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AspectRatioFixed->True, FontSize->18, CellTags->"Section 3.2"], Cell[TextData[{ "\tIn ", ButtonBox["Section 3.1", ButtonData:>{"ca0301.nb", None}, ButtonStyle->"Hyperlink"], "we showed that computing the derivative of complex functions written in a \ nice form such as ", Cell[BoxData[ \(f \((z)\) = z\^2\)]], " is a rather simple task. But life is not so easy, for many times we \ encounter complex functions written as ", Cell[BoxData[ \(f \((z)\) = u \((x, y)\) + \[ImaginaryI]\ v \((x, y)\)\)]], ". For example, suppose we had \n\n\t\t\t", Cell[BoxData[ \(f \((z)\) = \(f \((x, y)\) = \(u \((x, y)\) + \[ImaginaryI]\ v \((x, y)\) = \((x\^3 - 3\ x\ y\^2)\) + \[ImaginaryI]\ \((3\ x\^2\ y - y\^3)\)\)\)\)]], ".\n\nIs there some criterion---perhaps involving the partial derivatives \ for u, and v---that we can use to determine whether f is differentiable, and \ if so, to find the value of ", Cell[BoxData[ \(f \((z)\)\)]], "?\n\n\tThe answer to this question is ", StyleBox["yes", FontSlant->"Italic"], ", thanks in part to the independent discovery of two important equations \ relating the partial derivatives of u and v by the French mathematician ", ButtonBox["Augustin Louis Cauchy", ButtonData:>{ URL[ "http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Cauchy.\ html"], None}, ButtonStyle->"Hyperlink"], " and the German mathematician ", ButtonBox["Georg Friedrich Bernhard Riemann", ButtonData:>{ URL[ "http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Riemann.\ html"], None}, ButtonStyle->"Hyperlink"], ". " }], "Text"], Cell[TextData[{ "\tLet ", Cell[BoxData[ \(f \((z)\) = u \((x, y)\) + \[ImaginaryI]v \((x, y)\)\)]], " be a complex function that is differentiable at the point ", Cell[BoxData[ \(z\_0\)]], ". Then it is natural to seek a formula for computing ", Cell[BoxData[ \(f' \((z\_0)\)\)]], " in terms of the partial derivatives of u(x,y) and v(x,y). If we \ investigate this idea, then it is easy to find the required formula", "; ", " but we will find that there are special conditions that must be satisfied \ before it can be used. In addition, we will discover two important equations \ relating the partial derivatives of u and v, which were discovered \ independently by the French mathematician A. L Cauchy and the German \ mathematician G. F. B. Riemann." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["\nTheorem 3.4, Page 103. (", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ButtonBox["Cauchy-Riemann Equations", ButtonData:>{ URL[ "http://mathworld.wolfram.com/Cauchy-RiemannEquations.html"], None}, ButtonStyle->"Hyperlink"], StyleBox[")", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Suppose that \n\n\t\t\t", Cell[BoxData[ \(f \((z)\) = \(f \((x + \[ImaginaryI]y)\) = u \((x, y)\) + \[ImaginaryI]\ v \((x, y)\)\)\)]], " \n\nis differentiable at the point ", Cell[BoxData[ \(z\_0 = x\_0 + \[ImaginaryI]\ y\_0\)]], ". Then the partial derivatives of u and v exist at the point ", Cell[BoxData[ \(\((x\_0, y\_0)\)\)]], ", and we have \n\n\t\t\t", Cell[BoxData[ \(f' \((z\_0)\) = \(u\_x\) \((x\_0, y\_0)\) + \[ImaginaryI]\ \(v\_x\) \((x\_0, y\_0)\)\)]], ",\n\t\tand\n\t\t\t", Cell[BoxData[ \(f' \((z\_0)\) = \(v\_y\) \((x\_0, y\_0)\) - \[ImaginaryI]\ \(u\_y\) \((x\_0, y\_0)\)\)]], ". \n\nEquating the real and imaginary parts of these two versions for the \ derivatives gives the so-called ", StyleBox["Cauchy-Riemann Equations", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " \n\n\t\t", Cell[BoxData[ \(\(u\_x\) \((x\_0, y\_0)\) = \(v\_y\) \((x\_0, y\_0)\)\)]], " and ", Cell[BoxData[ \(\(u\_y\) \((x\_0, y\_0)\) = \(-v\_x\) \((x\_0, y\_0)\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[StyleBox["Proof of Theorem 3.4, see text Page 103. ", FontWeight->"Bold", FontColor->RGBColor[0, 1, 1]]], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["\n", FontSize->14, FontWeight->"Bold"], StyleBox["Derivation of the Cauchy-Riemann Equations with ", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], StyleBox[".", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution for the Cauchy-Riemann Equations.