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Inleiding - Historische motivatie", ButtonData:>"H1", ButtonStyle->"Hyperlink"], "\n\t", ButtonBox["Re", ButtonData:>"H1 D1", ButtonStyle->"Hyperlink"], ButtonBox["\[EDoubleDot]", ButtonData:>"H1 D1", ButtonStyle->"Hyperlink"], ButtonBox["le getallen alleen...", ButtonData:>"H1 D1", ButtonStyle->"Hyperlink"], "\n\t", ButtonBox["... zijn niet voldoende.", ButtonData:>"H1 D2", ButtonStyle->"Hyperlink"], "\n", ButtonBox["2. Constructie van de complexe getallen", ButtonData:>"H2", ButtonStyle->"Hyperlink"], "\n\t", ButtonBox["Punten in het vlak", ButtonData:>"H2 D1", ButtonStyle->"Hyperlink"], "\n\t", ButtonBox["Vermenigvuldiging", ButtonData:>"H2 D2", ButtonStyle->"Hyperlink"], "\n\t", ButtonBox["Het optellen", ButtonData:>"H2 D3", ButtonStyle->"Hyperlink"], "\n", ButtonBox[ "3. Vergelijking van de structuur van de complexe getallen en re", ButtonData:>"H3", ButtonStyle->"Hyperlink"], ButtonBox["\[EDoubleDot]", ButtonData:>"H3", ButtonStyle->"Hyperlink"], ButtonBox["le getallen - het lichaam", ButtonData:>"H3", ButtonStyle->"Hyperlink"], "\n", ButtonBox["4. Complexe functies", ButtonData:>"H4", ButtonStyle->"Hyperlink"], "\n\t", ButtonBox["Inleiding", ButtonData:>"H4 D1", ButtonStyle->"Hyperlink"], "\n\t", ButtonBox["Machtsverheffing", ButtonData:>"H4 D2", ButtonStyle->"Hyperlink"], "\n\t", ButtonBox["Wortels", ButtonData:>"H4 D3", ButtonStyle->"Hyperlink"], "\n\t", ButtonBox[ "Kennismaking met de werking van enkele complexe functies. 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Antwoorden op vraagstukken", ButtonData:>"H9", ButtonStyle->"Hyperlink"], "\n\t", ButtonBox[ "grafische illustratie van de wetten voor het optellen en vermenigvuldigen", ButtonData:>"H9 D1", ButtonStyle->"Hyperlink"], "\n\t", ButtonBox[ "de formules voor het multiplicatief invers en de tegengestelde van een \ complex getal", ButtonData:>"H9 D2", ButtonStyle->"Hyperlink"], "\n\t", ButtonBox["inductiebewijs - de wet van de Moivre.", ButtonData:>"H9 D3", ButtonStyle->"Hyperlink"] }], "Text", CellMargins->{{27, Inherited}, {Inherited, Inherited}}] }, Open ]], Cell[CellGroupData[{ Cell["Inleiding - Historische motivatie", "Section", CellTags->"H1"], Cell[CellGroupData[{ Cell[TextData[{ "Re", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, FontFamily->"Times New Roman"], "le getallen alleen..." }], "Subsection", CellDingbat->"\[FilledDiamond]", CellMargins->{{36, Inherited}, {Inherited, Inherited}}, CellTags->"H1 D1"], Cell[TextData[{ "\"God made the natural numbers. The others, were man-made\" \ (Weierstrass)...\n\n\t... en de ", StyleBox["natuurlijke getallen", FontColor->RGBColor[1, 0, 0]], " dienden om voorwerpen te tellen: ", StyleBox["1", FontSlant->"Italic"], " halssnoer, ", StyleBox["2", FontSlant->"Italic"], " halssnoeren, ", StyleBox["3 ", FontSlant->"Italic"], " halssnoeren... - heel \"natuurlijke\" toepassing.\n\t\n\tMaar zo \ eenvoudig is het leven toch niet: \"Ik had ", StyleBox["17", FontSlant->"Italic"], " halssnoeren. Een aantal daarvan heb ik aan mijn dochter gegeven. Nu heb \ ik nog maar twee halssnoeren over. Hoeveel heb ik er weggegeven?\" - \ dergelijke vragen hebben de mensen aangezet om de negatieve getallen uit te \ vinden (of gewoon \"ontdekken\"). Daarna konden de mensen niet meer zeggen: \ vergelijking ", Cell[BoxData[ FormBox[ StyleBox[\(x + 17 = 2\), FontSlant->"Italic"], TraditionalForm]]], " heeft geen oplossing. Zij heeft er \[EAcute]\[EAcute]n - in de \ zogenaamde ", StyleBox["gehele getallen", FontColor->RGBColor[1, 0, 0]], ".