonBox["Lars Valerian Ahlfors", ButtonData:>{ URL[ "http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Ahlfors.\ html"], None}, ButtonStyle->"Hyperlink"], ", ", StyleBox["Complex Analysis", FontSlant->"Italic"], ", 3rd ed. (New York: McGraw-Hill, 1979)." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n\tThere are two important facets of Theorem 6.1 that are worth \ mentioning. First, the theorem makes the problem of evaluating complex-valued \ functions along contours easy since it reduces our task to one that requires \ the evaluation of complex-valued functions over real intervals---a procedure \ we studied in ", ButtonBox["Section 6.1", ButtonData:>{"ca0601.nb", None}, ButtonStyle->"Hyperlink"], ". Second, according to the theorem this transformation yields the same \ answer regardless of the parametrization we choose for C." }], "Text"], Cell[TextData[{ "\n", StyleBox["Example 6.7, Page 212.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Give an exact the contour integral ", Cell[BoxData[ \(\[Integral]\_C\ \[ExponentialE]\^z\ \[DifferentialD]z\)]], " where C is a the line segment joining the point ", Cell[BoxData[ \(a\ = \(0\ \ \ to\ \ \ b = 2 + \(\[ImaginaryI]\ \[Pi]\)\/4\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 6.7.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ "Enter the function ", Cell[BoxData[ \(f[z] = \[ExponentialE]\^z\)], AspectRatioFixed->True], " and set up the parameterization for the computing a contour integral." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[a, b, t, u, v, z, Z];\)\ \), "\[IndentingNewLine]", \(\(Clear[f, int, val];\)\ \), "\n", \(\(a\ = \ 0;\)\ \), "\n", \(\(b\ = \ 2\ + \(\(\ \)\(\[ImaginaryI]\ \[Pi]\)\)\/4;\)\ \), "\n", \(\(z[t_]\ = \ \ a\ + \ \((b\ - \ a)\)\ t;\)\ \), "\n", \(\(F[Z_]\ = \ \[ExponentialE]\^Z;\)\ \), "\n", \(\(f[t_]\ = \ \[ExponentialE]\^z[t];\)\ \), "\n", \(\(u[t_]\ = \ ComplexExpand[Re[f[t]]];\)\ \), "\n", \(\(v[t_]\ = \ ComplexExpand[Im[f[t]]];\)\ \), "\n", \(\(Print["\< F[z] = \>", F[z]];\)\ \), "\n", \(\(Print["\< z[t] = \>", z[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\< F[z[t]] = \>", f[t]];\)\ \), "\n", \(\(Print["\< u[t] = \>", u[t]];\)\ \), "\n", \(\(Print["\< v[t] = \>", v[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\< z'[t] = \>", \(z'\)[t]];\)\ \), "\n", \(\(Print["\", f[t] \(z'\)[t]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Method (i).", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The first method uses a complex integrand f[z[t]] z'[t]." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[int, val];\)\ \), "\[IndentingNewLine]", \(\(int\ = \[Integral]f[t] \(z'\)[t] \[DifferentialD]t;\)\ \), "\n", \(\(val\ = \ \[Integral]\_0\%1 f[t] \(z'\)[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(V[t_] = ComplexExpand[{Re[z[t]], Im[z[t]]}];\)\ \), "\[IndentingNewLine]", \(\(Print["\", f[t] \(z'\)[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\< \[Integral]f[z[t]]z'[t]\[DifferentialD]t = \>", int];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%1\)f[z[t]]z'[t]\[DifferentialD]t = \ \>\"", val];\)\ \), "\[IndentingNewLine]", \(\(val = ComplexExpand[FullSimplify[val]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%1\)f[z[t]]z'[t]\[DifferentialD]t = \ \>\"", val];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%1\)f[z[t]]z'[t]\[DifferentialD]t = (\ \>\"", ComplexExpand[Re[val]], "\<) + \[ImaginaryI] (\>", ComplexExpand[Im[val]], "\<)\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%1\)f[z[t]]z'[t]\[DifferentialD]t = \ \>\"", N[val]];\)\ \), "\[IndentingNewLine]", \(\(ParametricPlot[Evaluate[V[t]], {t, 0, 1}, PlotRange \[Rule] {{0, 2}, {0, 1}}, AspectRatio \[Rule] 1\/2, Ticks \[Rule] {Range[\(-1\), 2, 1], Range[0, 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] Magenta];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_C\) \>\"", F[z], "\<\[DifferentialD]z = \>", ComplexExpand[FullSimplify[val]]];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_C\) \>\"", F[z], "\<\[DifferentialD]z = (\>", ComplexExpand[Re[val]], "\<) + \[ImaginaryI](\>", ComplexExpand[Im[val]], "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_C\) \>\"", F[z], "\<\[DifferentialD]z = \>", N[val]];\)\ \), "\[IndentingNewLine]", \(\(Print["\", z[t], "\<, for 0 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\ \[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\n", StyleBox["Method (ii).", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The second method uses a integrals of u[t] z'[t] and v[t] z'[t]." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[ImInt, ImVal, ReInt, ReVal, V, Val];\)\ \), "\[IndentingNewLine]", \(\(ReInt\ = \ \[Integral]u[t] \(z'\)[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(ImInt\ = \ \[Integral]v[t] \(z'\)[ t] \[DifferentialD]t;\)\ \), "\n", \(\(ReVal\ = \ \ \[Integral]\_0\%1 u[t] \(z'\)[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(ImVal\ = \ \ \[Integral]\_0\%1 v[t] \(z'\)[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(Val\ = \ \ ReVal\ + \ \[ImaginaryI]\ ImVal;\)\ \), "\ \[IndentingNewLine]", \(\(V[t_] = ComplexExpand[{Re[z[t]], Im[z[t]]}];\)\ \), "\[IndentingNewLine]", \(\(Print["\", u[t] \(z'\)[t]];\)\ \), "\n", \(\(Print["\", v[t] \(z'\)[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\<\[Integral]u[t]z'[t]\[DifferentialD]t + \[ImaginaryI] \ \[Integral]v[t]z'[t]\[DifferentialD]t\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", ReInt, "\< + \[ImaginaryI](\>", ImInt, "\<)\>"];\)\ \), "\n", \(\(Print["\< \>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%1\)u[t]z'[t]\[DifferentialD]t + \ \[ImaginaryI] \!