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[\[CapitalDelta]x, \[CapitalDelta]y, E1, E2, E3, E4, E5, f, L1, L2, u, v, x, y, z];\)\ \), "\[IndentingNewLine]", \(\(TagSet[x, Im[x], 0];\)\ \), "\[IndentingNewLine]", \(\(TagSet[y, Im[y], 0];\)\ \), "\[IndentingNewLine]", \(\(TagSet[u, Im[u[x, y]], 0];\)\ \), "\[IndentingNewLine]", \(\(TagSet[v, Im[v[x, y]], 0];\)\ \), "\[IndentingNewLine]", \(\(TagSet[x, Re[x], x];\)\ \), "\[IndentingNewLine]", \(\(TagSet[y, Re[y], y];\)\ \), "\[IndentingNewLine]", \(\(TagSet[u, Re[u[x, y]], u[x, y]];\)\ \), "\[IndentingNewLine]", \(\(TagSet[v, Re[v[x, y]], v[x, y]];\)\ \), "\n", \(\(f[x_\ + \ \[ImaginaryI]\ y_]\ := \ u[x, y]\ + \ \[ImaginaryI]\ v[x, y];\)\ \), "\n", \(\(L1\ = \ Limit[\(f[x + \[CapitalDelta]x + \[ImaginaryI]\ y] - f[x + \ \[ImaginaryI]\ y]\)\/\[CapitalDelta]x, \[CapitalDelta]x \[Rule] 0, Analytic \[Rule] True];\)\ \), "\n", \(\(L2\ = \ Limit[\(f[x + \[ImaginaryI] \((y + \[CapitalDelta]y)\)] - f[x + \ \[ImaginaryI]\ y]\)\/\(\[ImaginaryI]\ \[CapitalDelta]y\), \[CapitalDelta]y \ \[Rule] 0, Analytic \[Rule] True];\)\ \), "\n", \(\(Print["\< f(z) = \>", f[x\ + \ \[ImaginaryI]\ y]];\)\ \), "\n", \(\(Print["\", ComplexExpand[Re[f[x + \[ImaginaryI]\ y]]]];\)\ \), "\n", \(\(Print["\", ComplexExpand[Im[f[x + \[ImaginaryI]\ y]]]];\)\ \), "\n", \(\(Print["\< \>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(lim\+\(\[CapitalDelta]x \[Rule] 0\)\) \!\(\ \[CapitalDelta]f\/\[CapitalDelta]x\) = \>\"", L1];\)\ \), "\n", \(\(Print[\*"\"\<\!\(lim\+\(\[CapitalDelta]y \[Rule] 0\)\) \!\(\ \[CapitalDelta]f\/\(\[ImaginaryI]\\\ \[CapitalDelta]y\)\) = \>\"", L2];\)\ \), "\n", \(Print["\<\>"]\), "\n", \(\(E1\ = \ L1\ \[Equal] \ L2;\)\ \), "\n", \(\(Print[E1];\)\ \), "\n", \(\(Print["\< \>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\n", \(\(E2\ = \ ReplaceAll[ E1, {\[ImaginaryI] \[Rule] 0, \(-\[ImaginaryI]\) \[Rule] 0}];\)\ \), "\n", \(\(Print[E2];\)\ \), "\n", \(\(E3\ = \ Thread[\(-E1\), Equal];\)\ \), "\n", \(\(E4\ = \ Thread[E2 + E3, Equal];\)\ \), "\n", \(\(E5\ = \ ReplaceAll[ E4, {\[ImaginaryI] \[Rule] 1, \(-\[ImaginaryI]\) \[Rule] \(-1\)}];\)\ \), "\n", \(\(Print[E5];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nWe have used ", StyleBox["Mathematica", FontSlant->"Italic"], " to derive the Cauchy-Riemann equations ", Cell[BoxData[ RowBox[{\(\(u\_x\) \((x\_0, y\_0)\)\), "=", RowBox[{ RowBox[{\(v\_y\), \((x\_0, y\_0)\), " ", StyleBox["and", FontFamily->"Times New Roman"], " ", \(u\_y\), \((x\_0, y\_0)\)}], "=", \(\(-v\_x\) \((x\_0, y\_0)\)\)}]}]]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n\tNote carefully some of the implications of this theorem:\n\t\n", StyleBox["(i).", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], " If f is differentiable at ", Cell[BoxData[ \(z\_0\)]], ", then we know the Cauchy-Riemann equations will be satisfied at ", Cell[BoxData[ \(z\_0\)]], ", and we can use either equation in Theorem 3.4 to evaluate ", Cell[BoxData[ \(f' \((z)\)\)], AspectRatioFixed->True], ". \n\t\n", StyleBox["(ii).", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], " Taking the contrapositive, if the Cauchy-Riemann equations are ", StyleBox["not", FontSlant->"Italic"], " satisfied at ", Cell[BoxData[ \(z\_0\)]], ", then we know automatically that f ", StyleBox["is not", FontSlant->"Italic"], " differentiable at ", Cell[BoxData[ \(z\_0\)]], ". \n\n", StyleBox["(iii).", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], " On the other hand, just because the Cauchy-Riemann equations are \ satisfied at ", Cell[BoxData[ \(z\_0\)]], ", we cannot ", StyleBox["necessarily", FontSlant->"Italic"], " conclude that f is differentiable at ", Cell[BoxData[ \(z\_0\)]], "." }], "Text"], Cell[TextData[{ StyleBox["\nExample 3.4, Page 104.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " We know that ", Cell[BoxData[ \(f \((z)\) = \(z\^2 = x\^2 - y\^2 + \[ImaginaryI]\ 2\ x\ y\)\)]], " differentiable and that ", Cell[BoxData[ \(\(f'\)[z] = 2 z\)], AspectRatioFixed->True], ". We also have \n\n", Cell[BoxData[ \(f \((z)\) = \(z\^2 = \(\((x + \[ImaginaryI]\ y)\)\^2 = \((x\^2 - y\^2)\) + \[ImaginaryI]\ \((2\ x\ y)\)\)\)\)]], ". \n\nAs advertised, the Cauchy-Riemann equations are satisfied, because \ \n\n\t", Cell[BoxData[ \(\(u\_x\) \((x, y)\) = \(2 x = \(v\_y\) \((x, y)\)\)\)]], " and ", Cell[BoxData[ \(\(u\_y\) \((x, y)\) = \(\(-2\) y = \(-v\_x\) \((x, y)\)\)\)]], ". \n\nUsing the formulae in Theorem 3.4 to compute ", Cell[BoxData[ \(\(f'\)[z]\)], AspectRatioFixed->True], " gives\n\n\t\t", Cell[BoxData[ \(\(f'\)[ z] = \(\(u\_x\) \((x, y)\) + \[ImaginaryI]\ \(v\_x\) \((x, y)\) = \(2 x + \[ImaginaryI]\ 2 y = \(\(2\) \(z\)\(\ \)\)\)\)\)], AspectRatioFixed->True], ",\n\tand\n\t\t", Cell[BoxData[ \(\(f'\)[ z] = \(\(v\_y\) \((x, y)\) - \[ImaginaryI]\ \(u\_y\) \((x, y)\) = \(2 x - \[ImaginaryI] \((\(-2\) y)\) = 2 z\)\)\)], AspectRatioFixed->True], ",\nas expected." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 3.4.