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"\n\tIn ", ButtonBox["Section 2.2", ButtonData:>{"ca0202.nb", None}, ButtonStyle->"Hyperlink"], " we investigated linear functions. Here we examine power functions. \n\n\t\ For ", Cell[BoxData[ \(z = r\ \[ExponentialE]\^\[ImaginaryI]\[Theta]\)]], ", we can express the function ", Cell[BoxData[ \(w = \(f \((z)\) = z\^2\)\)]], " in polar coordinates by \n\n\t\t\t", Cell[BoxData[ \(w = \(f \((z)\) = \(z\^2 = \(r\^2\) \[ExponentialE]\^\[ImaginaryI]2\ \[Theta]\)\)\)]], ". \n\nIf we also use polar coordinates for ", Cell[BoxData[ \(w = \[Rho]\ \[ExponentialE]\^\[ImaginaryI]\[Phi]\)]], " in the w -plane, we can express this mapping by the system of equations \n\ \n\t\t\t", Cell[BoxData[ \(\[Rho] = \(r\^2\ \ and\ \ \[Phi] = 2 \[Theta]\)\)]], ". " }], "Text"], Cell[TextData[{ "\tSince an argument of the product zz is twice an argument of z, we say \ that f doubles angles at the origin. Points that lie on the ray ", Cell[BoxData[ \(r > 0\)]], ", ", Cell[BoxData[ \(\[Theta] = \[Alpha]\)]], " are mapped onto points that lie on the ray ", Cell[BoxData[ \(\[Rho] > 0\)]], " , ", Cell[BoxData[ \(\[Phi] = 2 \[Alpha]\)]], ". If we now restrict the domain of ", Cell[BoxData[ \(w = \(f \((z)\) = z\^2\)\)]], " to the region \n\n\t\t\t", Cell[BoxData[ \(A = {r\ \(\(\[ExponentialE]\^\[ImaginaryI]\[Theta]\) : \ r > 0\ and\ \(-\[Pi]\)\/2 < \ \[Theta] \[LessEqual] \[Pi]\/2\)}\ \)]], ", \n\nthen the image of A under the mapping ", Cell[BoxData[ \(w = z\^2\)]], " can be described by the set \n\n\t\t\t", Cell[BoxData[ \(B = {\[Rho]\ \(\(\[ExponentialE]\^\[ImaginaryI]\[Phi]\) : \ \[Rho] > 0\ and\ - \[Pi] < \ \[Phi] \[LessEqual] \[Pi]\)}\)]], ", \n\nwhich consists of all points in the w-plane except the point w=0." }], "Text"], Cell[TextData[{ "\tThe inverse mapping of f, which we shall denote by g, is then\n\n\t\t\t\ ", Cell[BoxData[ \(z = \(g \((w)\) = \(w\^\(1/2\) = \(\[Rho]\^\(1/ 2\)\) \[ExponentialE]\^\(\[ImaginaryI]\[Phi]/2\)\)\)\)]], ", where ", Cell[BoxData[ \(w \[Element] B\)]], ".\n\t\tThat is\n\t\t\t", Cell[BoxData[ \(z = \(g \((w)\) = \(w\^\(1/2\) = \(\(|\)\(w\)\( | \^\(1/ 2\)\)\(\[ExponentialE]\^\(\[ImaginaryI]Arg \((w)\)/ 2\)\)\)\)\)\)]], ", where ", Cell[BoxData[ \(w \[NotEqual] 0\)]], ". \n\t\t\t\n\tThe function g is so important that it is worth calling \ special attention to it:" }], "Text"], Cell[TextData[{ "\n", StyleBox["Definition 2.1 (principal ", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ButtonBox["Square Root", ButtonData:>{ URL[ "http://mathworld.wolfram.com/SquareRoot.html"], None}, ButtonStyle->"Hyperlink"], StyleBox[" function), Page 63.