\(\[Integral]\_0\%1\)v[t]z'[t]\[DifferentialD]t\>\""];\)\ \), \ "\[IndentingNewLine]", \(\(Print["\<= \>", ReVal, "\< + \[ImaginaryI](\>", ImVal, "\<)\>"];\)\ \), "\n", \(\(Print["\<= \>", ComplexExpand[Val]];\)\ \), "\n", \(\(Print["\<= \>", Apart[ComplexExpand[Val]]];\)\ \), "\[IndentingNewLine]", \(\(Val = ComplexExpand[FullSimplify[Val]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= \>", Val];\)\ \), "\[IndentingNewLine]", \(\(Print["\<= (\>", ComplexExpand[Re[Val]], "\<) + \[ImaginaryI] (\>", ComplexExpand[Im[Val]], "\<)\>"];\)\ \), "\n", \(\(Print["\<= \>", N[ComplexExpand[Val]]];\)\ \), "\[IndentingNewLine]", \(\(ParametricPlot[Evaluate[V[t]], {t, 0, 1}, PlotRange \[Rule] {{0, 2}, {0, 1}}, AspectRatio \[Rule] 1\/2, Ticks \[Rule] {Range[\(-1\), 2, 1], Range[0, 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] Magenta];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_C\) \>\"", F[z], "\<\[DifferentialD]z = \>", ComplexExpand[FullSimplify[val]]];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_C\) \>\"", F[z], "\<\[DifferentialD]z = (\>", ComplexExpand[Re[val]], "\<) + \[ImaginaryI](\>", ComplexExpand[Im[val]], "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_C\) \>\"", F[z], "\<\[DifferentialD]z = \>", N[val]];\)\ \), "\[IndentingNewLine]", \(\(Print["\", z[t], "\<, for 0 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\ \[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n", StyleBox["Example 6.8, Page 212.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Evaluate the contour integral ", Cell[BoxData[ \(\[Integral]\_C\ 1\/\(z\ - \ 2\)\ \[DifferentialD]z\)]], " where C is a the upper semicircle with radius 1 centered at ", Cell[BoxData[ \(x = 2\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 6.8.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ "Enter the function ", Cell[BoxData[ \(f[z] = 1\/\(z\ - \ 2\)\)], AspectRatioFixed->True], " and set up the parameterization and compute the contour integral." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[t, z];\)\ \), "\[IndentingNewLine]", \(\(Clear[f, int, val, V];\)\ \), "\n", \(\(f[z_]\ = \ 1\/\(z\ - \ 2\);\)\ \), "\n", \(\(z[t_]\ = \ 2\ + \ \[ExponentialE]\^\(\[ImaginaryI]\ t\);\)\ \), "\n", \(\(int\ = \ \[Integral]f[z[t]] \(z'\)[ t] \[DifferentialD]t;\)\ \), "\n", \(\(val\ = \ \[Integral]\_0\%\[Pi] f[z[t]] \(z'\)[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(V[t_] = ComplexExpand[{Re[z[t]], Im[z[t]]}];\)\ \), "\n", \(\(Print["\< f[z] = \>", f[z]];\)\ \), "\n", \(\(Print["\< z[t] = \>", z[t]];\)\ \), "\n", \(\(Print["\< f[z[t]] = \>", f[z[t]]];\)\ \), "\n", \(\(Print["\< z'[t] = \>", \(z'\)[ t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\< f[z[t]]z'[t] = \>", f[z[t]], \(z'\)[t]];\)\ \), "\n", \(\(Print["\< f[z[t]]z'[t] = \>", f[z[t]] \(z'\)[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\< \[Integral]f[z[t]]z'[t]\[DifferentialD]t = \>", int];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\< \!\(\[Integral]\_0\%\[Pi]\)f[z[t]]z'[t]\ \[DifferentialD]t = \>\"", val];\)\ \), "\n", \(\(ParametricPlot[Evaluate[V[t]], {t, 0, \[Pi]}, PlotRange \[Rule] {{0, 3}, {0, 1}}, AspectRatio \[Rule] 1\/3, Ticks \[Rule] {Range[\(-1\), 3, 1], Range[0, 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] Magenta];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_C\) \>\"", f[z], "\<\[DifferentialD]z = \>", val];\)\ \), "\[IndentingNewLine]", \(\(Print["\", z[t], "\<, for 0 \[LessEqual] t \[LessEqual] \[Pi].\>"];\)\ \), "\ \[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n\tSuppose ", Cell[BoxData[ \(f \((z)\) = u \((z)\) + \[ImaginaryI]\ v \((t)\)\)]], " , and ", Cell[BoxData[ \(z \((t)\) = x \((t)\) + \[ImaginaryI]\ y \((t)\)\)]], " is a parametrization for the contour C. Then\n\n\t\t", Cell[BoxData[ \(\[Integral]\_C\ f \((z)\)\ \[DifferentialD]z = \[Integral]\_a\%b f \((z \((t)\))\) z' \((t)\) \[DifferentialD]t\)]], " \n\n\t\t", Cell[BoxData[ \(\(\(=\)\(\[Integral]\_a\%b\((u \((z \((t)\))\) + \[ImaginaryI]\ v \ \((z \((t)\))\))\) \((x' \((t)\) + \[ImaginaryI]\ y' \((t)\))\) \ \[DifferentialD]t\)\)\)]], " \n\n\t\t", Cell[BoxData[ \(\(\(=\)\(\[Integral]\_a\%b\([u \((z \((t)\))\)\ x' \((t)\) - v \((z \((t)\))\)\ y' \((t)\)]\) \(\(\[DifferentialD]t\)\(\ \ \)\)\)\)\)]], " \n\n\t\t", Cell[BoxData[ \(\(\(\ \ \ \ \ \)\(\(+\ \[ImaginaryI]\)\ \(\[Integral]\_a\%b\([v \((z \ \((t)\))\) x' \((t)\) + u \((z \((t)\))\) y' \((t)\)]\) \[DifferentialD]t\)\)\)\)]], " \n\t\t\n\t\t", Cell[BoxData[ \(\(\(=\)\(\[Integral]\_a\%b\((u\ x' - v\ \ y')\) \[DifferentialD]t\ + \ \[ImaginaryI]\ \(\ \[Integral]\_a\%b\((v\ x' + u\ y')\) \[DifferentialD]t\)\)\)\)]], " \n\nwhere we are equating u with u(z(t)), x' with