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell["Enter the function f[z] and determine u[x,y] and v[x,y].", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[f, x, y, u, v, w, w1, z];\)\ \), "\n", \(\(f[z_]\ = \ z\^2;\)\ \), "\n", \(\(w\ = \ ComplexExpand[f[x + \[ImaginaryI]\ y]];\)\ \), "\n", \(\(u[x_, y_] = ComplexExpand[Re[f[x + \[ImaginaryI]\ y]]];\)\ \), "\n", \(\(v[x_, y_] = ComplexExpand[Im[f[x + \[ImaginaryI]\ y]]];\)\ \), "\n", \(\(Print["\< f[z] = \>", f[z]];\)\ \), "\n", \(\(Print["\< f[z] = \>", w];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\< f[z] = u[x,y] + \[ImaginaryI] v[x,y]\>"];\)\ \), "\ \[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", u[x, y]];\)\ \), "\n", \(\(Print["\", v[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell["\<\ Verify that the Cauchy-Riemann equations hold.\ \>", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Print[\*"\"\\""];\)\ \), "\[IndentingNewLine]", \(\(Print["\", \[PartialD]\_x\ u[x, y], "\< = \>", \[PartialD]\_y\ v[x, y], "\< ? \>", \[PartialD]\_x\ u[x, y] \[Equal] \[PartialD]\_y\ v[x, y]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\""];\)\ \), "\[IndentingNewLine]", \(\(Print["\", \[PartialD]\_y\ u[x, y], "\< = \>", \(-\[PartialD]\_x\ v[x, y]\), "\< ? \>", \[PartialD]\_y\ u[x, y] \[Equal] \(-\[PartialD]\_x\ v[x, y]\)];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nWe see that the Cauchy-Riemann equations hold.\n\nVerify that ", Cell[BoxData[ \(\(f'\)[z] = u\_x[x, y] + \[ImaginaryI]\ v\_x[x, y]\)], AspectRatioFixed->True], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(w1\ = \ \[PartialD]\_x\ u[x, y]\ \ + \ \ \[ImaginaryI]\ \[PartialD]\_x\ v[x, y];\)\ \), "\n", \(\(Print["\", \(f'\)[z]];\)\ \), "\[IndentingNewLine]", \(\(Print["\", \(f'\)[x + \[ImaginaryI]\ y]];\)\ \), "\n", \(\(Print["\", ComplexExpand[\(f'\)[x + \[ImaginaryI]\ y]]];\)\ \), "\n", \(\(Print["\< \>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\""];\)\ \), "\[IndentingNewLine]", \(\(Print["\", \[PartialD]\_x\ u[x, y], "\<) + \[ImaginaryI](\>", \[PartialD]\_x\ v[x, y], "\<)\>"];\)\ \), "\n", \(\(Print["\", w1];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n", StyleBox["Example 3.5, Page 105.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Verify that the complex function ", Cell[BoxData[ \(\(\(f \((z)\)\)\(=\)\(\(\(z\)\(\ \)\)\&_\)\(\ \)\)\)]], " is ", StyleBox["NOT", FontColor->RGBColor[1, 0, 1]], " analytic for any value of z. \nFurthermore, the Cauchy-Riemann are not \ satisfied at any point." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 3.5.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell["Enter the function f[z].", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[f, u, Ux, Ux0, Uy, Uy0, v, Vx, Vx0, Vy, Vy0, w, x, y];\)\ \), "\n", \(\(TagSet[x, Im[x], 0];\)\ \), "\n", \(\(TagSet[y, Im[y], 0];\)\ \), "\n", \(\(TagSet[x, Re[x], x];\)\ \), "\n", \(\(TagSet[y, Re[y], y];\)\ \), "\n", \(\(w\ = \ Conjugate[x + \[ImaginaryI]\ y];\)\ \), "\n", \(\(Print["\", w];\)\ \), "\n", \(\(w\ = \ ComplexExpand[w, TargetFunctions \[Rule] {Im, Re}];\)\ \), "\n", \(\(Print["\", w];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell["\<\ Find the real and imaginary parts of f[z].\ \>", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(u[x_, y_]\ = \ Together[ComplexExpand[Re[w]]];\)\ \), "\n", \(\(v[x_, y_]\ = \ Together[ComplexExpand[Im[w]]];\)\ \), "\n", \(\(u[0, 0]\ = \ 0;\)\ \), "\n", \(\(v[0, 0]\ = \ 0;\)\ \), "\n", \(\(f[x_, y_]\ = \ u[x, y]\ + \ \[ImaginaryI]\ v[x, y];\)\ \), "\n", \(\(f[0, 0]\ = \ 0;\)\ \), "\[IndentingNewLine]", \(\(Print["\< f[z] = u[x,y] + \[ImaginaryI] v[x,y]\>"];\)\ \), "\n", \(\(Print["\< f[z] = \>", f[x, y]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\", u[x, y]];\)\ \), "\n", \(\(Print["\", v[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell["\<\ Determine if the Cauchy-Riemann equations hold.\ \>", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Ux\ = \ Together[\[PartialD]\_x\ u[x, y]];\)\ \), "\n", \(\(Uy\ = \ Together[\[PartialD]\_y\ u[x, y]];\)\ \), "\n", \(\(Vx\ = \ Together[\[PartialD]\_x\ v[x, y]];\)\ \), "\n", \(\(Vy\ = \ Together[\[PartialD]\_y\ v[x, y]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(u\_x\)[x,y] = \>\"", Ux];\)\ \), "\n", \(\(Print[\*"\"\<\!\(u\_y\)[x,y] = \>\"", Uy];\)\ \), "\n", \(\(Print[\*"\"\<\!\(v\_x\)[x,y] = \>\"", Vx];\)\ \), "\n", \(\(Print[\*"\"\<\!\(v\_y\)[x,y] = \>\"", Vy];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(u\_x\)[x,y] - \!\(v\_y\)[x,y] = \>\"", Simplify[Ux - Vy]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(u\_y\)[x,y] + \!\(v\_x\)[x,y] = \>\"", Simplify[Uy + Vx]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nHence the Cauchy-Riemann do not hold at any point.\n\nThus, we have \ shown that ", Cell[BoxData[ \(\(\(f \((z)\)\)\(=\)\(\ \)\(\(\(z\)\(\ \)\)\&_\)\(\ \)\)\)]], " is ", StyleBox["NOT ", FontColor->RGBColor[1, 0, 1]], "analytic for any value of z." }], "Text", Evaluatable->False, AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n", StyleBox["Example 3.6, Page 105.