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Cell[BoxData[ \(\((\(-1\))\)\^\[ImaginaryI]\)]], " without appealing to a number system beyond the framework of complex \ numbers. We will do this by taking note of some rudimentary properties of the \ complex exponential and logarithm, and then using our imagination.\n" }], "Text"], Cell[TextData[{ "\tWe begin by generalizing Identity (5-15). Equations (5-12) and (5-14) \ show that ", Cell[BoxData[ \(log \((z)\)\)]], " can be expressed as the set ", Cell[BoxData[ RowBox[{\(log \((z)\)\), "=", RowBox[{"{", RowBox[{\(Log \((z)\)\), "+", RowBox[{"\[ImaginaryI]", " ", "2", " ", "n", " ", RowBox[{"\[Pi]", ":", " ", RowBox[{"n", " ", StyleBox["is", FontFamily->"Times New Roman", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}], StyleBox[" ", FontFamily->"Times New Roman", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}], StyleBox["an", FontFamily->"Times New Roman", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}], StyleBox[" ", FontFamily->"Times New Roman", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}], StyleBox["integer", FontFamily->"Times New Roman", FontSize->12, FontWeight->"Plain", FontSlant->"Plain", FontColor->GrayLevel[0], FontVariations->{"Underline"->False, "StrikeThrough"->False}]}]}]}]}], "}"}]}]]], ". It is easy to show (we leave as an exercise) that for ", Cell[BoxData[ \(z \[NotEqual] 0\)]], ", ", Cell[BoxData[ \(\[ExponentialE]\^\(\(log\_\[Alpha]\) \((z)\)\) = z\)]], ", where ", Cell[BoxData[ \(\(log\_\[Alpha]\) \((z)\)\)]], " is any branch of the function ", Cell[BoxData[ \(log \((z)\)\)]], ". But this means that for any ", Cell[BoxData[ \(\[Zeta] \[Element] log \((z)\)\)]], ", the identity ", Cell[BoxData[ \(\[ExponentialE]\^\[Zeta] = z\)]], " holds true. Since ", Cell[BoxData[ \(\[ExponentialE]\^\(log \((z)\)\)\)]], " denotes the set ", Cell[BoxData[ \({\(\[ExponentialE]\^\[Zeta]\) : \ \[Zeta] \[Element] log \((z)\)}\)]], ", we see that ", Cell[BoxData[ \(\[ExponentialE]\^\(\(log\_\[Alpha]\) \((z)\)\) = z\)]], ", for ", Cell[BoxData[ \(z \[NotEqual] 0\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\tNext, we note that identity (5-17) gives us ", Cell[BoxData[ \(log \((z\^n)\) = n\ log \((z)\)\)]], ", where n is any natural number, so that ", Cell[BoxData[ \(\[ExponentialE]\^\(log \((z\^n)\)\) = \(\[ExponentialE]\^\(n\ log \ \((z)\)\) = z\^n\)\)]], " for ", Cell[BoxData[ \(z \[NotEqual] 0\)]], ". With these preliminaries out of the way we are now in a position to \ come up with a definition of what is meant by a complex number raised to a \ complex power." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["Definition 5.4, Page 177.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Let c be a complex number. We define ", Cell[BoxData[ StyleBox[\(z\^c\), FontColor->RGBColor[0, 0, 1]]], FontWeight->"Bold"], " as follows \n\n\t\t\t", Cell[BoxData[ StyleBox[\(z\^c = \ \[ExponentialE]\^\(c\ log \((z)\)\)\), FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]]]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n\tThis definition makes sense, since if both z and c are real \ numbers with ", Cell[BoxData[ \(z > 0\)]], ", and it gives the familiar (real) definition for ", Cell[BoxData[ \(z\^c\)]], ".