x'(t), and so on. " }], "Text"], Cell[TextData[{ "\n\tIf we use the differentials, then the above equation can be written in \ terms of line integrals of the real-valued functions u and v, giving \n\n\t\t\ \t", Cell[BoxData[ \(\(\(\[Integral]\_C\ f \((z)\)\ \[DifferentialD]z\)\(=\)\)\)]], Cell[BoxData[ \(\(\(=\)\(\[Integral]\_a\%b u \[DifferentialD]x\ - v\ \[DifferentialD]y\ + \ \[ImaginaryI]\ \(\[Integral]\_a\%b v\ \[DifferentialD]x\)\ + u\ \[DifferentialD]y\)\)\)]], ",\n\nwhich is easy to remember if we recall that symbolically\n\n\t\t\t", Cell[BoxData[ \(f \((z)\)\ \[DifferentialD]z\ = \ \((u + \[ImaginaryI]\ v)\) \((\ \[DifferentialD]x + \[ImaginaryI] \[DifferentialD]y)\)\)]], ". " }], "Text"], Cell["\<\ \tWe emphasize that Equation (6-16) is merely a notational device \ for applying Theorem 6.1. You should carefully apply Theorem 6.1as \ illustrated in Examples 6.7 and 6.8 before using any shortcuts suggested by \ the latter.\ \>", "Text"], Cell[TextData[{ "\n", StyleBox["Example 6.9, Page 214.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Evaluate the contour integrals of ", Cell[BoxData[ \(f \((z)\) = z\)]], " starting at the points ", Cell[BoxData[ \(z\_1 = \(\(-1\) - i\ \ and\ \ z\_2 = 3 + i\)\)]], ". \n", StyleBox["(a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Use the line segment joining the points. ", StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Use a portion of a parabola joining the points." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 6.9 (a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ StyleBox["\n(a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Use the straight line segment connecting the points ", Cell[BoxData[ \(z\_1 = \(\(-1\) - i\ \ and\ \ z\_2 = 3 + i\)\)]], ", set up the parameterization and compute the contour integral. Enter \ the function ", Cell[BoxData[ \(f[z] = z\)], AspectRatioFixed->True], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[t, u, U, v, V, w, W, z];\)\ \), "\[IndentingNewLine]", \(\(Clear[f, U1, U2, V1, V2, W1, W2, z1, z2];\)\ \), "\n", \(\(f[z_]\ = \ z\ ;\)\ \), "\[IndentingNewLine]", \(\(z1[t_]\ = \ 2 t\ + \ 1\ + \ \[ImaginaryI]\ t;\)\ \), "\n", \(\(w[t_]\ = \ ComplexExpand[f[z1[t]]\ \(z1'\)[t]];\)\ \), "\n", \(\(u[t_]\ = \ ComplexExpand[Re[w[t]]];\)\ \), "\n", \(\(v[t_]\ = \ ComplexExpand[Im[w[t]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\< f[z] = \>", f[z]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_1\)[t] = \>\"", z1[t]];\)\ \), "\n", \(\(Print[\*"\"\< f[\!\(z\_1\)[t]] = \>\"", f[z1[t]]];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_1\)'[t] = \>\"", \(z1'\)[ t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\\"", w[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\< u[t] = \>", u[t]];\)\ \), "\n", \(\(Print["\< v[t] = \>", v[t]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Method (i).", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The integral can be computed using the real and imaginary parts." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(U1\ = \[Integral]\_\(-1\)\%1 u[ t] \[DifferentialD]t;\)\ \), "\n", \(\(V1\ = \[Integral]\_\(-1\)\%1 v[t] \[DifferentialD]t;\)\ \), "\n", \(\(W1\ = \ U1\ + \ \[ImaginaryI]\ V1;\)\ \), "\n", \(\(U[t_]\ = \ \[Integral]u[t] \[DifferentialD]t;\)\ \), "\n", \(\(V[ t_]\ = \ \[Integral]v[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(reint\ = \ \[Integral]u[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(imint\ = \ \[Integral]v[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(P1[t_] = ComplexExpand[{Re[z1[t]], Im[z1[t]]}];\)\ \), "\[IndentingNewLine]", \(\(Print["\", u[t]];\)\ \), "\n", \(\(Print["\", v[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\[Integral]u[t]\[DifferentialD]t + \[ImaginaryI] \ \[Integral]v[t]\[DifferentialD]t = \>", reint, "\< + \[ImaginaryI] (\>", imint, "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\", U[t], "\<, V[t] = \>", V[t]];\)\ \), "\n", \(\(Print["\", U[1], "\<, V[1] = \>", V[1]];\)\ \), "\n", \(\(Print["\", U[\(-1\)], "\<, V[-1] = \>", V[\(-1\)]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%1\)u[t]\[DifferentialD]t + \ \[ImaginaryI]\!\(\[Integral]\_\(-1\)\%1\)v[t]\[DifferentialD]t = U[1]-U[-1] + \ \[ImaginaryI](V[1]-V[-1])\>\""];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%1\)u[t]\[DifferentialD]t + \ \[ImaginaryI]\!\(\[Integral]\_\(-1\)\%1\)v[t]\[DifferentialD]t = \>\"", U[1] - U[\(-1\)], "\< + \[ImaginaryI] (\>", V[1] - V[\(-1\)], "\<)\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%1\)u[t]\[DifferentialD]t + \ \[ImaginaryI] \!\(\[Integral]\_\(-1\)\%1\)v[t]\[DifferentialD]t = \>\"", W1];\)\ \), "\n", \(\(ParametricPlot[Evaluate[P1[t]], {t, \(-1\), 1}, PlotRange \[Rule] {{\(-1\), 3}, {\(-1\), 1}}, AspectRatio \[Rule] 2\/3, Ticks \[Rule] {Range[\(-1\), 3, 1], Range[\(-1\), 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] Magenta];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_C\) \>\"", f[z], "\<\[DifferentialD]z = \>", W1];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"", z1[t], "\<, for -1 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\ \[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Method (ii).", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The integral can be computed using a complex integrand." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(W[ t_]\ = \[Integral]w[t] \[DifferentialD]t;\)\ \), "\n", \(\(W1\ = \[Integral]\_\(-1\)\%1 w[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(P1[t_] = ComplexExpand[{Re[z1[t]], Im[z1[t]]}];\)\ \), "\[IndentingNewLine]", \(\(Print["\", w[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", ComplexExpand[W[t]]];\)\ \), "\n", \(\(Print["\", W[1], "\<, W[-1] = \>", W[\(-1\)]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%1\)w[t]\[DifferentialD]t = \ W[1]-W[-1] = \>\"", W[1], "\< - (\>", W[\(-1\)], "\<)\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%1\)w[t]\[DifferentialD]t = \ \>\"", W1];\)\ \), "\[IndentingNewLine]", \(\(ParametricPlot[Evaluate[P1[t]], {t, \(-1\), 1}, PlotRange \[Rule] {{\(-1\), 3}, {\(-1\), 1}}, AspectRatio \[Rule] 2\/3, Ticks \[Rule] {Range[\(-1\), 3, 1], Range[\(-1\), 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] Magenta];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_C\) \>\"", f[z], "\<\[DifferentialD]z = \>", W1];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"", z1[t], "\<, for -1 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\ \[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 6.9 (b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Use the portion of a parabola connecting the points ", Cell[BoxData[ \(z\_1 = \(\(-1\) - i\ \ and\ \ z\_2 = 3 + i\)\)]], ", set up the parameterization and compute the contour integral." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(z2[t_]\ = \ t\^2\ + \ 2 t\ + \ \[ImaginaryI]\ t;\)\ \), "\n", \(\(w[t_]\ = \ ComplexExpand[f[z2[t]]\ \(z2'\)[t]];\)\ \), "\n", \(\(u[t_]\ = \ ComplexExpand[Re[w[t]]];\)\ \), "\n", \(\(v[t_]\ = \ ComplexExpand[Im[w[t]]];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_2\)[t] = \>\"", z2[t]];\)\ \), "\n", \(\(Print[\*"\"\< f[\!\(z\_2\)[t]] = \>\"", f[z2[t]]];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_2\)'[t] = \>\"", \(z2'\)[ t]];\)\ \), "\n", \(\(Print[\*"\"\\"", w[t]];\)\ \), "\n", \(\(Print["\< u[t] = \>", u[t]];\)\ \), "\n", \(\(Print["\< v[t] = \>", v[t]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Method (i).", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The integral can be computed using the real and imaginary parts." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(U2\ = \[Integral]\_\(-1\)\%1 u[ t] \[DifferentialD]t;\)\ \), "\n", \(\(V2\ = \[Integral]\_\(-1\)\%1 v[t] \[DifferentialD]t;\)\ \), "\n", \(\(W2\ = \ U2\ + \ \[ImaginaryI]\ V2;\)\ \), "\n", \(\(U[t_]\ = \ \[Integral]u[t] \[DifferentialD]t;\)\ \), "\n", \(\(V[ t_]\ = \ \[Integral]v[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(reint\ = \ \[Integral]u[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(imint\ = \ \[Integral]v[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(P2[t_] = ComplexExpand[{Re[z2[t]], Im[z2[t]]}];\)\ \), "\[IndentingNewLine]", \(\(Print["\", u[t]];\)\ \), "\n", \(\(Print["\", v[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\[Integral]u[t]\[DifferentialD]t + \[ImaginaryI] \ \[Integral]v[t]\[DifferentialD]t = \>", reint, "\< + \[ImaginaryI] (\>", imint, "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\", U[t], "\<, V[t] = \>", V[t]];\)\ \), "\n", \(\(Print["\", U[1], "\<, V[1] = \>", V[1]];\)\ \), "\n", \(\(Print["\", U[\(-1\)], "\<, V[-1] = \>", V[\(-1\)]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%1\)u[t]\[DifferentialD]t + \ \[ImaginaryI]\!\(\[Integral]\_\(-1\)\%1\)v[t]\[DifferentialD]t = U[1]-U[-1] + \ \[ImaginaryI](V[1]-V[-1])\>\""];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%1\)u[t]\[DifferentialD]t + \ \[ImaginaryI]\!\(\[Integral]\_\(-1\)\%1\)v[t]\[DifferentialD]t = \>\"", U[1] - U[\(-1\)], "\< + \[ImaginaryI] (\>", V[1] - V[\(-1\)], "\<)\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%1\)u[t]\[DifferentialD]t + \ \[ImaginaryI] \!\(\[Integral]\_\(-1\)\%1\)v[t]\[DifferentialD]t = \>\"", W2];\)\ \), "\n", \(\(ParametricPlot[Evaluate[P2[t]], {t, \(-1\), 1}, PlotRange \[Rule] {{\(-1\), 3}, {\(-1\), 1}}, AspectRatio \[Rule] 2\/3, Ticks \[Rule] {Range[\(-1\), 3, 1], Range[\(-1\), 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] Green];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_C\) \>\"", f[z], "\<\[DifferentialD]z = \>", W2];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"", z2[t], "\<, for -1 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\ \[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Method (ii).", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The integral can be computed using a complex integrand." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(W[ t_]\ = \[Integral]w[t] \[DifferentialD]t;\)\ \), "\n", \(\(W2\ = \[Integral]\_\(-1\)\%1 w[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(P2[t_] = ComplexExpand[{Re[z2[t]], Im[z2[t]]}];\)\ \), "\[IndentingNewLine]", \(\(Print["\", w[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", ComplexExpand[W[t]]];\)\ \), "\n", \(\(Print["\", W[1], "\<, W[-1] = \>", W[\(-1\)]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%1\)w[t]\[DifferentialD]t = \ W[1]-W[-1] = \>\"", W[1], "\< - (\>", W[\(-1\)], "\<)\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_C\) \>\"", f[z], "\<\[DifferentialD]z = \>", W2];\)\ \), "\[IndentingNewLine]", \(\(ParametricPlot[Evaluate[P2[t]], {t, \(-1\), 1}, PlotRange \[Rule] {{\(-1\), 3}, {\(-1\), 1}}, AspectRatio \[Rule] 2\/3, Ticks \[Rule] {Range[\(-1\), 3, 1], Range[\(-1\), 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] Green];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_C\) \!\(z\^2\) \[DifferentialD]z = \ \>\"", W2];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"", z2[t], "\<, for -1 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\ \[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nWhich is the same as the value along the other path because ", Cell[BoxData[ \(f \((z)\) = z\^2\)]], " is an analytic function" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(P1[t_] = ComplexExpand[{Re[z1[t]], Im[z1[t]]}];\)\ \), "\[IndentingNewLine]", \(\(ParametricPlot[Evaluate[P1[t]], {t, \(-1\), 1}, PlotRange \[Rule] {{\(-1\), 3}, {\(-1\), 1}}, AspectRatio \[Rule] 2\/3, Ticks \[Rule] {Range[\(-1\), 3, 1], Range[\(-1\), 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] Magenta];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_C\) \>\"", f[z], "\<\[DifferentialD]z = \>", W1];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"", z1[t], "\<, for -1 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\ \[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n", StyleBox["Extra Example (i) for Page 214.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Evaluate the contour integrals of ", Cell[BoxData[ \(f \((z)\) = z\^2\)]], " starting at the points ", Cell[BoxData[ \(z\_1 = \(\(-1\) - i\ \ and\ \ z\_2 = 3 + i\)\)]], ". \n", StyleBox["(a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Use the line segment joining the points. ", StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Use a portion of a parabola joining the points.\n", StyleBox["Remark.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The example in the text used the function ", Cell[BoxData[ \(f \((z)\) = z\)]], " which is difficult for hand computations but it is not a challenge for \ ", StyleBox["Mathematica", FontSlant->"Italic"], ", hence we choose to use the function ", Cell[BoxData[ \(f \((z)\) = z\^2\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution for Extra Example (i) Page 214 part (a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ StyleBox["\n(a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Use the straight line segment connecting the points ", Cell[BoxData[ \(z\_1 = \(\(-1\) - i\ \ and\ \ z\_2 = 3 + i\)\)]], ", set up the parameterization and compute the contour integral. Enter \ the function ", Cell[BoxData[ \(f[z] = z\^2\)], AspectRatioFixed->True], ". 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i\ \ and\ \ z\_2 = 3 + i\)\)]], ", set up the parameterization and compute the contour integral." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(TagSet[t, Im[t], 0];\)\ \), "\n", \(\(z2[t_]\ = \ t\^2\ + \ 2 t\ + \ \[ImaginaryI]\ t;\)\ \), "\n", \(\(w[t_]\ = \ ComplexExpand[f[z2[t]]\ \(z2'\)[t]];\)\ \), "\n", \(\(u[t_]\ = \ ComplexExpand[Re[w[t]]];\)\ \), "\n", \(\(v[t_]\ = \ ComplexExpand[Im[w[t]]];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_2\)[t] = \>\"", z2[t]];\)\ \), "\n", \(\(Print[\*"\"\< f[\!\(z\_2\)[t]] = \>\"", ComplexExpand[\ f[z2[t]]\ ]];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_2\)'[t] = \>\"", \(z2'\)[ t]];\)\ \), "\n", \(\(Print[\*"\"\\"", w[t]];\)\ \), "\n", \(\(Print["\< u[t] = \>", u[t]];\)\ \), "\n", \(\(Print["\< v[t] = \>", v[t]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Method (i).", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The integral can be computed using the real and imaginary parts." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(U2\ = \[Integral]\_\(-1\)\%1 u[ t] \[DifferentialD]t;\)\ \), "\n", \(\(V2\ = \[Integral]\_\(-1\)\%1 v[t] \[DifferentialD]t;\)\ \), "\n", \(\(W2\ = \ U2\ + \ \[ImaginaryI]\ V2;\)\ \), "\n", \(\(U[t_]\ = \ \[Integral]u[t] \[DifferentialD]t;\)\ \), "\n", \(\(V[ t_]\ = \ \[Integral]v[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(reint\ = \ \[Integral]u[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(imint\ = \ \[Integral]v[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(P2[t_] = ComplexExpand[{Re[z2[t]], Im[z2[t]]}];\)\ \), "\[IndentingNewLine]", \(\(Print["\", u[t]];\)\ \), "\n", \(\(Print["\", v[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\[Integral]u[t]\[DifferentialD]t + \[ImaginaryI] \ \[Integral]v[t]\[DifferentialD]t = \>", reint, "\< + \[ImaginaryI] (\>", imint, "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\", U[t], "\<, V[t] = \>", V[t]];\)\ \), "\n", \(\(Print["\", U[1], "\<, V[1] = \>", V[1]];\)\ \), "\n", \(\(Print["\", U[\(-1\)], "\<, V[-1] = \>", V[\(-1\)]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%1\)u[t]\[DifferentialD]t + \ \[ImaginaryI]\!