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Verify that the complex function ", Cell[BoxData[ \(f \((z)\) = \((\ \(\(z\)\(\ \)\)\&_\ )\)\^2\/z\)]], " is ", StyleBox["NOT", FontColor->RGBColor[1, 0, 1]], " analytic for any value of z. \nFurthermore, the Cauchy-Riemann equations \ hold at (0,0), but f(z) is not differentiable at z = 0." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 3.6.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell["Enter the function f[z].", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[f, u, Ux, Ux0, Uy, Uy0, v, Vx, Vx0, Vy, Vy0, w, x, y];\)\ \), "\n", \(\(TagSet[x, Im[x], 0];\)\ \), "\n", \(\(TagSet[y, Im[y], 0];\)\ \), "\n", \(\(TagSet[x, Re[x], x];\)\ \), "\n", \(\(TagSet[y, Re[y], y];\)\ \), "\n", \(\(w\ = \ Conjugate[x + \[ImaginaryI]\ y]\^2\/\(x + \[ImaginaryI]\ y\);\)\ \), \ "\n", \(\(Print["\", w];\)\ \), "\n", \(\(w\ = \ ComplexExpand[w, TargetFunctions \[Rule] {Im, Re}];\)\ \), "\n", \(\(Print["\", w];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell["\<\ Find the real and imaginary parts of f[z].\ \>", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(u[x_, y_]\ = \ Together[ComplexExpand[Re[w]]];\)\ \), "\n", \(\(v[x_, y_]\ = \ Together[ComplexExpand[Im[w]]];\)\ \), "\n", \(\(u[0, 0]\ = \ 0;\)\ \), "\n", \(\(v[0, 0]\ = \ 0;\)\ \), "\n", \(\(f[x_, y_]\ = \ u[x, y]\ + \ \[ImaginaryI]\ v[x, y];\)\ \), "\n", \(\(f[0, 0]\ = \ 0;\)\ \), "\[IndentingNewLine]", \(\(Print["\< f[z] = u[x,y] + \[ImaginaryI] v[x,y]\>"];\)\ \), "\n", \(\(Print["\< f[z] = \>", f[x, y]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\", u[x, y]];\)\ \), "\n", \(\(Print["\", v[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell["\<\ Determine where the Cauchy-Riemann equations hold.\ \>", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Ux\ = \ Together[\[PartialD]\_x\ u[x, y]];\)\ \), "\n", \(\(Uy\ = \ Together[\[PartialD]\_y\ u[x, y]];\)\ \), "\n", \(\(Vx\ = \ Together[\[PartialD]\_x\ v[x, y]];\)\ \), "\n", \(\(Vy\ = \ Together[\[PartialD]\_y\ v[x, y]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(u\_x\)[x,y] = \>\"", Ux];\)\ \), "\n", \(\(Print[\*"\"\<\!\(u\_y\)[x,y] = \>\"", Uy];\)\ \), "\n", \(\(Print[\*"\"\<\!\(v\_x\)[x,y] = \>\"", Vx];\)\ \), "\n", \(\(Print[\*"\"\<\!\(v\_y\)[x,y] = \>\"", Vy];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(u\_x\)[x,y] - \!\(v\_y\)[x,y] = 0 = \>\"", Simplify[Ux - Vy]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(u\_y\)[x,y] + \!\(v\_x\)[x,y] = 0 = \>\"", Simplify[Uy + Vx]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nHence the function f(z) is differentiable only when ", Cell[BoxData[ RowBox[{\(x\^2 - y\^2\), "=", RowBox[{ RowBox[{"0", " ", StyleBox["and", FontFamily->"Times New Roman"], " ", "x", " ", "y"}], "=", "0"}]}]]], " hence only at the origin. \nLet's check to see if the Cauchy-Riemann \ equations hold at (0,0)." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Ux0\ = \ Limit[\(u[x, 0] - u[0, 0]\)\/\(x - 0\), \ x \[Rule] 0, \ Analytic \[Rule] True];\)\ \), "\n", \(\(Vy0\ = \ Limit[\(v[0, y] - v[0, 0]\)\/\(y - 0\), \ y \[Rule] 0, \ Analytic \[Rule] True];\)\ \), "\n", \(\(Uy0\ = \ Limit[\(u[0, y] - u[0, 0]\)\/\(y - 0\), \ y \[Rule] 0, \ Analytic \[Rule] True];\)\ \), "\n", \(\(Vx0\ = \ Limit[\(v[x, 0] - v[0, 0]\)\/\(x - 0\), \ x \[Rule] 0, \ Analytic \[Rule] True];\)\ \), "\n", \(\(Print[\*"\"\<\!\(u\_x\)[0,0] = \>\"", Ux0];\)\ \), "\n", \(\(Print[\*"\"\<\!\(u\_y\)[0,0] = \>\"", Uy0];\)\ \), "\n", \(\(Print[\*"\"\<\!\(v\_x\)[0,0] = \>\"", Vx0];\)\ \), "\n", \(\(Print[\*"\"\<\!\(v\_y\)[0,0] = \>\"", Vy0];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\\"", Ux0 \[Equal] Vy0];\)\ \), "\n", \(\(Print[\*"\"\\"", Uy0 \[Equal] \(-Vx0\)];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nSo, ", Cell[BoxData[ RowBox[{\(\(u\_x\) \((0, 0)\)\), "=", RowBox[{ RowBox[{\(v\_y\), \((0, 0)\), " ", StyleBox["and", FontFamily->"Times New Roman"], " ", \(u\_y\), \((0, 0)\)}], "=", \(\(-v\_x\) \((0, 0)\)\)}]}]]], ". \nBUT f(z) is ", StyleBox["NOT ", FontColor->RGBColor[1, 0, 1]], "analytic at (0,0) because f(z) is ", StyleBox["NOT", FontColor->RGBColor[1, 0, 1]], " analytic in a neighborhood of (0,0).\n\nAlso, f(z) is ", StyleBox["NOT ", FontColor->RGBColor[1, 0, 1]], "analytic at (0,0) because the following two limits are distinct." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(dfdx\ = \ Limit[\(f[x, 0] - f[0, 0]\)\/x, x \[Rule] 0, Analytic \[Rule] True];\)\ \), "\n", \(\(dfdt\ = \ Limit[\(f[t, t] - f[0, 0]\)\/\(t + \[ImaginaryI]\ t\), t \[Rule] 0, Analytic \[Rule] True];\)\ \), "\n", \(\(Print["\"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(lim\+\(x \[Rule] 0\)\) \!\(df\/dx\) = \!\(lim\+\(x \ \[Rule] 0\)\) \!\(\(f[x, 0] - f[0, 0]\)\/x\)\>\""];\)\ \), "\n", \(\(Print[\*"\"\<\!\(lim\+\(x \[Rule] 0\)\) \!\(df\/dx\) = \!