\n\n\tSince ", Cell[BoxData[ \(log \((z)\)\)]], " is multivalued, the function ", Cell[BoxData[ \(z\^c\)]], " will in general be multivalued. The function f given by ", Cell[BoxData[ \(f \((z)\) = exp \((c\ Log \((z)\))\)\)]], " is called the ", StyleBox["principal branch", FontColor->RGBColor[0, 0, 1]], " of the multivalued function ", Cell[BoxData[ \(z\^c\)]], ". Note that the principal branch of ", Cell[BoxData[ \(z\^c\)]], " is obtained from the definition by replacing ", Cell[BoxData[ \(log \((z)\)\)]], " with the principal branch of the logarithm." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 5.6, Page 177.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find ", Cell[BoxData[ \(4\^\(1/2\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 5.6.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ "Enter ", Cell[BoxData[ \(z\_0 = 4\)]], " and find the principal value of ", Cell[BoxData[ \(4\^\(1/2\)\)]], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[w];\)\ \), "\n", \(\(Clear[e, pts, W];\)\ \), "\n", \(\(w\_0\ = \ 4\^\(1/2\);\)\ \), "\n", \(\(w\_1\ = \ \[ExponentialE]\^\(\(1\/2\) \((\ Log[4] + \[ImaginaryI]\ 2 \ n\ \[Pi])\)\);\)\ \), "\n", \(\(w\_2\ = \ \(e\^\(\(1\/2\) Log[4]\)\) \[ExponentialE]\^\(\(1\/2\) \[ImaginaryI]\ 2 n\ \ \[Pi]\);\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(w\_0\) = \!\(4\^\(1/2\)\)\>\""];\)\ \), "\n", \(\(Print[\*"\"\<\!\(w\_0\) = \>\"", w\_0];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(w\_1\) = \!\(4\^\(1/2\)\) = \>\"", w\_1];\)\ \), "\n", \(\(Print[\*"\"\<\!\(w\_1\) = \>\"", w\_1];\)\ \), "\n", \(\(w\_1\ = \ ComplexExpand[w\_1]\ ;\)\ \), "\n", \(\(Print[\*"\"\<\!\(w\_1\) = \>\"", w\_1];\)\ \), "\n", \(\(w\_1\ = \ Simplify[w\_1, Element[n, Integers]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(w\_1\) = \>\"", w\_1];\)\ \), "\n", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<\!\(w\_2\) = \!\(4\^\(1/2\)\) = \!\(e\^\(\(1\/2\) Log[4]\ \)\)\!\(\[ExponentialE]\^\(\(1\/2\) \[ImaginaryI]2n\[Pi]\)\)\>\""];\)\ \), "\n\ ", \(\(Print[\*"\"\<\!\(w\_2\) = \>\"", \(e\^\(\(1\/2\) Log[4]\)\) \[ExponentialE]\^\(\(1\/2\) \[ImaginaryI]\ 2 n\ \ \[Pi]\)];\)\ \), "\n", \(\(Print[\*"\"\<\!\(w\_2\) = \>\"", \(e\^\(\(1\/2\) Log[4]\)\) ComplexExpand[\[ExponentialE]\^\(\(1\/2\) \[ImaginaryI]\ 2 n\ \ \[Pi]\)]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(w\_2\) = \>\"", \(e\^\(\(1\/2\) Log[4]\)\) Simplify[\[ExponentialE]\^\(\(1\/2\) \[ImaginaryI]\ 2 n\ \[Pi]\), Element[n, Integers]]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(w\_2\) = \>\"", Simplify[e\^\(\(1\/2\) Log[4]\)] Simplify[\[ExponentialE]\^\(\(1\/2\) \[ImaginaryI]\ 2 n\ \[Pi]\), Element[n, Integers]]];\)\ \), "\n", \(\(Print[\*"\"\<\!\(w\_2\) = \>\"", Simplify[\[ExponentialE]\^\(\(1\/2\) \[ImaginaryI]\ 2 n\ \[Pi]\), Element[n, Integers]], "\< \>", Simplify[\[ExponentialE]\^\(\(1\/2\) Log[4]\)]];\)\ \)}], "Input"] }, Closed]], Cell[TextData[{ "\n", StyleBox["Example 5.7, Page 178. (a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the principal value of ", Cell[BoxData[ \(\((1 + \[ImaginaryI])\)\^\(1/2\)\)]], ", ", StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the principal value of ", Cell[BoxData[ \(\((\(-1\))\)\^\[ImaginaryI]\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 5.7.