\(\[Integral]\_\(-1\)\%1\)v[t]\[DifferentialD]t = U[1]-U[-1] + \ \[ImaginaryI](V[1]-V[-1])\>\""];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%1\)u[t]\[DifferentialD]t + \ \[ImaginaryI]\!\(\[Integral]\_\(-1\)\%1\)v[t]\[DifferentialD]t = \>\"", U[1] - U[\(-1\)], "\< + \[ImaginaryI] (\>", V[1] - V[\(-1\)], "\<)\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%1\)u[t]\[DifferentialD]t + \ \[ImaginaryI] \!\(\[Integral]\_\(-1\)\%1\)v[t]\[DifferentialD]t = \>\"", W2];\)\ \), "\n", \(\(ParametricPlot[Evaluate[P2[t]], {t, \(-1\), 1}, PlotRange \[Rule] {{\(-1\), 3}, {\(-1\), 1}}, AspectRatio \[Rule] 2\/3, Ticks \[Rule] {Range[\(-1\), 3, 1], Range[\(-1\), 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] Green];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_C\) \>\"", f[z], "\<\[DifferentialD]z = \>", W2];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"", z2[t], "\<, for -1 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\ \[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Method (ii).", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The integral can be computed using a complex integrand." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(W[ t_]\ = \[Integral]w[t] \[DifferentialD]t;\)\ \), "\n", \(\(W2\ = \[Integral]\_\(-1\)\%1 w[ t] \[DifferentialD]t;\)\ \), "\[IndentingNewLine]", \(\(P2[t_] = ComplexExpand[{Re[z2[t]], Im[z2[t]]}];\)\ \), "\[IndentingNewLine]", \(\(Print["\", w[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", ComplexExpand[W[t]]];\)\ \), "\n", \(\(Print["\", W[1], "\<, W[-1] = \>", W[\(-1\)]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%1\)w[t]\[DifferentialD]t = \ W[1]-W[-1] = \>\"", W[1], "\< - (\>", W[\(-1\)], "\<)\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_\(-1\)\%1\)w[t]\[DifferentialD]t = \ \>\"", W2];\)\ \), "\[IndentingNewLine]", \(\(ParametricPlot[Evaluate[P2[t]], {t, \(-1\), 1}, PlotRange \[Rule] {{\(-1\), 3}, {\(-1\), 1}}, AspectRatio \[Rule] 2\/3, Ticks \[Rule] {Range[\(-1\), 3, 1], Range[\(-1\), 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] Green];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_C\) \>\"", f[z], "\<\[DifferentialD]z = \>", W2];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"", z2[t], "\<, for -1 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\ \[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nWhich is ", StyleBox["different", FontColor->RGBColor[1, 0, 1]], " from the value for the straight line path because ", Cell[BoxData[ \(f \((z)\) = \(\(\ \)\(z\)\)\&_\)]], " is ", StyleBox["not", FontColor->RGBColor[1, 0, 1]], " an analytic function." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(P1[t_] = ComplexExpand[{Re[z1[t]], Im[z1[t]]}];\)\ \), "\[IndentingNewLine]", \(\(ParametricPlot[Evaluate[P1[t]], {t, \(-1\), 1}, PlotRange \[Rule] {{\(-1\), 3}, {\(-1\), 1}}, AspectRatio \[Rule] 2\/3, Ticks \[Rule] {Range[\(-1\), 3, 1], Range[\(-1\), 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] Magenta];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_C\) \>\"", f[z], "\<\[DifferentialD]z = \>", W1];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"", z1[t], "\"];\)\ \)}], \ "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n", StyleBox["Example 6.10, Page 215. (a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Show that ", Cell[BoxData[ \(\[Integral]\_\(C\_1\)\ \(\(\ \)\(z\)\)\&_\ \[DifferentialD]z = \(-\ \[Pi]\)\ \[ImaginaryI]\)]], " where ", Cell[BoxData[ \(C\_1\)]], " is the semicircular path from -1 to 1 in the upper half plane. ", StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Show that ", Cell[BoxData[ \(\[Integral]\_\(C\_2\)\ \(\(\ \)\(z\)\)\&_\ \[DifferentialD]z = \(-4\)\ \ \[ImaginaryI]\)]], " where ", Cell[BoxData[ \(C\_2\)]], " is the polygonal path from -1 to -1+i to 1+i to 1." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 6.10 (a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ StyleBox["(a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Use the semicircular path from -1 to 1 in the upper half plane, set up \ the parameterization and compute the contour integral. Enter the function \ ", Cell[BoxData[ \(f \((z)\) = \(\(\ \)\(z\)\)\&_\)], AspectRatioFixed->True], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[f, t, u, v, w, W, W1, z];\)\ \), "\n", \(\(f[z_]\ = \ Conjugate[z];\)\ \), "\n", \(\(Print["\< f[z] = \>", f[z]];\)\ \), "\n", \(\(Print["\", ComplexExpand[ f[x + \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(TagSet[t, Im[t], 0];\)\ \), "\n", \(\(z[ t_]\ = \ \ \(-\ Cos[t]\)\ + \ \[ImaginaryI]\ Sin[t];\)\ \), "\n", \(\(w[t_]\ = \ ComplexExpand[f[z[t]]\ \(z'\)[t]];\)\ \), "\n", \(\(u[t_]\ = \ ComplexExpand[Re[w[t]]];\)\ \), "\n", \(\(v[t_]\ = \ ComplexExpand[Im[w[t]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\< f[z] = \>", f[z]];\)\ \), "\n", \(\(Print["\", ComplexExpand[ f[x + \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\< z[t] = \>", z[t]];\)\ \), "\n", \(\(Print["\", ComplexExpand[f[z[t]]]];\)\ \), "\n", \(\(Print["\< z'[t] = \>", \(z'\)[t]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", w[t]];\)\ \), "\n", \(\(w[t_]\ = \ Simplify[w[t]];\)\ \), "\n", \(\(Print["\", w[t]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Method.