\(lim\+\(x \ \[Rule] 0\)\) (\>\"", f[x, 0] - f[0, 0], "\<)/(\>", x, "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(lim\+\(x \[Rule] 0\)\) \!\(df\/dx\) = \!\(lim\+\(x \ \[Rule] 0\)\) (\>\"", \((f[x, 0] - f[0, 0])\)/ x, "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(lim\+\(x \[Rule] 0\)\) \!\(df\/dx\) = \>\"", dfdx];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(lim\+\(t \[Rule] 0\)\) \!\(df\/dt\) = \!\(lim\+\(t \ \[Rule] 0\)\) \!\(\(f[t, t] - f[0, 0]\)\/\(t + \[ImaginaryI]\\\ \ t\)\)\>\""];\)\ \), "\n", \(\(Print[\*"\"\<\!\(lim\+\(t \[Rule] 0\)\) \!\(df\/dt\) = \!\(lim\+\(t \ \[Rule] 0\)\) (\>\"", f[t, t] - f[0, 0], "\<)/(\>", t + \[ImaginaryI]\ t, "\<)\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(lim\+\(t \[Rule] 0\)\) \!\(df\/dt\) = \!\(lim\+\(t \ \[Rule] 0\)\) (\>\"", \((f[t, t] - f[0, 0])\)/\((t + \[ImaginaryI]\ t)\), "\<)\>"];\)\ \), "\ \[IndentingNewLine]", \(\(Print[\*"\"\<\!\(lim\+\(t \[Rule] 0\)\) \!\(df\/dt\) = \>\"", dfdt];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nThus, we have shown that ", Cell[BoxData[ \(f \((z)\) = \((\ \(\(z\)\(\ \)\)\&_\ )\)\^2\/z\)]], " is ", StyleBox["NOT ", FontColor->RGBColor[1, 0, 1]], "analytic for any value of z." }], "Text", Evaluatable->False, AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n\tExample 3.6 reiterates that the mere satisfaction of the \ Cauchy-Riemann equations is not a sufficient criterion to guarantee the \ differentiability of a function. However, theorem 3.5 gives us sufficient \ conditions under which we can use equations (3-15) or (3-16) to compute the \ derivative ", Cell[BoxData[ \(f' \((z\_0)\)\)]], ". They are referred to as the ", StyleBox["Cauchy-Riemann conditions", FontSlant->"Italic"], " for differentiability." }], "Text"], Cell[TextData[{ StyleBox["\nTheorem 3.5 (Cauchy-Riemann conditions for differentiability), \ Page 106.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Let ", Cell[BoxData[ \(f \((z)\) = \(f \((x + \[ImaginaryI]\ y)\) = u \((x, y)\) + \[ImaginaryI]\ v \((x, y)\)\)\)]], " be a continuous function that is defined in some neighborhood of the \ point ", Cell[BoxData[ \(z\_0 = x\_0 + \[ImaginaryI]\ y\_0\)]], ". If all the partial derivatives ", Cell[BoxData[ RowBox[{\(u\_x\), ",", \(u\_y\), ",", RowBox[{\(v\_x\), " ", StyleBox["and", FontFamily->"Times New Roman", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}], " ", \(v\_y\)}]}]]], " are continuous at the point", Cell[BoxData[ \(\((x\_0, y\_0)\)\)]], " and if the Cauchy-Riemann equations ", Cell[BoxData[ \(\(u\_x\) \((x\_0, y\_0)\) = \(v\_y\) \((x\_0, y\_0)\)\)]], " and ", Cell[BoxData[ \(\(u\_y\) \((x\_0, y\_0)\) = \(-v\_x\) \((x\_0, y\_0)\)\)]], " hold, then f(z) is differentiable at ", Cell[BoxData[ \(z\_0\)]], " and the derivative ", Cell[BoxData[ \(f' \((z\_0)\)\)]], "can be computed with either formula \n\n\t\t\t", Cell[BoxData[ \(f' \((z\_0)\) = \(u\_x\) \((x\_0, y\_0)\) + \[ImaginaryI]\ \(v\_x\) \((x\_0, y\_0)\)\)]], ",\n\t\tor\n\t\t\t", Cell[BoxData[ \(f' \((z\_0)\) = \(v\_y\) \((x\_0, y\_0)\) - \[ImaginaryI]\ \(u\_y\) \((x\_0, y\_0)\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[StyleBox["Proof of Theorem 3.5, see text Page 106.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 1]]], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["\nExample 3.7, Page 108.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Show that the function ", Cell[BoxData[ \(f \((z)\) = x\^3 - 3\ x\ y\^2 + \[ImaginaryI]\ \((3\ x\^2\ y - y\^3)\)\)]], " is differentiable for all z and find its derivative. Verify that the \ Cauchy-Riemann equations hold and that ", Cell[BoxData[ \(\(f'\)[z] = u\_x[x, y] + \[ImaginaryI]\ v\_x[x, y]\)], AspectRatioFixed->True], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 3.7.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell["Enter the function f[z] and determine u[x,y] and v[x,y].", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[f, x, y, u, v, w, w1, z];\)\ \), "\n", \(\(f[z_]\ = \ z\^3;\)\ \), "\n", \(\(w\ = \ ComplexExpand[f[x + \[ImaginaryI]\ y]];\)\ \), "\n", \(\(u[x_, y_] = ComplexExpand[Re[f[x + \[ImaginaryI]\ y]]];\)\ \), "\n", \(\(v[x_, y_] = ComplexExpand[Im[f[x + \[ImaginaryI]\ y]]];\)\ \), "\n", \(\(Print["\< f[z] = \>", f[z]];\)\ \), "\n", \(\(Print["\< f[z] = \>", w];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print["\< f[z] = u[x,y] + \[ImaginaryI] v[x,y]\>"];\)\ \), "\ \[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", u[x, y]];\)\ \), "\n", \(\(Print["\", v[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell["\<\ Verify that the Cauchy-Riemann equations hold.