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ StyleBox["(a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Enter ", Cell[BoxData[ \(z\_0 = 1 + \[ImaginaryI]\)]], " and find the principal value of ", Cell[BoxData[ \(\((1 + \[ImaginaryI])\)\^\(1/2\)\)]], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[z, w];\)\ \), "\[IndentingNewLine]", \(\(Clear[e];\)\ \), "\n", \(\(z\_0\ = \ 1\ + \ \[ImaginaryI];\)\ \), "\n", \(\(w\_0\ = \ ComplexExpand[Log[z\_0]];\)\ \), "\n", \(\(w\_1\ = \ ComplexExpand[\[ExponentialE]\^\(1\/2\ Log[z\_0]\)];\)\ \), "\n", \(\(w\_2\ = \ ComplexExpand[\@z\_0];\)\ \), "\n", \(\(w\_3\ = N[\@z\_0];\)\ \), "\n", \(\(Print["\", z\_0, "\<] = \>", w\_0];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[e\^\(1\/2\ Log[z\_0]\), "\< = \>", w\_1];\)\ \), "\n", \(\(Print["\< \>"];\)\ \), "\n", \(\(Print["\< \>", \@z\_0, "\< = \>", w\_2];\)\ \), "\n", \(\(Print["\< \>", \@z\_0, "\< = \>", w\_3];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ StyleBox["\n(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the principal value of ", Cell[BoxData[ \(\((\(-1\))\)\^\[ImaginaryI]\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[w];\)\ \), "\[IndentingNewLine]", \(\(Clear[e];\)\ \), "\n", \(\(w\_1\ = \ \[ExponentialE]\^\(\[ImaginaryI]\ Log[\(-1\)]\);\)\ \), "\n\ ", \(\(w\_2\ = ComplexExpand[\((\(-1\))\)\^\[ImaginaryI]];\)\ \), "\n", \(\(w\_3\ = Chop[N[\((\(-1\))\)\^\[ImaginaryI]]];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\[ExponentialE]\^\(\[ImaginaryI]\\\ Log[\(-1\)]\)\) \ = \>\"", w\_1];\)\ \), "\n", \(\(Print["\< \>"];\)\ \), "\n", \(\(Print[\*"\"\< (-1\!\(\()\^\[ImaginaryI]\)\) = \>\"", w\_2];\)\ \), "\n", \(\(Print[\*"\"\< (-1\!\(\()\^\[ImaginaryI]\)\) = \>\"", w\_3];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"] }, Closed]], Cell[TextData[{ "\n\tLet us now consider the various possibilities that may arise in the \ definition of ", Cell[BoxData[ \(z\^c\)]], ". \n\n", StyleBox["Case (i).", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Suppose ", Cell[BoxData[ StyleBox[\(c = k\), FontColor->RGBColor[0, 0, 1]]]], " is an integer, i.e. ", Cell[BoxData[ StyleBox[\(c\ = \ \(k\ \ = \ ... \)\ , \(-2\), \(-1\), 0, 1, 2, ... \), FontColor->RGBColor[0, 0, 1]]]], " then ", Cell[BoxData[ RowBox[{ StyleBox[\(z\^k\), FontColor->RGBColor[0, 0, 1]], "=", " ", \(\(r\^k\) \((cos\ k\ \[Theta]\ + \ \[ImaginaryI]\ sin\ k\ \ \[Theta])\)\)}]], FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ". \n\nThis is easy to establish. If ", Cell[BoxData[ \(z = r\ \[ExponentialE]\^\(\[ImaginaryI]\ \[Theta]\)\)]], " then ", Cell[BoxData[ \(k\ log \((z)\) = k\ ln \((r)\) + \[ImaginaryI]\ k \((\[Theta] + 2\ \[Pi]\ n)\)\)]], ", where n is an integer, and we have \n", Cell[BoxData[ \(z\^k = \ \(\[ExponentialE]\^\(k\ log \((z)\)\) = \ \ \(\[ExponentialE]\^\(k\ ln \((r)\)\ + \ \[ImaginaryI]\ k \((\[Theta]\ + \ 2\ \ \[Pi]\ n)\)\) = \(\(\[ExponentialE]\^\(\(k\ ln \((r)\)\)\(\ \)\)\) \ \[ExponentialE]\^\(\[ImaginaryI]\ k\ \[Theta]\ + \ \[ImaginaryI]\ 2\ \[Pi]\ \ n\) = \(\(r\^k\) \[ExponentialE]\^\(\[ImaginaryI]\ k\ \[Theta]\) = \(r\^k\) \ \((cos\ k\ \[Theta]\ + \ \[ImaginaryI]\ sin\ k\[Theta])\)\)\)\)\)\)]], ". This is the single valued ", Cell[BoxData[ \(k\^th\)]], " power of z.