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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 67769, 1899]*) (*NotebookOutlinePosition[ 92128, 2784]*) (* CellTagsIndexPosition[ 92020, 2777]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell[TextData[{ StyleBox["COMPLEX ANALYSIS: ", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["Mathematica 4.1", FontSize->18, FontSlant->"Italic", FontColor->RGBColor[1, 0, 1]], StyleBox[" Notebooks\n(c) John H. Mathews, and\nRussell W. 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That is, a complex number z is ", StyleBox["defined to be", FontSlant->"Italic"], " \n\t\t\t", Cell[BoxData[ \(z = \((x, y)\)\)]], ", \nwhere x and y are both real numbers.\n\tThe reason we say ", StyleBox["ordered", FontSlant->"Italic"], " pair is because we are thinking of a point in the plane. The point (2, \ 3), for example, is not the same as (3, 2). The ", StyleBox["order", FontSlant->"Italic"], " in which we write x and y in the equation makes a difference. Clearly, \ then, two complex numbers are ", StyleBox["equal", FontSlant->"Italic"], " if and only if their x coordinates are and their y coordinates are equal. \ In other words,\n\t\t(x, y)=(u, v) iff x=u and y=v.\n\t(Throughout this text, \ iff means if and only if.)\n\n\tIf we are to have a meaningful number system, \ there needs to be a method for combining these ordered pairs. 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", StyleBox["(a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], StyleBox[" ", FontWeight->"Bold"], "Find ", Cell[BoxData[ \(z\_1 + z\_2\)]], " and ", StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], StyleBox[" ", FontWeight->"Bold"], Cell[BoxData[ \(z\_1 - z\_2\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 1.1. (a).", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ StyleBox["(a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], StyleBox[" ", FontWeight->"Bold"], "Compute ", Cell[BoxData[ \(z\_1 + z\_2\)]], "." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[V, w, z];\)\ \), "\[IndentingNewLine]", \(\(z\_1 = 3 + 7 \[ImaginaryI];\)\ \), "\[IndentingNewLine]", \(\(z\_2 = 5 - 6 \[ImaginaryI];\)\ \), "\n", \(\(w\_1 = z\_1 + z\_2;\)\ \), "\n", \(\(V\_1\ = \ Graphics[{Red, Arrow[{0, \ 0}, {Re[z\_1], Im[z\_1]}]}];\)\ \), "\n", \(\(V\_2\ = \ Graphics[{Green, Arrow[{Re[z\_1], Im[z\_1]}, {Re[w\_1], Im[w\_1]}]}];\)\ \), "\n", \(\(V\_3\ = \ Graphics[{Blue, Arrow[{0, \ 0}, {Re[w\_1], Im[w\_1]}]}];\)\ \), "\n", \(\(Show[V\_1, V\_2, V\_3, PlotRange \[Rule] {{0, 8}, {0, 7}}, \[IndentingNewLine]Ticks \[Rule] {Range[0, 8, 1], Range[0, 7, 1]}, \[IndentingNewLine]AspectRatio \[Rule] 7\/8, Axes \[Rule] True, AxesLabel \[Rule] {"\", "\"}];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_1\) = \>\""\ , \ z\_1];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_2\) = \>\""\ , \ z\_2];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_1\) + \!\(z\_2\) = \>\""\ , \ z\_1\ + \ z\_2];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 1.1. (b).", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ " ", StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], StyleBox[" ", FontWeight->"Bold"], "Compute ", Cell[BoxData[ \(z\_1 - z\_2\)]], "." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[w, z];\)\ \), "\[IndentingNewLine]", \(\(z\_1 = 3 + 7 \[ImaginaryI];\)\ \), "\[IndentingNewLine]", \(\(z\_2 = 5 - 6 \[ImaginaryI];\)\ \), "\n", \(\(w\_2 = z\_1 - z\_2;\)\ \), "\n", \(\(v1\ = \ Graphics[{Red, Arrow[{0, \ 0}, {Re[z\_1], Im[z\_1]}]}];\)\ \), "\n", \(\(v2\ = \ Graphics[{Green, Arrow[{Re[w\_2], Im[w\_2]}, {Re[z\_1], Im[z\_1]}]}];\)\ \), "\n", \(\(v3\ = \ Graphics[{Blue, Arrow[{0, \ 0}, {Re[w\_2], Im[w\_2]}]}];\)\ \), "\n", \(\(Show[v1, v2, v3, PlotRange \[Rule] {{\(-2\), 3}, {0, 13}}, Ticks \[Rule] {Range[\(-2\), 3, 1], Range[0, 13, 1]}, \[IndentingNewLine]AspectRatio \[Rule] 13\/5, Axes \[Rule] True, AxesLabel \[Rule] {"\", "\"}];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_1\) = \>\""\ , \ z\_1];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_2\) = \>\""\ , \ z\_2];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_1\) - \!