\ \>", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Print[\*"\"\\""];\)\ \), "\[IndentingNewLine]", \(\(Print["\", \[PartialD]\_x\ u[x, y], "\< = \>", \[PartialD]\_y\ v[x, y], "\< ? \>", \[PartialD]\_x\ u[x, y] \[Equal] \[PartialD]\_y\ v[x, y]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\""];\)\ \), "\[IndentingNewLine]", \(\(Print["\", \[PartialD]\_y\ u[x, y], "\< = \>", \(-\[PartialD]\_x\ v[x, y]\), "\< ? \>", \[PartialD]\_y\ u[x, y] \[Equal] \(-\[PartialD]\_x\ v[x, y]\)];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nWe see that the Cauchy-Riemann equations hold.\n\nVerify that ", Cell[BoxData[ \(\(f'\)[z] = u\_x[x, y] + \[ImaginaryI]\ v\_x[x, y]\)], AspectRatioFixed->True], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(w1\ = \ \[PartialD]\_x\ u[x, y]\ \ + \ \ \[ImaginaryI]\ \[PartialD]\_x\ v[x, y];\)\ \), "\n", \(\(Print["\", \(f'\)[z]];\)\ \), "\[IndentingNewLine]", \(\(Print["\", \(f'\)[x + \[ImaginaryI]\ y]];\)\ \), "\n", \(\(Print["\", ComplexExpand[\(f'\)[x + \[ImaginaryI]\ y]]];\)\ \), "\n", \(\(Print["\< \>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\""];\)\ \), "\[IndentingNewLine]", \(\(Print["\", \[PartialD]\_x\ u[x, y], "\<) + \[ImaginaryI](\>", \[PartialD]\_x\ v[x, y], "\<)\>"];\)\ \), "\n", \(\(Print["\", w1];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n", StyleBox["Example 3.8, Page 108.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Use the Cauchy-Riemann equations to show that ", Cell[BoxData[ \(f \((z)\) = \[ExponentialE]\^\(-y\)\ cos \((x)\) + \[ImaginaryI]\ \ \[ExponentialE]\^\(-y\)\ sin \((x)\)\)]], " is differentiable for all ", Cell[BoxData[ \(z = x + \[ImaginaryI]\ y\)]], ". Furthermore, f(z) is analytic for all z." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 3.8.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell["\<\ Enter the function f[z] and determine if the Cauchy-Riemann \ equations hold.\ \>", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[f, u, Ux, Uy, v, Vx, Vy, w1, x, y, z];\)\ \), "\n", \(\(u[x_, y_]\ = \ \[ExponentialE]\^\(-y\)\ Cos[x];\)\ \), "\n", \(\(v[x_, y_]\ = \ \(\[ExponentialE]\^\(-y\)\) Sin[x];\)\ \), "\n", \(\(Ux\ = \ Together[\[PartialD]\_x\ u[x, y]];\)\ \), "\n", \(\(Uy\ = \ Together[\[PartialD]\_y\ u[x, y]];\)\ \), "\n", \(\(Vx\ = \ Together[\[PartialD]\_x\ v[x, y]];\)\ \), "\n", \(\(Vy\ = \ Together[\[PartialD]\_y\ v[x, y]];\)\ \), "\[IndentingNewLine]", \(\(Print["\< f[z] = \>", u[x, y] + \[ImaginaryI]\ v[x, y]];\)\ \), "\[IndentingNewLine]", \(\(Print["\< f[z] = u[x,y] + \[ImaginaryI] v[x,y]\>"];\)\ \), "\n", \(\(Print["\< u[x,y] = \>", u[x, y]];\)\ \), "\n", \(\(Print["\< v[x,y] = \>", v[x, y]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(u\_x\)[x,y] = \>\"", Ux];\)\ \), "\n", \(\(Print[\*"\"\<\!\(u\_y\)[x,y] = \>\"", Uy];\)\ \), "\n", \(\(Print[\*"\"\<\!\(v\_x\)[x,y] = \>\"", Vx];\)\ \), "\n", \(\(Print[\*"\"\<\!\(v\_y\)[x,y] = \>\"", Vy];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\\"", Ux \[Equal] Vy];\)\ \), "\n", \(\(Print[\*"\"\\"", Uy \[Equal] \(-Vx\)];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\nThe Cauchy-Riemann equations hold everywhere, so that ", Cell[BoxData[ \(f \((z)\) = \[ExponentialE]\^\(-y\)\ cos \((x)\) + \[ImaginaryI]\ \ \[ExponentialE]\^\(-y\)\ sin \((x)\)\)]], " is analytic for all values of z." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Print["\< f[z] = f[x + \[ImaginaryI] y] = \>", u[x, y]\ + \ \[ImaginaryI]\ v[x, y]];\)\ \), "\[IndentingNewLine]", \(\(Print["\", \[PartialD]\_x\ u[x, y]\ + \ \[ImaginaryI]\ \[PartialD]\_x\ v[x, y]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["\nRemark.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " We can write this function as ", Cell[BoxData[ \(f \((z)\) = \(\[ExponentialE]\^\(\[ImaginaryI]\ z\) = \(\ \[ExponentialE]\^\(\[ImaginaryI]\ \((x + \[ImaginaryI]\ y)\)\) = \ \[ExponentialE]\^\(\(-y\) + \[ImaginaryI]\ x\)\)\)\)]], ", and investigate ", Cell[BoxData[ \(f' \((z)\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(f[ z_]\ = \ \[ExponentialE]\^\(\[ImaginaryI]\ z\);\)\ \), "\n", \(\(w1\ = \ \[PartialD]\_x\ u[x, y]\ \ + \ \ \[ImaginaryI]\ \[PartialD]\_x\ v[x, y];\)\ \), "\n", \(\(Print["\< f[z] = \>", f[z]];\)\ \), "\n", \(\(Print["\< f[x+\[ImaginaryI]y] = \>", f[x + \[ImaginaryI]\ y]];\)\ \), "\n", \(\(Print["\< f[x+\[ImaginaryI]y] = \>", ComplexExpand[ f[x\ + \ \[ImaginaryI]\ y]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\< f[x+\[ImaginaryI]y] = u[x,y] + \[ImaginaryI] v[x,y]\>"];\)\ \ \), "\n", \(\(Print["\< \>"];\)\ \), "\n", \(\(Print["\< f'[z] = \>", \(f'\)[z]];\)\ \), "\[IndentingNewLine]", \(\(Print["\", \(f'\)[ x + \[ImaginaryI]\ y]];\)\ \), "\n", \(\(Print["\", ComplexExpand[\(f'\)[x + \[ImaginaryI]\ y]]];\)\ \), "\n", \(\(Print["\< \>"];\)\ \), "\n", \(\(Print[\*"\"\\""];\)\ \), "\n", \(\(Print["\", w1];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nWe have shown that ", Cell[BoxData[ \(f \((z)\) = \[ExponentialE]\^\(-y\)\ cos \((x)\) + \[ImaginaryI]\ \ \[ExponentialE]\^\(-y\)\ sin \((x)\)\)]], " is differentiable and hence analytic for all z." }], "Text"] }, Closed]], Cell["\<\ \tThe Cauchy-Riemann Conditions are particularly useful in determining the \ set of points for which a function f is differentiable.