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontColor->RGBColor[0, 0, 1]], "The function", StyleBox[" \n\n\t\t\t", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \(z = \(g \((w)\) = \(w\^\(1/2\) = \(\(|\)\(w\)\( | \^\(1/ 2\)\)\(\[ExponentialE]\^\(\[ImaginaryI]Arg \((w)\)/ 2\)\)\)\)\)\)]], ", for ", Cell[BoxData[ \(w \[NotEqual] 0\)]], " \n\nis called the ", StyleBox["principal square root function", FontSlant->"Italic"], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\tIt is left as an exercise to show that f and g satisfy equations \ g(f(z))=z and f(g(w))=w, and thus they are inverses of each other that map \ the set A one-to-one and onto the set B and the set B one-to-one and onto the \ set A, respectively. \n\n\tWhat are the images of rectangles under the \ mapping ", Cell[BoxData[ \(w = z\^2\)]], "? To find out, we make use of the Cartesian form \n\n\t\t", Cell[BoxData[ \(w = \(u + \[ImaginaryI]\ v = \(f \((z)\) = z\^2\)\)\)]], Cell[BoxData[ \(\(\(\ \)\(\(=\)\(\(x\^2 - y\^2 + \[ImaginaryI]\ 2\ x\ y\)\(=\)\(\ \)\)\)\)\)]], Cell[BoxData[ \(\((x\^2 - y\^2, 2\ x\ y)\) = \((u, v)\)\)]], "\n\nand the resulting system of equations ", Cell[BoxData[ RowBox[{"u", "=", RowBox[{ RowBox[{\(x\^2\), "-", RowBox[{\(y\^2\), " ", StyleBox["and", FontFamily->"Times New Roman", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}], " ", "v"}]}], "=", \(2\ x\ y\)}]}]]], ". " }], "Text"], Cell[TextData[{ StyleBox["\nExample 2.12, Page 63.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " The transformation ", Cell[BoxData[ \(w = \(f \((z)\) = z\^2\)\)]], " usually maps vertical and horizontal lines onto lines or parabolas.\n", StyleBox["(a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the image of the vertical line ", Cell[BoxData[ \(x = a\)]], ". ", StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the image of the horizontal line ", Cell[BoxData[ \(y = b\)]], ". \n", StyleBox["(c)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the image of the ray ", Cell[BoxData[ \(r > 0, \[Theta] = \[Alpha]\)]], " and circle ", Cell[BoxData[ \(r = c\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 2.12.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell["\<\ Enter the function w = f[z] and determine the real and imaginary \ parts.\ \>", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[a, b, eq1, eq2, eq3, eq4, x, y, u, u1, u2, v, U, V];\)\ \), "\n", \(\(f[z_] := z\^2;\)\ \), "\n", \(\(eq1\ = \ u\ \[Equal] \ ComplexExpand[Re[f[x + \[ImaginaryI]\ y]]];\)\ \), "\n", \(\(eq2\ = \ v\ \[Equal] \ ComplexExpand[Im[f[x + \[ImaginaryI]\ y]]];\)\ \), "\n", \(\(Print["\", f[z]];\)\ \), "\n", \(\(Print["\", f[x + \[ImaginaryI]\ y]];\)\ \), "\n", \(\(Print["\", Expand[f[x + \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\", Expand[f[x + \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[eq1];\)\ \), "\n", \(\(Print[eq2];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["\n(a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the image of the vertical line ", Cell[BoxData[ \(x = a\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Print[ eqns\ = \ ReplaceAll[{eq1, eq2}, x \[Rule] a]];\)\ \), "\n", \(\(Print[eq3\ = \ Eliminate[eqns, {y}]];\)\ \), "\n", \(\(Print[solset1\ = \ Solve[eq3, u]];\)\ \), "\n", \(\(u1[v_]\ = \ Expand[solset1\_\(\(\[LeftDoubleBracket]\)\(1, 1, 2\)\(\ \[RightDoubleBracket]\)\)];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", u1[v]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nHence, the