\(z\_2\) = \>\""\ , \ z\_1\ - \ z\_2];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n\tAt this point, it is tempting to define the product ", Cell[BoxData[ \(z\_1\ z\_2\)]], " as ", Cell[BoxData[ \(z\_1\ z\_2\ = \((x\_1\ x\_2, y\_1\ y\_2)\)\)]], ". It turns out, however, that this is not a good definition, and you will \ be asked in the problem set for this section to explain why. How, then, \ should products be defined? Again, if we equate (x,y) with x+iy and assume, \ for the moment, that ", Cell[BoxData[ \(\[ImaginaryI] = \@\(-1\)\)]], " makes sense (so that ", Cell[BoxData[ \(\[ImaginaryI]\^2 = \(-1\)\)]], "), we have \n\n\t\t\t", Cell[BoxData[ \(z\_1\ z\_2\ = \((x\_1, \ y\_1)\) \((x\_2, y\_2)\)\)]], " \n\t\t\t", Cell[BoxData[ \(\(\(\ \ \ \ \ \ \ \ \ \ \)\(\(=\)\(\((x\_1 + \[ImaginaryI]\ y\_1)\) \ \((x\_2 + \[ImaginaryI]y\_2)\)\)\)\)\)]], " \n\t\t\t", Cell[BoxData[ \(\(\(\ \ \ \ \ \ \ \ \ \ \)\(\(=\)\(x\_1\ x\_2 + \[ImaginaryI]\ x\_1\ \ y\_2 + \[ImaginaryI]\ x\_2\ y\_1 + \[ImaginaryI]\^2\ y\_1\ y\_2\)\)\)\)]], " \n\t\t\t", Cell[BoxData[ \(\(\(\ \ \ \ \ \ \ \ \ \ \)\(\(=\)\(x\_1\ x\_2 - y\_1\ y\_2 + \[ImaginaryI]\ \((x\_1\ y\_2 + x\_2\ y\_1)\)\)\)\)\)]], " \n\t\t\t", Cell[BoxData[ \(\(\(\ \ \ \ \ \ \ \ \ \ \)\(\(=\)\((x\_1\ x\_2 - y\_1\ y\_2, x\_1\ y\_2 + x\_2\ y\_1)\)\)\)\)]], " \n\n\tThus, it appears we are forced into the following definition." }], "Text"], Cell[TextData[{ StyleBox["\nDefinition 1.3, (", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ButtonBox["Complex Multiplication", ButtonData:>{ URL[ "http://mathworld.wolfram.com/ComplexMultiplication.html"], None}, ButtonStyle->"Hyperlink"], StyleBox["), Page 8.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], StyleBox[" \nFormula (1.8)", FontColor->RGBColor[0, 0, 1]], " We can derive the general formula ", Cell[BoxData[ \(z\_1\ z\_2\ = \(\(\((x\_1, \ y\_1)\) \((x\_2, y\_2)\)\)\(=\)\)\)]], Cell[BoxData[ \(\((x\_1\ x\_2 - y\_1\ y\_2, x\_1\ y\_2 + x\_2\ y\_1)\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Derivation of Formula (1.8).", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[x, y, z];\)\ \), "\n", \(\(z\_1 = \ x\_1\ + \ \[ImaginaryI]\ y\_1;\)\ \), "\n", \(\(z\_2 = \ x\_2\ + \ \[ImaginaryI]\ y\_2;\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_1\) = \>\""\ , \ z\_1];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_2\) = \>\""\ , \ z\_2];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_1\) \!\(z\_2\) = \>\""\ , \ z\_1\ z\_2];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_1\) \!\(z\_2\) = \>\""\ , \ ComplexExpand[z\_1\ z\_2]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ StyleBox["\nExample 1.2, Page 8.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Given ", Cell[BoxData[ \(z\_1 = \(3 + 7 \[ImaginaryI]\ \ \ and\ \ \ z\_2 = 5 - 6 \[ImaginaryI]\)\)]], ". Find ", Cell[BoxData[ \(\(z\_1\) z\_2\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 1.2.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ "Compute ", Cell[BoxData[ \(\(z\_1\) z\_2\)]], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[V, w, z];\)\ \), "\[IndentingNewLine]", \(\(z\_1 = 3 + 7 \[ImaginaryI];\)\ \), "\[IndentingNewLine]", \(\(z\_2 = 5 - 6 \[ImaginaryI];\)\ \), "\n", \(\(w\_3 = \(z\_1\) z\_2;\)\ \), "\n", \(\(V\_1\ = \ Graphics[{Red, Arrow[{0, \ 0}, {Re[z\_1], Im[z\_1]}]}];\)\ \), "\n", \(\(V\_2\ = \ Graphics[{Green, Arrow[{0, 0}, {Re[z\_2], Im[z\_2]}]}];\)\ \), "\n", \(\(V\_3 = \ Graphics[{Blue, Arrow[{0, \ 0}, {Re[w\_3], Im[w\_3]}]}];\)\ \), "\n", \(\(Show[V\_1, V\_2, V\_3, PlotRange \[Rule] {{0, 57}, {\(-6\), 17}}, Ticks \[Rule] {Range[0, 57, 5], Range[\(-6\), 17, 2]}, \[IndentingNewLine]AspectRatio \[Rule] 1, Axes \[Rule] True, AxesLabel \[Rule] {"\", "\"}];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_1\) = \>\""\ , \ z\_1];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_2\) = \>\""\ , \ z\_2];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_1\) \!\(z\_2\) = \>\""\ , \ \(z\_1\) z\_2];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\tTo motivate our definition for division, we will proceed along the same \ lines as we did for multiplication, assuming ", Cell[BoxData[ \(z\_2 \[NotEqual] 0\)]], ": \n\n\t\t\t", Cell[BoxData[ \(z\_1\/z\_2 = \(\((x\_1, \ y\_1)\)\/\((x\_2, y\_2)\) = \(x\_1 + \ \[ImaginaryI]\ y\_1\)\/\(x\_2 + \[ImaginaryI]y\_2\)\)\)]], " \n\t\n\tAt this point we need to figure out a way to be able to write \ the preceding quantity in the form x+iy. To do this, we use a standard trick \ and multiply the numerator and denominator by ", Cell[BoxData[ \(x\_2 + \[ImaginaryI]y\_2\)]], ". This gives \n\n\t\t\t", Cell[BoxData[ \(z\_1\/z\_2 = \(\((x\_1 + \[ImaginaryI]\ y\_1)\) \((x\_2 - \ \[ImaginaryI]\ y\_2)\)\)\/\(\((x\_2 + \[ImaginaryI]\ y\_2)\) \((x\_2 - \ \[ImaginaryI]\ y\_2)\)\)\)]], " \n\t\t\t", Cell[BoxData[ \(\(\(\ \ \ \ \ \ \)\(\(=\)\(\(x\_1\ x\_2 + y\_1\ y\_2 + \[ImaginaryI]\ \((\(-x\_1\)\ y\_2 + x\_2\ y\_1)\)\)\/\(x\_2\%2 + y\_2\%2\)\)\)\)\)]], " \n\t\t\t", Cell[BoxData[ \(\(\(\ \ \ \ \ \ \)\(\(=\)\(\(x\_1\ x\_2 + y\_1\ y\_2\)\/\(x\_2\%2 + y\ \_2\%2\) + \[ImaginaryI] \(\(\ \)\(\(-x\_1\)\ y\_2 + x\_2\ \ y\_1\)\)\/\(x\_2\%2 + y\_2\%2\)\)\)\)\)]], "\n\t\t\t ", Cell[BoxData[ \(\(\(\ \ \ \ \)\(\(=\)\((\(x\_1\ x\_2 + y\_1\ y\_2\)\/\(x\_2\%2 + y\_2\ \%2\), \(\(\ \)\(\(-x\_1\)\ y\_2 + x\_2\ y\_1\)\)\/\(x\_2\%2 + y\_2\%2\))\)\)\ \)\)]], ". \n\nThus, we finally arrive at a rather odd definition." }], "Text"], Cell[TextData[{ "\n", StyleBox["Definition 1.4, (", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ButtonBox["Complex Division", ButtonData:>{ URL[ "http://mathworld.wolfram.com/ComplexDivision.html"], None}, ButtonStyle->"Hyperlink"], StyleBox["), Page 9.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], StyleBox[" \nFormula (1.9)", FontColor->RGBColor[0, 0, 1]], " We can derive the general formula ", Cell[BoxData[ \(z\_1\/z\_2 = \(\(\((x\_1, \ y\_1)\)\/\((x\_2, y\_2)\)\)\(=\)\)\)]], Cell[BoxData[ \(\((\(x\_1\ x\_2 + y\_1\ y\_2\)\/\(x\_2\%2 + y\_2\%2\), \(\(\ \)\(\(-x\ \_1\)\ y\_2 + x\_2\ y\_1\)\)\/\(x\_2\%2 + y\_2\%2\))\)\)]], ", for ", Cell[BoxData[ \(z\_2 \[NotEqual] 0\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Derivation of Formula (1.9).", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[x, y, z];\)\ \), "\n", \(\(z\_1 = \ x\_1\ + \ \[ImaginaryI]\ y\_1;\)\ \), "\n", \(\(z\_2 = \ x\_2\ + \ \[ImaginaryI]\ y\_2;\)\ \), "\[IndentingNewLine]", \(\(n = Expand[\(z\_1\) \((x\_2 - \[ImaginaryI]\ y\_2)\)];\)\ \), "\ \[IndentingNewLine]", \(\(d = Expand[\(z\_2\) \((x\_2 - \[ImaginaryI]\ y\_2)\)];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_1\) = \>\""\ , \ z\_1];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_2\) = \>\""\ , \ z\_2];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_1\/z\_2\) = \>\""\ , \ z\_1\/z\_2];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_1\/z\_2\) = \>\""\ , n\/d];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(z\_1\/z\_2\) = \>\""\ , \(ComplexExpand[Re[n]] + \ \[ImaginaryI]\ ComplexExpand[Im[n]]\)\/d];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"] }, Closed]], Cell[TextData[{ StyleBox["\nExample 1.3, Page 9.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Given ", Cell[BoxData[ \(z\_1 = \(3 + 7 \[ImaginaryI]\ \ \ and\ \ \ z\_2 = 5 - 6 \[ImaginaryI]\)\)]], ". Find ", Cell[BoxData[ \(z\_1\/z\_2\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 1.3.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ "Compute ", Cell[BoxData[ \(z\_1\/z\_2\)]], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[V, w, z];\)\ \), "\[IndentingNewLine]", \(\(z\_1 = 3 + 7 \[ImaginaryI];\)\ \), "\[IndentingNewLine]", \(\(z\_2 = 5 - 6 \[ImaginaryI];\)\ \), "\n", \(\(w\_4 = z\_1\/z\_2;\)\ \), "\n", \(\(V\_1 = \ Graphics[{Red, Arrow[{0, \ 0}, {Re[z\_1], Im[z\_1]}]}];\)\ \), "\n", \(\(V\_2\ = \ Graphics[{Green, Arrow[{0, 0}, {Re[z\_2], Im[z\_2]}]}];\)\ \), "\n", \(\(V\_3\ = \ Graphics[{Blue, Arrow[{0, \ 0}, {Re[w\_4], Im[w\_4]}]}];\)\ \), "\n", \(\(Show[V\_1, V\_2, V\_3, PlotRange \[Rule] {{\(-1\), 5}, {\(-6\), 7}}, Ticks \[Rule] {Range[\(-1\), 5, 1], Range[\(-6\), 7, 1]}, \[IndentingNewLine]AspectRatio \[Rule] 13\/6, Axes \[Rule] True, AxesLabel \[Rule] {"\", "\"}];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_1\) = \>\""\ , \ z\_1];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\< \!\(z\_2\) = \>\""\ , \ z\_2];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_1\/z\_2\) = \>\""\ , \ z\_1\/z\_2];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n\tThe technique most mathematicians would use to perform operations on \ complex numbers is to appeal to the notation x+iy and perform the algebraic \ manipulations, as we did here, rather than to apply the complicated looking \ definitions we gave for those operations on ordered pairs. This is a valid \ procedure since the x+iy notation was used as a guide to see how we should \ define the operations in the first place. It is important to remember, \ however, that the x+iy notation is nothing more than a convenient bookkeeping \ device for keeping track of how to manipulate ordered pairs. It is the \ ordered pair algebraic definitions that really form the foundation on which \ our complex number system is based. In fact, if you were to program a \ computer to do arithmetic on complex numbers, your program would perform \ calculations on ordered pairs, using exactly the definitions that we gave.