\n\n", StyleBox["Case (ii).", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Suppose ", Cell[BoxData[ StyleBox[\(c = 1\/k\), FontColor->RGBColor[0, 0, 1]]]], " where k is a positive integer, i.e. ", Cell[BoxData[ StyleBox[\(c\ = \ \(1\/k\ \ = \ 1\/1\), 1\/2, 1\/3, 1\/4, ... \), FontColor->RGBColor[0, 0, 1]]]], " then ", Cell[BoxData[ RowBox[{ StyleBox[\(z\^\(1\/k\)\), FontColor->RGBColor[0, 0, 1]], "=", " ", \(\(r\^\(1\/k\)\) \((cos\ \(\[Theta]\ + \ 2 \[Pi]\ n\)\/k\ + \ \ \[ImaginaryI]\ sin\ \(\[Theta]\ + \ 2 \[Pi]\ n\)\/k)\)\)}]], FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " for ", Cell[BoxData[ StyleBox[\(n\ = \ 0, 1, 2, ... , k - 1\), FontColor->RGBColor[0, 0, 1]]]], ". \n\nThis is easy to establish. If ", Cell[BoxData[ \(z = r\ \[ExponentialE]\^\(\[ImaginaryI]\ \[Theta]\)\)]], " then ", Cell[BoxData[ \(1\/k\ log \((z)\) = 1\/k\ ln \((r)\) + \(\[ImaginaryI]\ \((\[Theta]\ + \ 2\ \[Pi]\ \ n)\)\)\/k\)]], ", where n is an integer, and we have \n", Cell[BoxData[ \(z\^\(1\/k\) = \ \(\[ExponentialE]\^\(1\/k\ log \((z)\)\) = \ \(\ \[ExponentialE]\^\(1\/k\ ln \((r)\)\ + \ \[ImaginaryI]\ \((\(\[Theta]\ + \ \ 2\ \[Pi]\ n\)\/k)\)\) = \(\(\[ExponentialE]\^\(\(1\/k\ ln \((r)\)\)\(\ \)\)\) \ \[ExponentialE]\^\(\[ImaginaryI]\ \((\(\[Theta]\ + \ 2\ \[Pi]\ n\)\/k)\)\) = \ \(r\^\(1\/k\)\) \((cos\ \(\[Theta]\ + \ 2 \[Pi]\ n\)\/k\ + \ \[ImaginaryI]\ \ sin\ \(\[Theta]\ + \ 2 \[Pi]\ n\)\/k)\)\)\)\)\)]], " for ", Cell[BoxData[ StyleBox[\(n\ = \ 0, 1, 2, ... , k - 1\), FontColor->GrayLevel[0]]]], ". This is the multi valued ", Cell[BoxData[ \(k\^th\)]], " root of z." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["Case (iii).", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Suppose ", Cell[BoxData[ StyleBox[\(c = j\/k\), FontColor->RGBColor[0, 0, 1]]]], " where j and k are a positive integers that have no common factors, i.e. \ then ", Cell[BoxData[ RowBox[{ StyleBox[\(z\^\(j\/k\)\), FontColor->RGBColor[0, 0, 1]], "=", " ", \(\(r\^\(j\/k\)\) \((cos\ \(\((\[Theta]\ + \ 2 \[Pi]\ n)\) j\)\ \/k\ + \ \[ImaginaryI]\ sin\ \(\((\[Theta]\ + \ 2 \[Pi]\ n)\) \ j\)\/k)\)\)}]], FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " for ", Cell[BoxData[ StyleBox[\(n\ = \ 0, 1, 2, ... , k - 1\), FontColor->RGBColor[0, 0, 1]]]], ". \n\nThis is easy to establish. If ", Cell[BoxData[ \(z = r\ \[ExponentialE]\^\(\[ImaginaryI]\ \[Theta]\)\)]], " then ", Cell[BoxData[ \(j\/k\ log \((z)\) = j\/k\ ln \((r)\) + \[ImaginaryI]\ \((\(\((\[Theta]\ + \ 2 \[Pi]\ n)\ \) j\)\/k)\)\)]], ", where n is an integer, and we have \n", Cell[BoxData[ \(z\^\(j\/k\) = \ \(\[ExponentialE]\^\(j\/k\ log \((z)\)\) = \ \(\(\ \[ExponentialE]\^\(\(j\/k\ ln \((r)\)\)\(\ \)\)\) \[ExponentialE]\^\(\ \[ImaginaryI]\ \((\(\((\[Theta]\ + \ 2 \[Pi]\ n)\) j\)\/k)\)\) = \ \(r\^\(j\/k\)\) \((cos\ \(\((\[Theta]\ + \ 2 \[Pi]\ n)\) j\)\/k\ + \ \ \[ImaginaryI]\ sin\ \(\((\[Theta]\ + \ 2 \[Pi]\ n)\) j\)\/k)\)\)\)\)]], " for ", Cell[BoxData[ StyleBox[\(n\ = \ 0, 1, 2, ... , k - 1\), FontColor->GrayLevel[0]]]], ". \n\n", StyleBox["Case (iv).", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Suppose ", Cell[BoxData[ StyleBox["c", FontColor->RGBColor[0, 0, 1]]]], " is not a rational number, then there are infinitely many values for ", StyleBox[" ", FontWeight->"Bold"], Cell[BoxData[ StyleBox[\(z\^c\), FontColor->RGBColor[0, 0, 1]]], FontWeight->"Bold"], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 5.8, Page 179.