image of the vertical line ", Cell[BoxData[ \(x = a\)]], " is the parabola ", Cell[BoxData[ \(u = a\^2 - v\^2\/\(4\ a\^2\)\)]], ". \n\n", StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the image of the horizontal line ", Cell[BoxData[ \(y = b\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Print[ eqns\ = \ ReplaceAll[{eq1, eq2}, y \[Rule] b]];\)\ \), "\n", \(\(Print[eq4\ = \ Eliminate[eqns, {x}]];\)\ \), "\n", \(\(Print[solset2\ = \ Solve[eq4, u]];\)\ \), "\n", \(\(u2[v_]\ = \ Expand[solset2\_\(\(\[LeftDoubleBracket]\)\(1, 1, 2\)\(\ \[RightDoubleBracket]\)\)];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", u2[v]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nHence, the image of the horizontal line ", Cell[BoxData[ \(y = b\)]], " is the parabola ", Cell[BoxData[ \(u = \(-b\^2\) + v\^2\/\(4\ b\^2\)\)]], ". \nUse ", StyleBox["Mathematica", FontSlant->"Italic"], " to make a graph of the mapping." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[f, Iden, z];\)\ \), "\n", \(\(Iden[z_]\ = \ z;\)\ \), "\n", \(\(f[z_]\ = \ z\^2;\)\ \), "\n", \(\(CartesianMap[ Iden, \ {0, 0.6, 0.1}, {0, 2, 0.1}, \[IndentingNewLine]PlotRange \[Rule] {{\(-0.05\), 0.65}, {\(-0.1\), 2.1}}, AspectRatio \[Rule] 2.2\/0.7, \[IndentingNewLine]Ticks \[Rule] {Range[0, 0.5, 0.5], Range[0, 2, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {Magenta, Pink}];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \ \), "\n", \(\(CartesianMap[ f, \ {0, 0.6, 0.1}, {0, 2.0, 0.1}, \[IndentingNewLine]PlotRange \[Rule] {{\(-4\), 0.5}, {\(-0.2\), 2.4}}, AspectRatio \[Rule] 2.6\/4.5, \[IndentingNewLine]Ticks \[Rule] {Range[\(-4\), 0, 1], Range[0, 2, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {Magenta, Pink}];\)\ \ \), "\n", \(\(Print[\*"\"\\""];\)\ \), "\n", \(\(Print["\", f[z]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["\n(c)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the image of the ray ", Cell[BoxData[ \(r > 0, \[Theta] = \[Alpha]\)]], " and circle ", Cell[BoxData[ \(r = c\)]], ". \nWe will use ", StyleBox["Mathematica", FontSlant->"Italic"], " to graph these images." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(PolarMap[ Iden, {0, 1.4, 0.1}, {0, \(\(\ \)\(\[Pi]\)\)\/2, \ \(\(\ \)\(\[Pi]\)\)\/24}, \ \[IndentingNewLine]PlotRange \[Rule] {{0, 1.5}, {0, 1.5}}, AspectRatio \[Rule] 1, Ticks \[Rule] {{0, 1\/2, 1, \@2}, {0, 1\/2, 1, \@2}}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {DarkGreen, Green}];\)\ \), "\[IndentingNewLine]", \(\(Print["\ 0, \[Theta] = \[Alpha], and circles r = c.\>"];\ \)\ \), "\n", \(\(PolarMap[ f\ , {0, 1.4, 0.1}, {0, \(\(\ \)\(\[Pi]\)\)\/2, \ \(\(\ \)\(\[Pi]\)\)\/24}, \ \[IndentingNewLine]PlotRange \[Rule] {{\(-2\), 2}, {0, 2}}, AspectRatio \[Rule] 1\/2, Ticks \[Rule] {Range[\(-1\), 2, 1], Range[0, 2, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {DarkGreen, Green}];\)\ \), "\n", \(\(Print[\*"\"\0, \[Theta]=2\[Alpha], and circles r=\!