\n\n\ \tIt turns out that our algebraic definitions give complex numbers all the \ properties we normally ascribe to the real number system. Taken together, \ they describe what algebraists call a ", ButtonBox["Field", ButtonData:>{ URL[ "http://mathworld.wolfram.com/Field.html"], None}, ButtonStyle->"Hyperlink"], ". In formal terms, a field is a set (in this case, the complex numbers) \ together with two binary operations (in this case, addition and \ multiplication) with the following properties:" }], "Text"], Cell[TextData[{ "\n", StyleBox["(P1) Commutative Law for Addition.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " \n", StyleBox["Property (P1), Page 10.", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontColor->RGBColor[1, 0, 1]], "We can derive the general formula ", Cell[BoxData[ \(z\_1 + z\_2\ = \ z\_2 + z\_1\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Derivation of Property (P1).", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[x, y, z];\)\ \), "\n", \(\(z\_1\ = \ x\_1\ + \ \[ImaginaryI]\ y\_1;\)\ \), "\n", \(\(z\_2\ = \ x\_2\ + \ \[ImaginaryI]\ y\_2;\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_1\) = \>\""\ , z\_1];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_2\) = \>\""\ , \ z\_2];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_1\) + \!\(z\_2\) = \>\""\ , \ z\_1 + z\_2];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_2\) + \!\(z\_1\) = \>\""\ , \ z\_2 + z\_1];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\\"", z\_1 + z\_2\ \[Equal] \ z\_2 + z\_1];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n", StyleBox["(P2) Associative Law for Addition.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " \n", StyleBox["Property (P2), Page 10.", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontColor->RGBColor[1, 0, 1]], "We can derive the general formula ", Cell[BoxData[ \(z\_1 + \((z\_2 + z\_3)\) = \((z\_1 + z\_2)\) + \ z\_3\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["(P3) Additive Identity.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], "\n", StyleBox["Property (P3), Page 10.", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontColor->RGBColor[1, 0, 1]], "There is a complex number ", Cell[BoxData[ \(\[Omega]\)]], " such that ", Cell[BoxData[ \(z + \[Omega] = z\)]], " for all complex numbers z. The number ", Cell[BoxData[ \(\[Omega]\)]], " is obviously the ordered pair (0,0). " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["(P4) Additive Inverses.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], "\n", StyleBox["Property (P4), Page 10.", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontColor->RGBColor[1, 0, 1]], "Given any complex number z, there is a complex number ", Cell[BoxData[ \(\[Eta]\)]], " (depending on z) with the property that ", Cell[BoxData[ \(z + \[Eta] = \((0, 0)\)\)]], ". Obviously, if ", Cell[BoxData[ \(z = \(\((x, y)\) = x + \[ImaginaryI]\ y\)\)]], ", the number ", Cell[BoxData[ \(\[Eta]\)]], " will be ", Cell[BoxData[ \(\[Eta] = \(\((\(-x\), \(-y\))\) = \(\(-x\) - \[ImaginaryI]\ y = \ \(-z\)\)\)\)]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["(P5) Commutative Law for Multiplication.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " \n", StyleBox["Property (P5), Page 10.", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontColor->RGBColor[1, 0, 1]], "We can derive the general formula ", Cell[BoxData[ \(\(z\_1\) z\_2\ = \ \(z\_2\) z\_1\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["(P6) Associative Law for Addition.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " \n", StyleBox["Property (P6), Page 10.", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontColor->RGBColor[1, 0, 1]], "We can derive the general formula ", Cell[BoxData[ \(z\_1 + \((\(z\_2\) z\_3)\) = \((\(z\_1\) z\_2)\) + \ z\_3\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["(P7) Multiplicative Identity.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], "\n", StyleBox["Property (P7), Page 10.", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontColor->RGBColor[1, 0, 1]], "There is a complex number ", Cell[BoxData[ \(\[Zeta]\)]], " such that ", Cell[BoxData[ \(\[Zeta]z = z\)]], " for all complex numbers z. As one might expect, it turns out that (1,0) \ is the unique complex number ", Cell[BoxData[ \(\[Zeta]\)]], " with this property. You will be asked to verify this in the problem set \ for this section. " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["(P8) Multiplicative Inverses.