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find all the values of ", Cell[BoxData[ \(2\^\(1/9\ + \ \[ImaginaryI]/50\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 5.8.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ StyleBox["(a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Enter ", Cell[BoxData[ \(w\_0 = 2\^\(1/9\ + \ \[ImaginaryI]/50\)\)]], " and find the principal value. " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[w];\)\ \), "\[IndentingNewLine]", \(\(Clear[e, pts, W];\)\ \), "\n", \(\(W\ = \ e\^\(\((1\/9 + \[ImaginaryI]\/50)\) Log[2]\);\)\ \), "\ \[IndentingNewLine]", \(\(w\_0\ = \ 2\^\(1/9\ + \ \[ImaginaryI]/50\);\)\ \), "\[IndentingNewLine]", \(\(w\_1\ = \ \[ExponentialE]\^\(\((1\/9\ + \(\(\ \)\(\[ImaginaryI]\)\)\ \/50)\)\ Log[2]\);\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(w\_0\) = \>\"", W];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(w\_0\) = \>\"", ComplexExpand[w\_0]];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(w\_0\) = \>\"", \ N[w\_0]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(w\_1\) = \>\"", w\_1];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(w\_1\) = \>\"", N[w\_1]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\"", w\_0 \[Equal] w\_1];\)\ \), "\n", \(\)}], "Input"], Cell[TextData[{ "\n", StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Enter ", Cell[BoxData[ \(w\_0 = 2\^\(1/9\ + \ \[ImaginaryI]/50\)\)]], " and find some of its values. " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[w];\)\ \), "\[IndentingNewLine]", \(\(w\_0 = \[ExponentialE]\^\(\((1\/9\ + \(\(\ \ \)\(\[ImaginaryI]\)\)\/50)\)\ \((Log[2] + 2 \[Pi]\ \[ImaginaryI]\ n)\)\);\)\ \ \), "\[IndentingNewLine]", \(\(table = Table[ComplexExpand[w\_0], {n, \(-9\), 9, 1}];\)\ \), "\[IndentingNewLine]", \(\(table = Prepend[Append[ table, "\<...\>"], "\<...\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(2\^\(1/9 + \[ImaginaryI]/50\)\) = \>\"", w\_0];\)\ \), "\[IndentingNewLine]", \(\(Print["\< = \>", table];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nUse ", StyleBox["Mathematica", FontSlant->"Italic"], " to plot some of the values of ", Cell[BoxData[ \(2\^\(1/9\ + \ \[ImaginaryI]/50\)\)]], ". " }], "Text"], Cell[BoxData[{ \(\n\(Remove[w];\)\ \), "\[IndentingNewLine]", \(\(w = \[ExponentialE]\^\(\((1\/9\ + \(\(\ \)\(\[ImaginaryI]\)\)\/50)\)\ \ \((Log[2] + 2 \[Pi]\ \[ImaginaryI]\ n)\)\);\)\ \), "\[IndentingNewLine]", \(\(pts = Table[{Re[w], Im[w]}, {n, \(-9\), 9, 1}];\)\ \), "\n", \(\(ListPlot[pts, PlotRange \[Rule] {{\(-2\), 3.5}, {\(-2\), 3.5}}, Ticks \[Rule] {Range[\(-2\), 3, 1], Range[\(-2\), 3, 1]}, AxesLabel \[Rule] {"\", "\"}, AspectRatio \[Rule] 1, Prolog \[Rule] {PointSize[0.02], Red}];\)\ \), "\n", \(\(Print[\*"\"\\""];\)\ \), "\n", \(\(Print[NumberForm[N[table], 3]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"] }, Closed]], Cell[TextData[{ "\n", StyleBox["The Rules for exponents, Page 180.