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The integral can be computed using a complex integrand." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(W[ t_]\ = \[Integral]w[t] \[DifferentialD]t;\)\ \), "\n", \(\(W1\ = \[Integral]\_0\%\[Pi] w[t] \[DifferentialD]t;\)\ \), "\n", \(\(Print["\", W[t]];\)\ \), "\n", \(\(Print["\", W[\[Pi]], "\<, W[0] = \>", W[0]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%\[Pi]\)w[t]\[DifferentialD]t = W[\ \[Pi]]-W[0] = \>\"", W[\[Pi]] - W[0]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%\[Pi]\)w[t]\[DifferentialD]t = \ \>\"", W1];\)\ \), "\[IndentingNewLine]", \(\(V[t_] = ComplexExpand[{Re[z[t]], Im[z[t]]}];\)\ \), "\n", \(\(ParametricPlot[Evaluate[V[t]], {t, 0, \[Pi]}, PlotRange \[Rule] {{\(-1\), 1}, {0, 1}}, AspectRatio \[Rule] 1\/2, Ticks \[Rule] {Range[\(-1\), 1, 1], Range[0, 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] Magenta];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_\(C\_1\)\) \>\"", f[z], "\<\[DifferentialD]z = \>", W1];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(C\_1\): z[t] = \>\"", z[t], "\<, for 0 \[LessEqual] t \[LessEqual] \[Pi].\>"];\)\ \), "\ \[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 6.10 (b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Use the polygonal path from -1 to -1+i to 1+i to 1, set up the \ parameterization and compute the contour integral. First, find a \ parameterization of the polygonal path ", Cell[BoxData[ \(C\_1\)]], " from -1 to 1 consisting of the three line segments:\nLine segment 1 from \ -1 to -1+i,\nLine segment 2 from -1+i to 1+i,\nLine segment 3 from 1+i to 1." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[t, V1, V2, V3, x0, x1, y0, y1, z1, Z, Z1];\)\ \), "\n", \(\(z0\ = \ \(-\ 1\);\)\ \), "\n", \(\(z1\ = \ \(-\ 1\)\ + \ \[ImaginaryI];\)\ \), "\n", \(x0\ = \ Re[z0]; \ y0\ = \ Im[z0]; \ x1\ = \ Re[z1]; \ y1\ = \ Im[z1];\ \), "\n", \(\(Z1[t_]\ = \ x0\ + \ \((x1 - x0)\) t\ + \ \[ImaginaryI] \((y0\ + \ \((y1 - y0)\) t)\);\)\ \), "\n", \(\(z0\ = \ \(-\ 1\)\ + \ \[ImaginaryI];\)\ \), "\n", \(\(z1\ = \ 1\ + \ \[ImaginaryI];\)\ \), "\n", \(x0\ = \ Re[z0]; \ y0\ = \ Im[z0]; \ x1\ = \ Re[z1]; \ y1\ = \ Im[z1];\ \), "\n", \(\(Z2[t_]\ = \ x0\ + \ \((x1 - x0)\) t\ + \ \[ImaginaryI] \((y0\ + \ \((y1 - y0)\) t)\);\)\ \), "\n", \(\(z0\ = \ 1\ + \ \[ImaginaryI];\)\ \), "\n", \(\(z1\ = \ 1;\)\ \), "\n", \(x0\ = \ Re[z0]; \ y0\ = \ Im[z0]; \ x1\ = \ Re[z1]; \ y1\ = \ Im[z1];\ \), "\n", \(\(Z3[t_]\ = \ x0\ + \ \((x1 - x0)\) t\ + \ \[ImaginaryI] \((y0\ + \ \((y1 - y0)\) t)\);\)\ \), "\n", \(\(Print[\*"\"\<\!\(Z\_1\)[t] = \>\"", Z1[t]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(Z\_2\)[t] = \>\"", Z2[t]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(Z\_3\)[t] = \>\"", Z3[t]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Method.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The integral can be computed using a three complex integrands." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Wset\ = \ {f[Z1[t]] \(Z1'\)[t], f[Z2[t]] \(Z2'\)[t], f[Z3[t]] \(Z3'\)[t]};\)\ \), "\n", \(\(funset\ = \ MapAll[ComplexExpand, Wset];\)\ \), "\n", \(\(Wint\ = \[Integral]funset \[DifferentialD]t;\)\ \), "\n", \(\(val\ = \[Integral]\_0\%1 funset \[DifferentialD]t;\)\ \), "\n", \(\(val2\ = ReplaceAll[val\ , List \[Rule] Plus];\)\ \), "\n", \(\(Print[\*"\"\<{\!\(w\_1\)[t],\!\(w\_2\)[t],\!\(w\_3\)[t]}\>\""]\ ;\)\ \ \), "\n", \(\(Print[\*"\"\<= \ {f[\!\(Z\_1\)[t]]\!\(Z\_1\)'[t],f[\!\(Z\_2\)[t]]\!\(Z\_2\)'[t],f[\!\(Z\_3\)[t]\ ]\!\(Z\_3\)'[t]}\>\""];\)\ \), "\n", \(\(Print["\<= \>", funset];\)\ \), "\n", \(\(Print["\< \>"];\)\ \), "\n", \(\(Print[\*"\"\<\[Integral]{\!\(w\_1\)[t],\!\(w\_2\)[t],\!\(w\_3\)[t]}\ \[DifferentialD]t = \>\""\ , Wint];\)\ \), "\n", \(\(Print["\< \>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%1\){\!\(w\_1\)[t],\!\(w\_2\)[t],\!\(\ w\_3\)[t]}\[DifferentialD]t = \>\"", val];\)\ \), "\[IndentingNewLine]", \(\(Print["\< \>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%1\)w[t]\[DifferentialD]t =\!\(\ \[Integral]\_0\%1\)\!\(w\_1\)[t]\[DifferentialD]t \ +\!\(\[Integral]\_0\%1\)\!\(w\_2\)[t]\[DifferentialD]t +\!\(\[Integral]\_0\%1\ \)\!\(w\_3\)[t]\[DifferentialD]t\>\""];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%1\)w[t]\[DifferentialD]t = (\>\"", val\_\(\(\[LeftDoubleBracket]\)\(1\)\(\[RightDoubleBracket]\)\), "\<) \ + (\>", val\_\(\(\[LeftDoubleBracket]\)\(2\)\(\[RightDoubleBracket]\)\), "\<) \ + (\>", val\_\(\(\[LeftDoubleBracket]\)\(3\)\(\[RightDoubleBracket]\)\), "\<)\ \>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[Integral]\_0\%1\)w[t]\[DifferentialD]t = \>\"", val2];\)\ \), "\[IndentingNewLine]", \(\(V1[t_] = ComplexExpand[{Re[Z1[t]], Im[Z1[t]]}];\)\ \), "\n", \(\(V2[t_] = ComplexExpand[{Re[Z2[t]], Im[Z2[t]]}];\)\ \), "\n", \(\(V3[t_] = ComplexExpand[{Re[Z3[t]], Im[Z3[t]]}];\)\ \), "\n", \(\(ParametricPlot[Evaluate[{V1[t], V2[t], V3[t]}], {t, 0, 1}, Ticks \[Rule] {Range[\(-1\), 3, 1], Range[\(-1\), 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {Red, Blue, Green}];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_\(C\_2\)\) \>\"", f[z], "\<\[DifferentialD]z = \>", W2];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(C\_2\): \!