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], "\n", StyleBox["Property (P8), Page 10.", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontColor->RGBColor[1, 0, 1]], "Given any number z other than the number (0, 0), there is a complex number \ (depending on z) which we shall denote by ", Cell[BoxData[ \(z\^\(-1\)\)]], " with the property that ", Cell[BoxData[ \(z\ z\^\(-1\) = \(\((1, 0)\) = 1\)\)]], ". Given our definition for division, it seems reasonable that the number \ ", Cell[BoxData[ \(z\^\(-1\)\)]], " would be ", Cell[BoxData[ \(z\^\(-1\) = \(\((1, 0)\)\/z = 1\/z\)\)]], ". You will be asked to confirm this in the problem set for this section. \ " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["(P9) The Distributive Law.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " \n", StyleBox["Property (P 9), Page 10.", FontColor->RGBColor[0, 0, 1]], " We can derive the general formula ", Cell[BoxData[ \(\(z\_1\) \((z\_2 + z\_3)\) = \(z\_1\) \((z\_2 + z\_3)\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Derivation of Property (P 9).", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[x, y, z];\)\ \), "\n", \(\(z\_1 = x\_1 + \[ImaginaryI]\ y\_1;\)\ \), "\n", \(\(z\_2 = x\_2 + \[ImaginaryI]\ y\_2;\)\ \), "\n", \(\(z\_3 = x\_3 + \[ImaginaryI]\ y\_3;\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_1\) = \>\""\ , \ z\_1];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_2\) = \>\""\ , \ z\_2];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_3\) = \>\""\ , \ z\_3];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_1\)(\!\(z\_2\) + \!\(z\_3\)) = \>\""\ , \ \(z\_1\) \((z\_2 + z\_3)\)];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_1\) \!\(z\_2\) + \!\(z\_1\) \!\(z\_3\) = \>\""\ , \ \(z\_1\) z\_2 + z\_1\ z\_3];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_1\)(\!\(z\_2\) + \!\(z\_3\)) = \>\""\ , \ Expand[\(z\_1\) \((z\_2 + z\_3)\)]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_1\) \!\(z\_2\) + \!\(z\_1\) \!\(z\_3\) = \>\""\ , \ \ Expand[z\_1\ z\_2 + z\_1\ z\_3]];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\\"", Expand[\(z\_1\) \((z\_2 + z\_3)\)]\ \ \[Equal] \ \ Expand[ z\_1\ z\_2 + z\_1\ z\_3]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell["\<\ \tNone of these properties is difficult to prove. Most of the proofs make use \ of corresponding facts in the real number system. Thus, we may move freely between the notations x+iy and (x,y), depending on \ which is more convenient for the context in which we are working.\ \>", "Text"], Cell[TextData[{ "\n", StyleBox["Definition 1.5, (Real Part of z), Page 12.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontColor->RGBColor[0, 0, 1]], "The real part of z=x+iy, denoted by Re(z), is the real number x." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Definition 1.6, (Imaginary Part of z), Page 12.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontColor->RGBColor[0, 0, 1]], "The imaginary part of z=x+iy, denoted by Im(z), is the real number y." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Definition 1.7, (Conjugate of z)", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], StyleBox[", Page 12.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontColor->RGBColor[0, 0, 1]], "The conjugate of z=x+iy, denoted by ", Cell[BoxData[ \(\(\(z\)\(\ \)\)\&_\)]], ", is the complex number (x,-y) = x - iy." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ \tThe following theorem gives some important facts relating to these \ operations. You will be asked for a proof in the exercises\ \>", "Text"], Cell[TextData[{ StyleBox["\nExample 1.4, Page 12.