\ \>", "Text"], Cell[TextData[{ "\n", StyleBox["Example 3.9, Page 109.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Show that the complex function ", Cell[BoxData[ \(f \((z)\) = x\^3 + 3\ x\ y\^2 + \[ImaginaryI]\ \((3\ x\^2\ y + y\^3)\)\)]], " is differentiable only at points that lie on the coordinate axes." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 3.9.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell["\<\ Enter the functions u[x,y] and v[x,y] and use them to form f[z]. \ Then determine where the Cauchy-Riemann equations hold.\ \>", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[u, Ux, Uy, v, Vx, Vy, x, y];\)\ \), "\n", \(\(u[x_, y_]\ = \ x\^3\ + \ 3\ x\ y\^2;\)\ \), "\n", \(\(v[x_, y_]\ = \ y\^3\ + \ 3\ x\^2\ y;\)\ \), "\n", \(\(Ux\ = \ Together[\[PartialD]\_x\ u[x, y]];\)\ \), "\n", \(\(Uy\ = \ Together[\[PartialD]\_y\ u[x, y]];\)\ \), "\n", \(\(Vx\ = \ Together[\[PartialD]\_x\ v[x, y]];\)\ \), "\n", \(\(Vy\ = \ Together[\[PartialD]\_y\ v[x, y]];\)\ \), "\[IndentingNewLine]", \(\(Print["\< f[z] = \>", u[x, y]\ + \ \[ImaginaryI]\ v[x, y]];\)\ \), "\[IndentingNewLine]", \(\(Print["\< f[z] = u[x,y] + \[ImaginaryI] v[x,y]\>"];\)\ \), "\n", \(\(Print["\< u[x,y] = \>", u[x, y]];\)\ \), "\n", \(\(Print["\< v[x,y] = \>", v[x, y]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(u\_x\)[x,y] = \>\"", Ux];\)\ \), "\n", \(\(Print[\*"\"\<\!\(u\_y\)[x,y] = \>\"", Uy];\)\ \), "\n", \(\(Print[\*"\"\<\!\(v\_x\)[x,y] = \>\"", Vx];\)\ \), "\n", \(\(Print[\*"\"\<\!\(v\_y\)[x,y] = \>\"", Vy];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\\"", Ux \[Equal] Vy];\)\ \), "\n", \(\(Print[\*"\"\\"", Uy \[Equal] \(-Vx\)];\)\ \), "\n", \(\(Print[\*"\"\\"", Uy + Vx \[Equal] 0];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nThe Cauchy-Riemann equations hold only if x y = 0 which occurs along \ the coordinate axes. Furthermore, all of the partial derivatives ", Cell[BoxData[ \(\[PartialD]\_x\ u[x, y] = 3\ x\^2 + 3 y\^2, \[PartialD]\_y\ u[x, y] = 6\ x\ y, \[PartialD]\_x\ v[x, y] = 6\ x\ y, \ and \[PartialD]\_y\ v[x, y] = 3\ x\^2 + 3 y\^2\)]], " are continuous. Hence, ", Cell[BoxData[ \(f \((z)\) = x\^3 + 3\ x\ y\^2 + \[ImaginaryI]\ \((3\ x\^2\ y + y\^3)\)\)]], " is differentiable only when x=0 or y=0, which occurs at points that lie \ on the coordinate axes. ", StyleBox["Remark.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " In the next section we will learn that ", Cell[BoxData[ \(f \((z)\) = x\^3 + 3\ x\ y\^2 + \[ImaginaryI]\ \((3\ x\^2\ y + y\^3)\)\)]], " is ", StyleBox["nowhere", FontColor->RGBColor[1, 0, 1]], " analytic." }], "Text", Evaluatable->False, AspectRatioFixed->True] }, Closed]], Cell[TextData[{ StyleBox["\nTheorem 3.6 (Polar Form of C-R equations), Page 109.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Let ", Cell[BoxData[ \(f \((z)\) = \(f \((r\ \[ExponentialE]\^\(\[ImaginaryI]\ \[Theta]\))\) \ = U \((r, \[Theta])\) + \[ImaginaryI]\ V \((r, \[Theta])\)\)\)]], " be a continuous function that is defined in some neighborhood of the \ point ", Cell[BoxData[ \(z\_0 = \(r\_0\) \[ExponentialE]\^\(\[ImaginaryI]\ \[Theta]\_0\)\)]], ". If all of the first order partial derivatives ", Cell[BoxData[ RowBox[{\(U\_r\), ",", \(U\_\[Theta]\), ",", RowBox[{\(V\_x\), " ", StyleBox["and", FontFamily->"Times New Roman", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}], " ", \(V\_\[Theta]\)}]}]]], " are continuous at the point ", Cell[BoxData[ \(\((r\_0, \[Theta]\_0)\)\)]], " and if the Cauchy-Riemann equations \n\n\t\t", Cell[BoxData[ \(\(U\_r\) \((r\_0, \[Theta]\_0)\) = \(1\/r\_0\) \(V\_\[Theta]\) \ \((r\_0, \[Theta]\_0)\)\)]], " and ", Cell[BoxData[ \(\(V\_r\) \((r\_0, \[Theta]\_0)\) = \(\(-1\)\/r\_0\) \(U\_\[Theta]\) \ \((r\_0, \[Theta]\_0)\)\)]], " hold, \n\nthen f(z) is differentiable at ", Cell[BoxData[ \(z\_0\)]], ", and we can compute the derivative ", Cell[BoxData[ \(f \((z\_0)\)\)]], " by using either polar formulae \n\n\t\t\t", Cell[BoxData[ \(f' \((z\_0)\) = \(\(f' \((r\_0\ \[ExponentialE]\^\(\[ImaginaryI]\ \ \[Theta]\_0\))\)\)\(=\)\(\ \)\)\)]], Cell[BoxData[ \(\(\(\ \)\(\(\[ExponentialE]\^\(\(-\[ImaginaryI]\)\ \[Theta]\_0\)\) \ \((\(U\_r\) \((r\_0, \[Theta]\_0)\) + \[ImaginaryI]\ \(V\_r\) \((r\_0, \ \[Theta]\_0)\))\)\)\)\)]], " \n\t\tor \n\t\t\t", Cell[BoxData[ \(f' \((z\_0)\) = \(\(f' \((r\_0\ \[ExponentialE]\^\(\[ImaginaryI]\ \ \[Theta]\_0\))\)\)\(=\)\(\ \)\)\)]], Cell[BoxData[ \(\(1\/r\_0\) \(\[ExponentialE]\^\(\(-\[ImaginaryI]\)\ \[Theta]\_0\)\) \ \((\(V\_\[Theta]\) \((r\_0, \[Theta]\_0)\) - \[ImaginaryI]\ \(U\_\[Theta]\) \ \((r\_0, \[Theta]\_0)\))\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[StyleBox["Proof of Theorem 3.4, see text Page 109.