\(Z\_1\)[t] = \>\"", Z1[t], "\<, for 0 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\n\ ", \(\(Print[\*"\"\< \!\(Z\_2\)[t] = \>\"", Z2[t], "\<, for 0 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\n", \(\(Print[\*"\"\< \!\(Z\_3\)[t] = \>\"", Z3[t], "\<, for 0 \[LessEqual] t \[LessEqual] 1.\>"];\)\ \), "\ \[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nWhich is ", StyleBox["different", FontColor->RGBColor[1, 0, 1]], " from the value for the other path ", "because ", Cell[BoxData[ \(f \((z)\) = \(\(\ \)\(z\)\)\&_\)]], " is ", StyleBox["not", FontColor->RGBColor[1, 0, 1]], " an analytic function." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(V[t_] = ComplexExpand[{Re[z[t]], Im[z[t]]}];\)\ \), "\n", \(\(ParametricPlot[Evaluate[V[t]], {t, 0, \[Pi]}, PlotRange \[Rule] {{\(-1\), 1}, {0, 1}}, AspectRatio \[Rule] 1\/2, Ticks \[Rule] {Range[\(-1\), 1, 1], Range[0, 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] Magenta];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[Integral]\_\(C\_1\)\) \>\"", f[z], "\<\[DifferentialD]z = \>", W1];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(C\_1\): z[t] = \>\"", z[t], "\<, for 0 \[LessEqual] t \[LessEqual] \[Pi].\>"];\)\ \), "\ \[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell["\<\ \tWe now give a few important inequalities relating to complex \ integrals.\ \>", "Text"], Cell[TextData[{ StyleBox["\nTheorem 6.2 (Integral Triangle Inequality), Page 217.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " If ", Cell[BoxData[ \(f \((t)\) = u \((t)\) + \[ImaginaryI]\ v \((t)\)\)]], " is a continuous function of the real parameter t, then \n\n\t\t\t", Cell[BoxData[ \(\(\(|\)\(\[Integral]\_a\%b f \((t)\) \[DifferentialD]t\)\(|\)\(\ \)\(\(\[LessEqual]\)\(\ \ \)\(\[Integral]\_a\%b\)\)\(\ \)\(|\)\(f \((t)\)\)\(|\)\(\[DifferentialD]t\)\)\ \)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[StyleBox["Proof of Theorem 6.2, see text Page 217.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 1]]], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["\nTheorem 6.2 (ML - Inequality), Page 218.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " If ", Cell[BoxData[ \(f \((z)\) = u \((x, y)\) + \[ImaginaryI]\ v \((x, y)\)\)]], " is continuous on the contour C, then \n\n\t\t\t", Cell[BoxData[ \(\[Integral]\_C\ \(\(|\)\(f \((z)\)\)\(|\)\(\ \)\(\[DifferentialD]z\ \ \[LessEqual] \ M\ L\)\)\)]], ". 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Write a report on contour integrals.", FontSize->16, FontWeight->"Bold", FontColor->RGBColor[0.996109, 0, 0.996109]], "\n\n", StyleBox["1.", FontWeight->"Bold"], " Baker, I. N., and P. J. Rippon (1985), ''A Note on Complex Iteration (in \ The Teaching of Math.),'' Am. Math. M., Vol. 92, No. 7. (Aug. - Sep., 1985), \ pp. 501-504, ", StyleBox[ButtonBox["Jstor.", ButtonData:>{ URL[ "http://www.jstor.org/"], None}, ButtonStyle->"Hyperlink"], FontWeight->"Bold"], " \n\n", StyleBox["2.", FontWeight->"Bold"], " Boas, Mary L. and R. P. Boas ''Simplification of Some Contour \ Integrations (in The Teaching of Math.),'' Am. Math. M., Vol. 92, No. ", "3,", " (Mar., 1985), pp. 212-213, ", StyleBox[ButtonBox["Jstor.", ButtonData:>{ URL[ "http://www.jstor.org/"], None}, ButtonStyle->"Hyperlink"], FontWeight->"Bold"], " \n\n", StyleBox["3.", FontWeight->"Bold"], " Firey, William J., ''Line Integrals of Exact Differentials (in Classroom \ Notes),'' Am. Math. M., Vol. 68, No. 1. (Jan., 1961), pp. 57-59, ", StyleBox[ButtonBox["Jstor.", ButtonData:>{ URL[ "http://www.jstor.org/"], None}, ButtonStyle->"Hyperlink"], FontWeight->"Bold"], " \n\n", StyleBox["4.", FontWeight->"Bold"], " Glaister, P., (1991), ''A Method for Determining Some Integrals,'' Math. \ and Comp. Ed., V. 25, No. 1, pp. 31-32.\n\n", StyleBox["5.", FontWeight->"Bold"], " Gluchoff, Alan, ''A Simple Interpretation of the Complex Contour Integral \ (in The Teaching of Math.),'' Am. Math. M., Vol. 98, No. ", StyleBox["7.", FontWeight->"Bold"], " (Aug. - Sep., 1991), pp. 641-644, ", StyleBox[ButtonBox["Jstor.", ButtonData:>{ URL[ "http://www.jstor.org/"], None}, ButtonStyle->"Hyperlink"], FontWeight->"Bold"], " \n\n", StyleBox["6.", FontWeight->"Bold"], " Braden, Bart, (1987), ''Polya's Geometric Picture of Complex Contour \ Integrals, Math. Mag., V. 60, No. 5, pp. 321-327.\n\n", StyleBox["7.", FontWeight->"Bold"], " Sachdeva, Baldev K. and Bertram Ross''Evaluation of Certain Real \ Integrals by Contour Integration (in Notes),'' Am. Math. M., Vol. 89, No. 4. \ (Apr., 1982), pp. 246-249, ", StyleBox[ButtonBox["Jstor.", ButtonData:>{ URL[ "http://www.jstor.org/"], None}, ButtonStyle->"Hyperlink"], FontWeight->"Bold"], " \n\n", StyleBox["8.", FontWeight->"Bold"], " Trahan, Donald H., (1965)''A New Approach to Integration for Functions of \ a Complex Variable'',. Math. Mag., Vol. 38, pp. 132-140.\n\n", StyleBox["9.", FontWeight->"Bold"], " Zimmerberg, H. J., ''Independence of Path for Line Integrals (in \ Classroom Notes),'' Am. Math. M., Vol. 66, No. 4. 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