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Given ", Cell[BoxData[ \(z\_1 = \(\(-3\) + 7 \[ImaginaryI]\ \ \ and\ \ \ z\_2 = 9 + 4 \[ImaginaryI]\)\)]], ". \n", StyleBox["1.4", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " ", StyleBox["(a) ", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], "Find ", Cell[BoxData[ \(Re[z\_1]\ \ and\ \ \ Re[z\_2]\)]], ". \n", StyleBox["1.4", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " ", StyleBox["(b) ", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], "Find ", Cell[BoxData[ \(Im[z\_1]\ \ and\ \ \ Im[z\_2]\)]], ". \n", StyleBox["1.4", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " ", StyleBox["(c) ", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], "Find ", Cell[BoxData[ \(\(z\_1\)\&_\ \ \ and\ \ \ \(z\_2\)\&_\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 1.4 (a).", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ StyleBox["(a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Compute ", Cell[BoxData[ \(Re[z\_1]\)]], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[V, z];\)\ \), "\n", \(\(z\_1\ = \ \(-3\)\ + \ 7 \[ImaginaryI];\)\ \), "\n", \(\(V\_1 = \ Graphics[{Red, Arrow[{0, \ 0}, {Re[z\_1], Im[z\_1]}]}];\)\ \), "\n", \(\(V\_2\ = \ Graphics[{Blue, Arrow[{0, \ 0}, {Re[z\_1], 0}]}];\)\ \), "\n", \(\(Show[V\_1, V\_2, PlotRange \[Rule] {{\(-3\), 0}, {0, 7}}, \[IndentingNewLine]Ticks \[Rule] {Range[\(-3\), 0, 1], Range[0, 7, 1]}, \[IndentingNewLine]AspectRatio \[Rule] 7\/3, Axes \[Rule] True, AxesLabel \[Rule] {"\", "\"}];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_1\) = \>\""\ , \ z\_1];\)\ \), "\n", \(\(Print[\*"\"\\""\ , \ Re[z\_1]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Compute ", Cell[BoxData[ \(Re[z\_2]\)]], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[V, z];\)\ \), "\n", \(\(z\_2\ = \ 9\ + \ 4 \[ImaginaryI];\)\ \), "\n", \(\(V\_1\ = \ Graphics[{Red, Arrow[{0, \ 0}, {Re[z\_2], Im[z\_2]}]}];\)\ \), "\n", \(\(V\_2\ = \ Graphics[{Blue, Arrow[{0, \ 0}, {Re[z\_2], 0}]}];\)\ \), "\n", \(\(Show[V\_1, V\_2, PlotRange \[Rule] {{0, 9}, {0, 4}}, \[IndentingNewLine]Ticks \[Rule] {Range[0, 9, 1], Range[0, 4, 1]}, \[IndentingNewLine]AspectRatio \[Rule] 4\/9, Axes \[Rule] True, AxesLabel \[Rule] {"\", "\"}];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_2\) = \>\""\ , \ z\_2];\)\ \), "\n", \(\(Print[\*"\"\\""\ , \ Re[z\_2]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 1.4 (b).", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ StyleBox["(a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Compute ", Cell[BoxData[ \(Im[z\_1]\)]], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[V, z];\)\ \), "\[IndentingNewLine]", \(\(z\_1\ = \ \(-3\)\ + \ 7 \[ImaginaryI];\)\ \), "\n", \(\(V\_1\ = \ Graphics[{Red, Arrow[{0, \ 0}, {Re[z\_1], Im[z\_1]}]}];\)\ \), "\n", \(\(V\_2\ = \ Graphics[{Green, Arrow[{0, \ 0}, {0, Im[z\_1]}]}];\)\ \), "\n", \(\(Show[V\_1, V\_2, PlotRange \[Rule] {{\(-3\), 0.1}, {0, 7}}, \[IndentingNewLine]Ticks \[Rule] {Range[\(-3\), 0, 1], Range[0, 7, 1]}, \[IndentingNewLine]AspectRatio \[Rule] 7\/3, Axes \[Rule] True, AxesLabel \[Rule] {"\", "\"}];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_1\) = \>\""\ , \ z\_1];\)\ \), "\n", \(\(Print[\*"\"\\""\ , \ Im[z\_1]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Compute ", Cell[BoxData[ \(Im[z\_2]\)]], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[V, z];\)\ \), "\[IndentingNewLine]", \(\(z\_2\ = \ 9\ + \ 4 \[ImaginaryI];\)\ \), "\n", \(\(V\_1\ = \ Graphics[{Red, Arrow[{0, \ 0}, {Re[z\_2], Im[z\_2]}]}];\)\ \), "\n", \(\(V\_2\ = \ Graphics[{Green, Arrow[{0, \ 0}, {0, Im[z\_2]}]}];\)\ \), "\n", \(\(Show[V\_1, V\_2, PlotRange \[Rule] {{0, 9}, {0, 4}}, \[IndentingNewLine]Ticks \[Rule] {Range[0, 9, 1], Range[0, 4, 1]}, \[IndentingNewLine]AspectRatio \[Rule] 4\/9, Axes \[Rule] True, AxesLabel \[Rule] {"\", "\"}];\)\ \), "\n", \(\(Print[\*"\"\< \!\(z\_2\) = \>\""\ , \ z\_2];\)\ \), "\n", \(\(Print[\*"\"\\""\ , \ Im[z\_2]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 1.4 (c).", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ StyleBox["(a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Compute ", Cell[BoxData[ \(\(\(\(z\_1\)\&_\)\(\ \)\)\)]], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[V, z];\)\ \), "\[IndentingNewLine]", \(\(z\_1\ = \ \(-3\)\ + \ 7 \[ImaginaryI];\)\ \), "\n", \(\(V\_1\ = \ Graphics[{Red, Arrow[{0, \ 0}, {Re[z\_1], Im[z\_1]}]}];\)\ \), "\n", \(\(V\_2\ = \ Graphics[{Blue, Arrow[{0, \ 0}, {Re[z\_1], \(-Im[z\_1]\)}]}];\)\ \), "\n", \(\(Show[V\_1, V\_2, PlotRange \[Rule] {{\(-3\), 0}, {\(-7\), 7}}, \[IndentingNewLine]Ticks \[Rule] {Range[\(-3\), 0, 1], Range[\(-7\), 7, 1]}, \[IndentingNewLine]AspectRatio \[Rule] 7\/3, Axes \[Rule] True, AxesLabel \[Rule] {"\", "\"}];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_1\) = \>\""\ , \ z\_1];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\(z\_1\)\&_\) = \>\""\ , Conjugate[z\_1]\ ];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Compute ", Cell[BoxData[ \(\(z\_2\)\&_\)]], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[V, z];\)\ \), "\[IndentingNewLine]", \(\(z\_2\ = \ 9\ + \ 4 \[ImaginaryI];\)\ \), "\n", \(\(V\_1\ = \ Graphics[{Red, Arrow[{0, \ 0}, {Re[z\_2], Im[z\_2]}]}];\)\ \), "\n", \(\(V\_2\ = \ Graphics[{Blue, Arrow[{0, \ 0}, {Re[z\_2], \(-Im[z\_2]\)}]}];\)\ \), "\n", \(\(Show[V\_1, V\_2, PlotRange \[Rule] {{0, 9}, {\(-4\), 4}}, \[IndentingNewLine]Ticks \[Rule] {Range[0, 8, 1], Range[\(-4\), 4, 1]}, \[IndentingNewLine]AspectRatio \[Rule] 8\/9, Axes \[Rule] True, AxesLabel \[Rule] {"\", "\"}];\)\ \), "\n", \(\(Print[\*"\"\<\!