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Some of the rules for exponents carry over from the real case. If c \ and d are complex numbers and ", Cell[BoxData[ \(z \[NotEqual] 0\)]], ", then \n\n\t\t\t", Cell[BoxData[ \(z\^\(-c\) = 1\/z\^c\)]], " ,\n\n\t\t\t", Cell[BoxData[ \(\(z\^c\) z\^d = z\^\(c + d\)\)]], ",\n\n\t\t\t", Cell[BoxData[ \(z\^c\/z\^d = z\^\(c - d\)\)]], ",\n\n\t\t\t", Cell[BoxData[ \(\((z\^c)\)\^n = z\^\(c\ n\)\)]], ", where n is an integer. \n\n\n\tThe following example shows that the \ last identity does not hold if n is replaced with an arbitrary complex \ value." }], "Text"], Cell[TextData[{ "\n", StyleBox["Example 5.9, Page 180. (a)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find all the values of ", Cell[BoxData[ \(\((\[ImaginaryI]\^2)\)\^\[ImaginaryI]\)]], ", ", StyleBox["(b)", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], StyleBox[" ", FontColor->RGBColor[1, 0, 0]], "Find all the values of ", Cell[BoxData[ \(\[ImaginaryI]\^\(2 \[ImaginaryI]\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 5.9.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text"], Cell[TextData[{ "First, enter ", Cell[BoxData[ RowBox[{\(w\_1\), "=", RowBox[{ RowBox[{\(\[ExponentialE]\^\(\[ImaginaryI]\ Log[\[ImaginaryI]\^2]\)\ \), " ", StyleBox["and", FontFamily->"Times New Roman"], " ", \(w\_2\)}], "=", \(\[ExponentialE]\^\(2 \[ImaginaryI]\ \ Log[\[ImaginaryI]]\)\)}]}]]], " and find the principal values." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[t, w];\)\ \), "\n", \(\(w\_1\ = \ \[ExponentialE]\^\(\[ImaginaryI]\ Log[\[ImaginaryI]\^2]\);\ \)\ \), "\n", \(\(w\_2\ = \ \[ExponentialE]\^\(2 \[ImaginaryI]\ Log[\[ImaginaryI]]\);\ \)\ \), "\n", \(\(w\_3\ = N[\[ExponentialE]\^\(\[ImaginaryI]\ Log[\[ImaginaryI]\^2]\)];\)\ \), \ "\n", \(\(w\_4\ = \ N[\[ExponentialE]\^\(2 \[ImaginaryI]\ Log[\[ImaginaryI]]\)];\)\ \), \ "\n", \(\(Print[\((\[ImaginaryI]\^2)\)\^\[ImaginaryI], "\< = \>", w\_1];\)\ \), "\n", \(\(Print["\< \>", \[ImaginaryI]\^\(2 \[ImaginaryI]\), "\< = \>", w\_2];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\((\[ImaginaryI]\^2)\)\^\[ImaginaryI], "\< = \>", w\_3];\)\ \), "\n", \(\(Print["\< \>", \[ImaginaryI]\^\(2 \[ImaginaryI]\), "\< = \>", w\_4];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell["\<\ Second, find the other values in the solution sets.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(t\_1\ = \ Table[\[ExponentialE]\^\(\[ImaginaryI]\ \((Log[\[ImaginaryI]\^2] + 2 \ \[Pi]\ \[ImaginaryI]\ n)\)\), {n, \(-4\), 4, 1}];\)\ \), "\n", \(\(t\_2\ = \ Table[\[ExponentialE]\^\(2 \[ImaginaryI] \((\ Log[\[ImaginaryI]]\ \ \ + 2 \[Pi]\ \[ImaginaryI]\ n)\)\), {n, \(-4\), 4, 1}];\)\ \), "\[IndentingNewLine]", \(\(set\_1\ = \ Append[Prepend[ t\_1, "\<...\>"], "\<...\>"];\)\ \), "\[IndentingNewLine]", \(\(set\_2\ = \ Append[Prepend[t\_2, "\<...\>"], "\<...\>"];\)\ \), "\n", \(\(Print[\((\[ImaginaryI]\^2)\)\^\[ImaginaryI], "\< = \>", set\_1];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\< \>", \[ImaginaryI]\^\(2 \[ImaginaryI]\), "\< = \>", set\_2];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ StyleBox["\nRemark.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], " The two solution sets are different. 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