\ \(c\^2\),\>\""];\)\ \), "\n", \(\(Print["\", f[z]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n\tWe can use knowledge of the inverse mapping ", Cell[BoxData[ \(w = z\^\(1/2\)\)], AspectRatioFixed->True], " to get further insight into how the mapping ", Cell[BoxData[ \(w = z\^\(1/2\)\)], AspectRatioFixed->True], " acts on rectangles. If we let ", Cell[BoxData[ \(z = x + \[ImaginaryI]\ y \[NotEqual] 0\)]], ", then \n\n\t\t\t", Cell[BoxData[ \(z = \(w\^2\ = u\^2 - v\^2 + \[ImaginaryI]\ 2\ u\ v\)\)]], ", \n\nand we see that the point z=x+iy in the z-plane is related to the \ point ", Cell[BoxData[ \(w = z\^\(1/2\)\)], AspectRatioFixed->True], " in the w-plane by the system of equations \n\n\t\t\t", Cell[BoxData[ RowBox[{"x", "=", RowBox[{ RowBox[{\(u\^2\), "-", RowBox[{\(v\^2\), " ", StyleBox["and", FontFamily->"Times New Roman", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}], " ", "y"}]}], "=", \(2\ u\ v\)}]}]]], ". " }], "Text"], Cell[TextData[{ "\n", StyleBox["Example 2.13, Page 65.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " The transformation ", Cell[BoxData[ \(w = z\^\(1/2\)\)], AspectRatioFixed->True], " usually maps vertical and horizontal lines onto lines or hyperbolas.\n", StyleBox["(a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the image of the vertical line ", Cell[BoxData[ \(x = a\)]], ". ", StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the image of the horizontal line ", Cell[BoxData[ \(y = b\)]], ". \n", StyleBox["(c)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the image of the ray ", Cell[BoxData[ \(r > 0, \[Theta] = \[Alpha]\)]], " and circle ", Cell[BoxData[ \(r = c\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 2.13.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell["Before we start, make x, y, u and v real variables.", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[u, v, x, y, z];\)\ \), "\[IndentingNewLine]", \(\(Clear[f, F, X, Y, U, V, Z];\)\ \), "\n", \(\(TagSet[x, Im[x], 0];\)\ \), "\n", \(\(TagSet[y, Im[y], 0];\)\ \), "\n", \(\(TagSet[u, Im[u[x, y]], 0];\)\ \), "\n", \(\(TagSet[v, Im[v[x, y]], 0];\)\ \), "\n", \(\(TagSet[x, Re[x], x];\)\ \), "\n", \(\(TagSet[y, Re[y], y];\)\ \), "\n", \(\(TagSet[u, Re[u[x, y]], u[x, y]];\)\ \), "\n", \(\(TagSet[v, Re[v[x, y]], v[x, y]];\)\ \)}], "Input"], Cell[TextData[{ "\nThe mapping ", Cell[BoxData[ \(w = z\^\(1/2\)\)]], " can be manipulated as follows. " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(eq0 = \@z \[Equal] w;\)\ \), "\n", \(\(eq1 = z \[Equal] w\^2;\)\ \), "\n", \(\(eq2 = ReplaceAll[ eq1, {z \[Rule] x + \[ImaginaryI]\ y, w \[Rule] u + \[ImaginaryI]\ v}];\)\ \), "\n", \(\(eq3 = ComplexExpand[eq2];\)\ \), "\n", \(\(eq4 = ComplexExpand[Map[Re, eq3]];\)\ \), "\n", \(\(eq5 = Thread[\(-eq4\), Equal];\)\ \), "\n", \(\(eq6 = Thread[eq5 + eq3, Equal];\)\ \), "\n", \(\(eq7 = ComplexExpand[Map[Im, eq6]];\)\ \), "\[IndentingNewLine]", \(\(Print[eq0];\)\ \), "\[IndentingNewLine]", \(\(Print[eq1];\)\ \), "\[IndentingNewLine]", \(\(Print[eq2];\)\ \), "\[IndentingNewLine]", \(\(Print[eq3];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[eq4];\)\ \), "\[IndentingNewLine]", \(\(Print[eq7];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\n", StyleBox["(a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the image of the vertical line ", Cell[BoxData[ \(x = a\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Print["\"];\)\ \), "\ \[IndentingNewLine]", \(\(Print[ eq4\_\(\(\[LeftDoubleBracket]\)\(1\)\(\[RightDoubleBracket]\)\) \ \[Equal] a];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), \ "\[IndentingNewLine]", \(\(Print[ eq4\_\(\(\[LeftDoubleBracket]\)\(2\)\(\[RightDoubleBracket]\)\) \ \[Equal] a];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\n", StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the image of the horizontal line ", Cell[BoxData[ \(y = b\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Print["\"];\)\ \), "\ \[IndentingNewLine]", \(\(Print[ eq7\_\(\(\[LeftDoubleBracket]\)\(1\)\(\[RightDoubleBracket]\)\) \ \[Equal] b];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), \ "\[IndentingNewLine]", \(\(Print[ eq7\_\(\(\[LeftDoubleBracket]\)\(2\)\(\[RightDoubleBracket]\)\) \ \[Equal] b];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nUse ", StyleBox["Mathematica", FontSlant->"Italic"], " to make a graph of the mapping." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[f, Iden];\)\ \), "\n", \(\(Iden[z_] = z;\)\ \), "\n", \(\(f[z_] = z\^\(1/2\);\)\ \), "\n", \(\(CartesianMap[ Iden, \ {\(-9\), 9, 1}, {0, 9, 1}, \[IndentingNewLine]PlotRange \[Rule] {{\(-9\), 10}, {\(-0.5\), 10}}, AspectRatio \[Rule] 1\/2, \[IndentingNewLine]Ticks \[Rule] {Range[\(-9\), 9, 3], Range[0, 9, 3]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {Cyan, Blue}];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \ \), "\n", \(\(CartesianMap[ f, \ {\(-9\), 9, 1}, {0, 9, 1}, \[IndentingNewLine]PlotRange \[Rule] {{\(-0.3\), 3.3}, {\(-0.3\), 3.3}}, AspectRatio \[Rule] 1, Ticks \[Rule] {Range[0, 3, 1], Range[0, 3, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {Cyan, Blue}];\)\ \), "\n", \(\(Print[\*"\"\\""];\)\ \), "\n", \(\(Print["\", f[z]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["(c)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the image of the ray ", Cell[BoxData[ \(r > 0, \[Theta] = \[Alpha]\)]], " and circle ", Cell[BoxData[ \(r = c\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(PolarMap[ Iden, {0, 9, 0.5}, {0, \ \[Pi], \ \(\(\ \)\(\[Pi]\)\)\/12}, \ \[IndentingNewLine]PlotRange \[Rule] {{\(-9\), 9}, {0, 9}}, AspectRatio \[Rule] 1\/2, Ticks \[Rule] {Range[\(-9\), 9, 1], Range[0, 9, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {DarkGreen, Green}];\)\ \), "\[IndentingNewLine]", \(\(Print["\ 0, \[Theta] = \[Alpha], and circles r = c.\>"];\ \)\ \), "\n", \(\(PolarMap[ f\ , {0, 9, 0.5}, {0, \ \[Pi], \ \(\(\ \)\(\[Pi]\)\)\/12}, \ \[IndentingNewLine]PlotRange \[Rule] {{0, 3}, {0, 3}}, AspectRatio \[Rule] 1, \[IndentingNewLine]Ticks \[Rule] {Range[0, 3, 1], Range[0, 3, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {DarkGreen, Green}];\)\ \), "\n", \(\(Print[\*"\"\0, \[Theta]=\!\(\[Alpha]\/2\), and \ circles r=\!