\(z\_2\) = \>\""\ , \ z\_2];\)\ \), "\n", \(\(Print[\*"\"\<\!\(\(z\_2\)\&_\) = \>\""\ , \ Conjugate[z\_2]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input", AspectRatioFixed->True] }, Closed]], Cell[TextData[{ "\n", StyleBox["Theorem 1.1, Page 12.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Suppose ", Cell[BoxData[ RowBox[{\(z\_1\), ",", RowBox[{\(z\_2\), " ", StyleBox["and", FontFamily->"Times New Roman", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}], " ", \(z\_3\)}]}]]], " are arbitrary complex numbers. Then if ", Cell[BoxData[ \(z\_2 \[NotEqual] 0\)]], " \n\n\t\t", Cell[BoxData[GridBox[{ {\(\((1.10)\)\(\ \ \ \ \ \)\), \(\(\(\(\(\(z\)\(\ \)\)\&_\)\(\ \)\)\ \&_\)\(=\)\(z\)\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)\)}, {\(\((1.11)\)\(\ \ \ \ \ \)\), \(\(z\_1 + z\_2\)\&_\ = \ \(\(\(z\_1\)\&_\)\(+\)\(\(z\_2\)\&_\)\(\ \ \ \ \)\)\)}, {\(\((1.12)\)\(\ \ \ \ \ \)\), \(\(\(z\_1\) z\_2\)\&_\ = \ \(\(\(z\_1\)\&_\)\(\ \ \)\(\(z\_2\)\&_\)\(\ \ \ \ \ \ \ \ \ \ \)\)\)}, {\(\((1.13)\)\(\ \ \ \ \ \)\), RowBox[{\(\(z\_1\/z\_2\)\&_\), " ", "=", RowBox[{ RowBox[{\(\(z\_1\)\&_\/\(\(\ \)\(\(\(z\_2\)\(\ \)\)\&_\)\(\ \ \)\)\), StyleBox["if", FontFamily->"Times New Roman", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}], " ", \(z\_2\)}], "\[NotEqual]", "0"}]}]}, {\(\((1.14)\)\(\ \ \ \ \ \)\), \(\(Re[ z]\)\(=\)\(\(\(\ \)\(z\ + \ \ \(\(z\)\(\ \)\)\&_\)\)\/2\)\(\ \ \ \ \ \ \ \ \)\)}, {\(\((1.15)\)\(\ \ \ \ \ \)\), \(\(Im[ z]\)\(=\)\(\(\(\ \)\(z\ - \ \ \(\(z\)\(\ \)\)\&_\)\)\/\(2 \ \[ImaginaryI]\)\)\(\ \ \ \ \ \ \ \ \)\)}, {\(\((1.16)\)\(\ \ \ \ \ \)\), \(Re[\[ImaginaryI]\ z] = \(-\ Im[z]\)\)}, {\(\((1.17)\)\(\ \ \ \ \ \)\), \(\(Im[\[ImaginaryI]\ z]\)\(=\)\(Re[ z]\)\(\ \ \ \)\)} }]]], "\n" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n\tBecause of what it erroneously connotes, it is a shame that the term \ ", StyleBox["imaginary", FontSlant->"Italic"], " is used in definition the definition of complex numbers. It was coined by \ the brilliant mathematician and philosopher ", ButtonBox["Ren\[EAcute] Descartes", ButtonData:>{ URL[ "http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Descartes.\ html"], None}, ButtonStyle->"Hyperlink"], " (1596-1650) during an era when quantities such as ", Cell[BoxData[ \(\@\(-1\)\)]], " were thought to be just that. Gauss, who was successful in getting \ mathematicians to adopt the phrase ", StyleBox["complex number", FontSlant->"Italic"], " rather than ", StyleBox["imaginary number", FontSlant->"Italic"], ", also suggested that we use ", StyleBox["lateral part", FontSlant->"Italic"], " of z in place of ", StyleBox["imaginary part", FontSlant->"Italic"], " of z. Unfortunately, this suggestion never caught on, and it appears we \ are stuck with what history has handed down to us." }], "Text"], Cell[CellGroupData[{ Cell[TextData[StyleBox["Library Research Experience for Undergraduates", FontSize->16, FontWeight->"Bold", FontColor->RGBColor[0.500008, 0, 0.996109]]], "Text"], Cell[TextData[{ StyleBox["Project I. Write a report about complex numbers.", FontSize->16, FontWeight->"Bold", FontColor->RGBColor[0.996109, 0, 0.996109]], "\n\n", StyleBox["1.", FontWeight->"Bold"], " Apostol, Tom M., Gulbank D. Chakerian, Geraldine C. Darden and John D. \ Neff (Editors), (1977), Selected Papers on Precalculus: Chapter 2(c) Complex \ Numbers, The Math. Asso. of Am.\n\n", StyleBox["2.", FontWeight->"Bold"], " Ballantine, J. P. ''Equiangular Complex Numbers (in Classroom Notes),'' \ Am. Math. M., Vol. 67, No. 7. (Aug. - Sep., 1960), pp. 674-678, ", StyleBox[ButtonBox["Jstor.", ButtonData:>{ URL[ "http://www.jstor.org/"], None}, ButtonStyle->"Hyperlink"], FontWeight->"Bold"], " \n\n", StyleBox["3.", FontWeight->"Bold"], " Baker, John A. and Willem Sluis, (1997), ''Arg,'' Math. Mag., Vol. 70, \ No. 4, p. 291.\n\n", StyleBox["4.", FontWeight->"Bold"], " Blake, Robert G. (1962), ''Plane Geometry and Complex Numbers'', Math. \ Mag., Vol. 35 , pp. 239-242.