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 1]]], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 3.10, Page 110.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Show that if ", Cell[BoxData[ \(f \((z)\) = \(z\^\(1/2\) = r\^\(1/2\)\ \((cos \((\[Theta]\/2)\) + \[ImaginaryI]\ sin \((\ \[Theta]\/2)\))\)\)\)]], " \n\nwhere the domain is restricted to be ", Cell[BoxData[ \(r > 0\ \ and\ \ - \[Pi] < \[Theta] < \[Pi]\)]], ", then the derivative is given by \n\n\t\t", Cell[BoxData[ \(f' \((z)\) = \(1\/\(2 z\^\(1/2\)\) = r\^\(\(-1\)/2\)\/2\ \((cos \((\[Theta]\/2)\) - \[ImaginaryI]\ sin \ \((\[Theta]\/2)\))\)\)\)]], " \n\nwhere ", Cell[BoxData[ \(r > 0\ \ and\ \ - \[Pi] < \[Theta] < \[Pi]\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 3.10.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ "Enter ", Cell[BoxData[ RowBox[{\(u[r, \[Theta]]\), " ", StyleBox["and", FontFamily->"Times New Roman"], " ", \(v[r, \[Theta]]\)}]]], " and use them to form f[z]. Then verify that polar form of the \ Cauchy-Riemann equations hold." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[r, \[Theta], u, v];\)\ \), "\n", \(\(u[r_, \[Theta]_] = \(r\^\(1/2\)\) Cos[\[Theta]\/2];\)\ \), "\n", \(\(v[r_, \[Theta]_] = \(r\^\(1/2\)\) Sin[\[Theta]\/2];\)\ \), "\n", \(\(Ur\ = \[PartialD]\_r\ u[r, \[Theta]];\)\ \), "\n", \(\(U\[Theta]\ = \[PartialD]\_\[Theta]\ u[r, \[Theta]];\)\ \), "\n", \(\(Vr\ = \[PartialD]\_r\ v[r, \[Theta]];\)\ \), "\n", \(\(V\[Theta]\ = \[PartialD]\_\[Theta]\ v[r, \[Theta]];\)\ \), "\[IndentingNewLine]", \(\(Print["\", u[r, \[Theta]] + \[ImaginaryI]\ v[ r, \[Theta]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\"];\)\ \), "\n", \(\(Print["\", u[r, \[Theta]]];\)\ \), "\n", \(\(Print["\", u[r, \[Theta]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[PartialD]\_r\) u[r,\[Theta]] = \>\"", \ \[PartialD]\_r\ u[r, \[Theta]]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[PartialD]\_\[Theta]\) u[r,\[Theta]] = \>\"", \ \[PartialD]\_\[Theta]\ u[r, \[Theta]]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[PartialD]\_r\) v[r,\[Theta]] = \>\"", \ \[PartialD]\_r\ v[r, \[Theta]]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\[PartialD]\_\[Theta]\) v[r,\[Theta]] = \>\"", \ \[PartialD]\_\[Theta]\ v[r, \[Theta]]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\\"", \[PartialD]\_r\ u[r, \[Theta]] \[Equal] 1\/r\ \[PartialD]\_\[Theta]\ v[r, \[Theta]]];\)\ \), "\n", \(\(Print[\*"\"\\"", \[PartialD]\_r\ v[r, \[Theta]] \[Equal] \(-1\)\/r\ \[PartialD]\_\[Theta]\ u[r, \[Theta]]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ " \n Second, find the derivative using polar formulae ", Cell[BoxData[ \(f' \((z)\) = \(\[ExponentialE]\^\(\(-\[ImaginaryI]\)\ \[Theta]\)\) \ \((\[PartialD]\_r\ u[r, \[Theta]] + \[ImaginaryI]\ \[PartialD]\_r\ v[r, \[Theta]])\)\)]], " and ", Cell[BoxData[ \(f' \((z)\) = \(\[ExponentialE]\^\(\(-\[ImaginaryI]\)\ \[Theta]\)\) \ \((\[PartialD]\_r\ u[r, \[Theta]] + \[ImaginaryI]\ \[PartialD]\_r\ v[r, \[Theta]])\)\)]], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Print["\< f(z) = \>", u[r, \[Theta]] + \[ImaginaryI]\ v[ r, \[Theta]]];\)\ \), "\[IndentingNewLine]", \(\(Print["\< f(z) = u[r,\[Theta]] + \[ImaginaryI] v[r,\[Theta]]\>"];\)\ \ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\""];\)\ \), "\ \[IndentingNewLine]", \(\(Print["\", \(\[ExponentialE]\^\(\(-\[ImaginaryI]\)\ \ \[Theta]\)\) \((\[PartialD]\_r\ u[r, \[Theta]] + \[ImaginaryI]\ \[PartialD]\_r\ v[r, \[Theta]])\)];\)\ \), "\n", \(\(Print["\", ComplexExpand[\[ExponentialE]\^\(\(-\[ImaginaryI]\)\ \[Theta]\)] \((\ \[PartialD]\_r\ u[r, \[Theta]] + \[ImaginaryI]\ \[PartialD]\_r\ v[r, \[Theta]])\)];\)\ \), "\n", \(\(Print["\", Simplify[ ComplexExpand[\[ExponentialE]\^\(\(-\[ImaginaryI]\)\ \[Theta]\)] \ \((\[PartialD]\_r\ u[r, \[Theta]] + \[ImaginaryI]\ \[PartialD]\_r\ v[r, \[Theta]])\)]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nWe have shown that if ", Cell[BoxData[ \(f \((z)\) = \(z\^\(1/2\) = r\^\(1/2\)\ \((cos \((\[Theta]\/2)\) + \[ImaginaryI]\ sin \((\ \[Theta]\/2)\))\)\)\)]], " where the domain is restricted to be ", Cell[BoxData[ \(r > 0\ \ and\ \ - \[Pi] < \[Theta] < \[Pi]\)]], ", then the derivative is given by ", Cell[BoxData[ \(f' \((z)\) = \(1\/\(2 z\^\(1/2\)\) = r\^\(\(-1\)/2\)\/2\ \((cos \((\[Theta]\/2)\) - \[ImaginaryI]\ sin \ \((\[Theta]\/2)\))\)\)\)]], " where ", Cell[BoxData[ \(r > 0\ \ and\ \ - \[Pi] < \[Theta] < \[Pi]\)]], ". 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