\(\@c\),\>\""];\)\ \), "\n", \(\(Print["\", f[z]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n\tWe can easily extend what we've done to integer powers higher than \ two. Let n be a positive integer and consider the function ", Cell[BoxData[ \(w = \(f \((z)\) = z\^n\)\)]], ", for ", Cell[BoxData[ \(z = r\ \[ExponentialE]\^\[ImaginaryI]\[Theta] \[NotEqual] 0\)]], ". This can be expressed in the polar coordinate form \n\n\t\t\t", Cell[BoxData[ \(w = \(f \((z)\) = \(z\^n = \(r\^n\) \[ExponentialE]\^\[ImaginaryI]n\ \[Theta]\)\)\)]], ". \n\nIf we use polar coordinates for ", Cell[BoxData[ \(w = \[Rho]\ \[ExponentialE]\^\[ImaginaryI]\[Phi]\)]], " in the w -plane, we can express this mapping by the system of equations \n\ \n\t\t\t", Cell[BoxData[ \(\[Rho] = \(r\^n\ \ and\ \ \[Phi] = n\[Theta]\)\)]], ". \n\n\tWe see that the image of the ray ", Cell[BoxData[ \(r > 0\)]], ", ", Cell[BoxData[ \(\[Theta] = \[Alpha]\)]], " is the ray ", Cell[BoxData[ \(\[Rho] > 0\)]], " , ", Cell[BoxData[ \(\[Phi] = n\[Alpha]\)]], " and that angles at the origin are increased by the factor n. Since the \ functions ", Cell[BoxData[ \(cos\ n\[Theta]\ and\ sin\ n\[Theta]\)]], " are periodic with period ", Cell[BoxData[ \(\(2 \[Pi]\)\/n\)]], ", we see that f is in general an n-to-one function", "; ", " that is, n points in the z-plane are mapped onto each non zero point in \ the w-plane." }], "Text"], Cell[TextData[{ "\tIf we now restrict the domain of ", Cell[BoxData[ \(w = \(f \((z)\) = z\^n\)\)]], " to the region \n\n\t\t\t", Cell[BoxData[ \(E = {r\ \(\(\[ExponentialE]\^\[ImaginaryI]\[Theta]\) : \ r > 0\ and\ \(-\[Pi]\)\/n < \ \[Theta] \[LessEqual] \[Pi]\/n\)}\ \)]], ", \n\nthen the image of E under the mapping ", Cell[BoxData[ \(w = z\^n\)]], " can be described by the set \n\n\t\t\t", Cell[BoxData[ \(F = {\[Rho]\ \(\(\[ExponentialE]\^\[ImaginaryI]\[Phi]\) : \ \[Rho] > 0\ and\ - \[Pi] < \ \[Phi] \[LessEqual] \[Pi]\)}\)]], ", \n\nwhich consists of all points in the w-plane except the point w=0. \ The inverse mapping of f, which we shall denote by g, is then\n\n\t\t\t", Cell[BoxData[ \(z = \(g \((w)\) = \(w\^\(1/n\) = \(\[Rho]\^\(1/ n\)\) \[ExponentialE]\^\(\[ImaginaryI]\[Phi]/n\)\)\)\)]], ", where ", Cell[BoxData[ \(w \[Element] F\)]], ".\n\t\tThat is\n\t\t\t", Cell[BoxData[ \(z = \(g \((w)\) = \(w\^\(1/n\) = \(\(|\)\(w\)\( | \^\(1/ n\)\)\(\[ExponentialE]\^\(\[ImaginaryI]Arg \((w)\)/ n\)\)\)\)\)\)]], ", where ", Cell[BoxData[ \(w \[NotEqual] 0\)]], ". \n\nAs with the principle square root function, we make an analogous \ definition for ", Cell[BoxData[ \(TraditionalForm\`n\^th\)]], " roots." }], "Text"], Cell[TextData[{ "\n", StyleBox["Definition 2.2 (principal n-th root function), Page 67.