\n\n", StyleBox["5.", FontWeight->"Bold"], " Bosch, W. and Krajkiewicz, P., (1970), ''A Categorical System of Axioms \ for the Complex Numbers,'' Math. Mag., Vol. 43, pp. 67-70.\n\n", StyleBox["6.", FontWeight->"Bold"], " Fuchs, L. and T. Szele, ''Introduction of Complex Numbers as Vectors of \ the Plane (in Classroom Notes),'' Am. Math. M., Vol. 59, No. 9. (Nov., 1952), \ pp. 628-631, ", StyleBox[ButtonBox["Jstor.", ButtonData:>{ URL[ "http://www.jstor.org/"], None}, ButtonStyle->"Hyperlink"], FontWeight->"Bold"], " \n\n", StyleBox["7.", FontWeight->"Bold"], " Holden, Herbert L., (1981), ''Applying Complex Arithmetic,'' T. Y. C. \ Math. J., V. 12, No. 3, pp. 190-194.\n\n", StyleBox["8.", FontWeight->"Bold"], " Jackson, Margaret, (1987), ''Complex Numbers and Pythagorean Triples,'' \ The Math. Gazette, V. 71, No. 456, p. 127\n\n", StyleBox["9.", FontWeight->"Bold"], " Kimberling, Clark, (1987), ''Power of Complex Numbers,'' The Math. \ Teach., V. 80, No. 1, pp. 63-67.\n\n", StyleBox["10.", FontWeight->"Bold"], " Kittappa, R., ''Complex Variables and Line-Coordinates, (1968), '' Math. \ Mag., Vol.. 41, pp. 269-272.\n\n", StyleBox["11.", FontWeight->"Bold"], " McGaughey, A. W., ''The Imaginary Number Problem (in Classroom Notes),'' \ Am. Math. M., Vol. 64, No. 3. (Mar., 1957), pp. 193-194., ", StyleBox[ButtonBox["Jstor.", ButtonData:>{ URL[ "http://www.jstor.org/"], None}, ButtonStyle->"Hyperlink"], FontWeight->"Bold"], " \n\n", StyleBox["12.", FontWeight->"Bold"], " Phillips, George M., ''Archimedes and the Complex Plane,'' Am. Math. M., \ Vol. 91, No. 2. (Feb., 1984), pp. 108-114, ", StyleBox[ButtonBox["Jstor.", ButtonData:>{ URL[ "http://www.jstor.org/"], None}, ButtonStyle->"Hyperlink"], FontWeight->"Bold"], " \n\n", StyleBox["13.", FontWeight->"Bold"], " Pineda, Anton and David C. Arney, (1991), ''Extending the Complex Number \ System,'' Math. and Comp. Ed., V. 25, No. 1, pp. 10-16.\n\n", StyleBox["14.", FontWeight->"Bold"], " Potts, D. H., ''Axioms for Complex Numbers (in Math. Notes) Am. Math. M., \ Vol. 70, No. 9. (Nov., 1963), pp. 974-975, ", StyleBox[ButtonBox["Jstor.", ButtonData:>{ URL[ "http://www.jstor.org/"], None}, ButtonStyle->"Hyperlink"], FontWeight->"Bold"], " \n\n", StyleBox["15.", FontWeight->"Bold"], " Richmond, D. E., ''Complex Numbers and Vector Algebra (in Classroom \ Notes),'' Am. Math. M., Vol. 58, No. 9. (Nov., 1951), pp. 622-628, ", StyleBox[ButtonBox["Jstor.", ButtonData:>{ URL[ "http://www.jstor.org/"], None}, ButtonStyle->"Hyperlink"], FontWeight->"Bold"], " \n\n", StyleBox["16.", FontWeight->"Bold"], " Richmond, D. E., ''Complex Numbers and Trigonometry,'' Am. Math. M., Vol. \ 64, No. 7. (Aug. - Sep., 1957), pp. 478-485, ", StyleBox[ButtonBox["Jstor.", ButtonData:>{ URL[ "http://www.jstor.org/"], None}, ButtonStyle->"Hyperlink"], FontWeight->"Bold"], " \n\n", StyleBox["17.", FontWeight->"Bold"], " Schelkunoff, S. A., ''A Note on Geometrical Applications of Complex \ Numbers (in Questions and Discussions),'' Am. Math. M., Vol. 37, No. 6. (Jun. \ - Jul., 1930), pp. 301-303, ", StyleBox[ButtonBox["Jstor.", ButtonData:>{ URL[ "http://www.jstor.org/"], None}, ButtonStyle->"Hyperlink"], FontWeight->"Bold"], " \n\n", StyleBox["18.", FontWeight->"Bold"], " Small, Lloyd L., ''Some Geometrical Applications of Complex Numbers,'' \ Am. Math. M., Vol. 36, No. 10. (Dec., 1929), pp. 504-511., ", StyleBox[ButtonBox["Jstor.", ButtonData:>{ URL[ "http://www.jstor.org/"], None}, ButtonStyle->"Hyperlink"], FontWeight->"Bold"], " \n\n", StyleBox["19.", FontWeight->"Bold"], " Subramaniam, K. B., (1979), ''On Euler's Formula ", Cell[BoxData[ \(\[ExponentialE]\^\(\[ImaginaryI]\ \[CapitalTheta]\) = cos\ \[CapitalTheta]\ + \ \[ImaginaryI]\ sin\ \[CapitalTheta]\)]], ",'' Int. J. of Math. Ed. in Sci. and Tech., V. 10, No. 2, pp. 279.\n\n", StyleBox["20.", FontWeight->"Bold"], " Webb, H. E., ''Discussion: Complex Numbers in Advanced Algebra (in \ Questions and Discussions),'' Am. Math. M., Vol. 27, No. 11. (Nov., 1920), \ pp. 411-413, ", StyleBox[ButtonBox["Jstor.", ButtonData:>{ URL[ "http://www.jstor.org/"], None}, ButtonStyle->"Hyperlink"], FontWeight->"Bold"], " \n\n", StyleBox["21.", FontWeight->"Bold"], " Weimer, Richard C., (1976), ''Can the Complex Numbers Be Ordered?,'' \ College Math. J., Vol. 7, No. 4, pp.10-12\n\n", StyleBox["22.", FontWeight->"Bold"], " Willson, William Wynne, (1970), ''An Approach to Complex Numbers,'' The \ Math. 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