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontColor->RGBColor[0, 0, 1]], "The function", StyleBox[" \n\n\t\t\t", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \(z = \(g \((w)\) = \(w\^\(1/n\) = \(\(|\)\(w\)\( | \^\(1/ n\)\)\(\[ExponentialE]\^\(\[ImaginaryI]Arg \((w)\)/ n\)\)\)\)\)\)]], ", for ", Cell[BoxData[ \(w \[NotEqual] 0\)]], " \n\nis 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Cell[StyleData["Subsubtitle", "Printout"], CellMargins->{{2, 10}, {8, 10}}, FontSize->14] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Section"], CellDingbat->"\[FilledSquare]", CellMargins->{{25, Inherited}, {8, 24}}, CellGroupingRules->{"SectionGrouping", 30}, PageBreakBelow->False, CounterIncrements->"Section", CounterAssignments->{{"Subsection", 0}, {"Subsubsection", 0}}, FontFamily->"Helvetica", FontSize->16, FontWeight->"Bold"], Cell[StyleData["Section", "Presentation"], CellMargins->{{40, 10}, {11, 32}}, LineSpacing->{1, 0}, FontSize->24], Cell[StyleData["Section", "Condensed"], CellMargins->{{18, Inherited}, {6, 12}}, FontSize->12], Cell[StyleData["Section", "Printout"], CellMargins->{{13, 0}, {7, 22}}, FontSize->14] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subsection"], CellDingbat->"\[FilledSmallSquare]", CellMargins->{{22, Inherited}, {8, 20}}, CellGroupingRules->{"SectionGrouping", 40}, PageBreakBelow->False, CounterIncrements->"Subsection", CounterAssignments->{{"Subsubsection", 0}}, FontSize->14, FontWeight->"Bold"], Cell[StyleData["Subsection", "Presentation"], CellMargins->{{36, 10}, {11, 32}}, LineSpacing->{1, 0}, FontSize->22], Cell[StyleData["Subsection", "Condensed"], CellMargins->{{16, Inherited}, {6, 12}}, FontSize->12], Cell[StyleData["Subsection", "Printout"], CellMargins->{{9, 0}, {7, 22}}, FontSize->12] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subsubsection"], CellDingbat->"\[FilledSmallSquare]", CellMargins->{{22, Inherited}, {8, 18}}, CellGroupingRules->{"SectionGrouping", 50}, PageBreakBelow->False, CounterIncrements->"Subsubsection", FontWeight->"Bold"], Cell[StyleData["Subsubsection", "Presentation"], CellMargins->{{34, 10}, {11, 26}}, LineSpacing->{1, 0}, FontSize->18], Cell[StyleData["Subsubsection", "Condensed"], CellMargins->{{17, Inherited}, {6, 12}}, FontSize->10], Cell[StyleData["Subsubsection", "Printout"], CellMargins->{{9, 0}, {7, 14}}, FontSize->11] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Body Text", "Section"], Cell[CellGroupData[{ Cell[StyleData["Text"], CellMargins->{{12, 10}, {7, 7}}, LineSpacing->{1, 3}, CounterIncrements->"Text"], Cell[StyleData["Text", "Presentation"], CellMargins->{{24, 10}, {10, 10}}, LineSpacing->{1, 5}], Cell[StyleData["Text", "Condensed"], CellMargins->{{8, 10}, {6, 6}}, LineSpacing->{1, 1}], Cell[StyleData["Text", "Printout"], CellMargins->{{2, 2}, {6, 6}}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SmallText"], CellMargins->{{12, 10}, {6, 6}}, LineSpacing->{1, 3}, CounterIncrements->"SmallText", FontFamily->"Helvetica", FontSize->9], Cell[StyleData["SmallText", "Presentation"], CellMargins->{{24, 10}, {8, 8}}, LineSpacing->{1, 5}, FontSize->12], Cell[StyleData["SmallText", "Condensed"], CellMargins->{{8, 10}, {5, 5}}, LineSpacing->{1, 2}, FontSize->9], Cell[StyleData["SmallText", "Printout"], CellMargins->{{2, 2}, {5, 5}}, FontSize->7] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles 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