\n\t\n\tHandel drijven - kopen en verkopen. \"Ik geef je ", StyleBox["1", FontSlant->"Italic"], " halssnoer voor ", StyleBox["3 ", FontSlant->"Italic"], " vissen. Als je nu de ", StyleBox["10", FontSlant->"Italic"], " vissen die je vandaag gevangen hebt, aan me kwijt wil, krijg je van me \ ", StyleBox["3", FontSlant->"Italic"], " halssnoeren en wees blij dat je al naar huis kan!\" Hoeveel vissen \ krijgt nu de koper voor ", StyleBox["1", FontSlant->"Italic"], " halssnoer? Wel, dat kan je weeral niet beschrijven met behulp van de \ getallen die we reeds kennen... Toen hebben de mensen de ", StyleBox["rationale getallen", FontColor->RGBColor[1, 0, 0]], " bedacht: breuken. De vergelijking ", Cell[BoxData[ FormBox[ StyleBox[\(3\ x = 10\), FontSlant->"Italic"], TraditionalForm]]], " heeft een oplossing in de nieuwe getallenwereld. Ooit dacht men er zelfs \ niet aan om de getallen, die als \"gehele\" eenheden werden beschouwd, te \ \"breken\". Men kan zeggen: \"De nood is de moeder der uitvindingen\".\n\t\n\t\ Verder hebben de mensen vastgesteld dat de verhouding van de omtrek van een \ cirkel tot zijn straal niet als breuk kan uitgedrukt worden. Dat soort \ getallen, die klaarblijkelijk ook al in de natuur bestonden, werden ", StyleBox["\"irrationaal\"", FontColor->RGBColor[1, 0, 0]], " genoemd - getallen, die niet als ratio - verhouding - van twee gehele \ getallen kunnen uitgedrukt worden." }], "Text", CellMargins->{{36, Inherited}, {Inherited, Inherited}}, TextAlignment->Left] }, Open ]], Cell[CellGroupData[{ Cell["... zijn niet voldoende.", "Subsection", CellDingbat->"\[FilledDiamond]", CellMargins->{{36, Inherited}, {Inherited, Inherited}}, CellTags->"H1 D2"], Cell[TextData[{ "\tHet leven ging zijn gewone gang, en de geleerde mensen hielden zich \ bezig met het vinden van formules om de vergelijking type ", Cell[BoxData[ \(TraditionalForm\`a\ x\^2 + b\ x + c = 0\)], FontSlant->"Italic"], " op te lossen. Die formules zijn algemeen gekend.\nHun grote nut heeft \ mensen als Cardano, Tartaglia, Ferrari en hun tijdgenoten (zestiende eeuw) \ aangezet tot de poging om gelijkaardige formules voor derdegraads \ vergelijkingen type ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(a\ x\^3 + b\ x\^2 + c\ x + d = 0\), FontSlant->"Italic"], " "}], TraditionalForm]]], " te vinden.\n\tVele mensen denken dat de motivatie om complexe getallen te \ defini", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, FontFamily->"Times New Roman"], "ren het verlangen was om de vergelijking ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ StyleBox[\(x\^2 + 1\), FontSlant->"Italic"], StyleBox["=", FontSlant->"Italic"], RowBox[{ StyleBox["0", FontSlant->"Italic"], " ", "op", " ", "te", " ", "lossen"}]}], ","}], " "}], TraditionalForm]]], "anders gezegd - het bestaan van vierkantswortels van negatieve getallen. \ Dit is wel het resultaat van het defini", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, FontFamily->"Times New Roman"], "ren van complexe getallen, maar niet de reden daarvan. De \"uitvinding\" \ van de complexe getallen was namelijk het neveneffect van het zoeken naar de \ formule voor de oplossingen van de derdegraadsvergelijking, waarvan mensen \ reeds WISTEN, dat ze minstens ", StyleBox["1", FontSlant->"Italic"], " oplossing had (omdat de polynomen ", Cell[BoxData[ FormBox[ StyleBox[\(f(x) = a\ x\^3 + bx\^2 + c\ x + d\), FontSlant->"Italic"], TraditionalForm]]], " negatieve waarden bereiken voor negatieve ", Cell[BoxData[ \(TraditionalForm\`x\)]], " met grote absolute waarden, en positieve waarden voor grote positieve ", Cell[BoxData[ \(TraditionalForm\`x\)]], " indien ", Cell[BoxData[ \(TraditionalForm\`a > 0\)], FontSlant->"Italic"], " en omgekeerd in het geval van ", Cell[BoxData[ FormBox[ StyleBox[\(a < 0\), FontSlant->"Italic"], TraditionalForm]]], ". Het is dus - minstens intuit\[IDoubleDot]ef - evident, dat ergens \ onderweg de waarde nul moet bereikt worden). Daarom was het zoeken naar de \ formule voor die oplossing wel verantwoord in tegenstelling tot het zoeken \ naar de oplossing van ", Cell[BoxData[ FormBox[ StyleBox[\(x\^2 + 1 = 0\), FontSlant->"Italic"], TraditionalForm]]], ", die voor die mensen gewoonweg NIET BESTOND (evenmin als de oplossing \ van ", Cell[BoxData[ FormBox[ StyleBox[\(x + 17 = 2\), FontSlant->"Italic"], TraditionalForm]]], " voor de mensen die geen negatieve getallen kenden. In geval van ", Cell[BoxData[ FormBox[ StyleBox[\(x\^2 + 1 = 0\), FontSlant->"Italic"], TraditionalForm]]], " ontbrak er zelfs het \"praktische\" nut van het bestaan van zulke \ oplossingen...).\n\tDe formule ziet er wel indrukwekkend uit - na substitutie \ ", Cell[BoxData[ FormBox[ StyleBox[\(x = y - b\/\(3\ a\)\), FontSlant->"Italic"], TraditionalForm]]], " in de vergelijking ", Cell[BoxData[ FormBox[ StyleBox[\(a\ x\^3 + b\ x\^2 + c\ x + d = 0\), FontSlant->"Italic"], TraditionalForm]]], " en het delen door ", Cell[BoxData[ \(TraditionalForm\`a\)]], ", krijgen we de oplossing van de vergelijking met de nieuwe coefficienten \ (die we voor grotere overzichtelijkheid ", Cell[BoxData[ \(TraditionalForm\`\(p\ \)\)]], " en ", Cell[BoxData[ \(TraditionalForm\`q\)]], " gaan noemen) ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox[\(y\^3 + p\ y + q\), FontSlant->"Italic"], StyleBox["=", FontSlant->"Italic"], RowBox[{ StyleBox["0", FontSlant->"Italic"], ":"}]}], " "}], TraditionalForm]]], "\n\t\n", Cell[BoxData[ \(TraditionalForm\`\(\t\t\t\t y = \(\@\(\(-\(q\/2\)\) + \@\(\((q\/2)\)\^2 + \((p\/3)\)\^3\)\)\%3 + \)\)\)]], Cell[BoxData[ \(TraditionalForm \`\@\(\(-\(q\/2\)\) - \@\(\((q\/2)\)\^2 + \((p\/3)\)\^3\)\)\%3\)]], ".\n\nMen wist dus, dat elke derdegraadsvergelijking minimum 1 re", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, FontFamily->"Times New Roman"], "le wortel heeft en dat we die kunnen vinden door de vorige formule. \ Problemen ontstonden als de uitdrukking onder de vierkantswortel negatief \ was. Dan moest men de derdegraadswortel van een \"onre", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, FontFamily->"Times New Roman"], "el\" getal trekken. De som van de twee derdegraadswortels was al wel een \ re", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, FontFamily->"Times New Roman"], "el getal, maar er moest iets bedacht worden om die \"tussenstap\" te \ verantwoorden. Er werd dus ", Cell[BoxData[ \(TraditionalForm\`\(\@\(-1\)\ gedefinieerd, \ die\ \)\)]], "echter uitsluitend als een soort \"formele fictie\" behandeld werd.\n\t\ Later, in 1777, heeft Euler notatie \[ImaginaryI] en -\[ImaginaryI] \ ingevoerd voor de twee verschillende wortels van ", StyleBox["-1", FontSlant->"Italic"], ". Dat werd - in tegenstelling tot de ", StyleBox["\"re", FontColor->RGBColor[1, 0, 0]], StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, FontFamily->"Times New Roman", FontColor->RGBColor[1, 0, 0]], StyleBox["le getallen\"", FontColor->RGBColor[1, 0, 0]], ", wiens recht van bestaan men kon verantwoorden op basis van de boven \ aangehaalde voorbeelden van het re", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, FontFamily->"Times New Roman"], "le leven - een imaginair getal genoemd - pure fictie dus. \n\tKunnen we \ dat echt als puur fictieve verschijnselen behandelen? Zou er geen beeldrijke \ interpretatie van de nieuw gedefinieerde getallen zijn? Daar gaan wij het in \ het volgende hoofdstuk over hebben." }], "Text", CellMargins->{{36, Inherited}, {Inherited, Inherited}}] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Constructie van de complexe getallen", "Section", CellTags->"H2"], Cell[CellGroupData[{ Cell["Punten in het vlak", "Subsection", CellDingbat->"\[FilledDiamond]", CellMargins->{{36, Inherited}, {Inherited, Inherited}}, CellTags->"H2 D1"], Cell[TextData[{ "Deze constructie is aan Gauss (1799) en Hamilton (1833) te danken.\n\nWe \ gaan eventjes de gedachtenstroom van die grote wiskundigen proberen te \ \"reconstrueren\". Onze bedoeling nu is het getallenstelsel te omschrijven, \ dat een uitbreiding van de re", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, FontFamily->"Times New Roman"], "le getallen is (dat wil zeggen - dat de re", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, FontFamily->"Times New Roman"], "le getallen bevat en waarin de gewone bewerking van optellen en \ vermenigvuldigen van re", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, FontFamily->"Times New Roman"], "le getallen uitgebreid worden op nieuwe getallen maar onveranderd blijven \ voor de re", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, FontFamily->"Times New Roman"], "le getallen), en waarin ", Cell[BoxData[ FormBox[ RowBox[{\(het\ getal\ \ \[ImaginaryI]\ \ bestaat\), ",", " ", RowBox[{ RowBox[{ RowBox[{ "dat", " ", "een", " ", "oplossing", " ", "is", " ", "van", " ", "de", " ", "vergelijking", " ", StyleBox[\(x\^2\), FontSlant->"Italic"]}], StyleBox["+", FontSlant->"Italic"], StyleBox["1", FontSlant->"Italic"]}], StyleBox["=", FontSlant->"Italic"], StyleBox[ RowBox[{ StyleBox["0", FontSlant->"Italic"], "."}]]}]}], TraditionalForm]]], "\nWe weten, dat we die oplossing niet onder de re", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, FontFamily->"Times New Roman"], "le getallen moeten zoeken. Als we dat nieuwe stelsel grafisch willen \ voorstellen, moeten we onze \[ImaginaryI] buiten de getallenas zoeken! Laat \ ons het gewone Cartesische co\[ODoubleDot]rdinatenstelsel nemen en de ", StyleBox["x", FontSlant->"Italic"], "-as als de re", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, FontFamily->"Times New Roman"], "le as beschouwen. Het punt ", StyleBox["(0,1)", FontSlant->"Italic"], " noemen we \[ImaginaryI]. Nu gaan we het vermenigvuldigen en optellen \ van de punten in het vlak defini", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, FontFamily->"Times New Roman"], "ren rekening houdend met de opgesomde voorwaarden. Elk punt van het vlak \ heeft poolco\[ODoubleDot]rdinaten. De ligging wordt bepaald door zijn afstand \ ", StyleBox["r", FontSlant->"Italic"], " van het punt ", StyleBox["(0,0)", FontSlant->"Italic"], " (straallengte) en het hoek \[CurlyPhi] tussen de straal en het positieve \ deel van de ", StyleBox["x", FontSlant->"Italic"], "-as." }], "Text", CellMargins->{{36, Inherited}, {Inherited, Inherited}}], Cell[CellGroupData[{ Cell["\<\ De tekening hieronder toont het duidelijk. 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StyleBox["(r,\[CurlyPhi])", FontSize->14, FontSlant->"Italic", FontColor->RGBColor[1, 0, 1]], " ) zijn ", StyleBox["(r cos\[CurlyPhi], r sin\[CurlyPhi])", FontSize->14, FontSlant->"Italic", FontColor->RGBColor[1, 0, 1]], ". Dit volgt uit de simpele toepassing van de goniometrie.\nOmgekeerd: \ voor elk punt ", Cell[BoxData[ \(TraditionalForm\`\((x, y)\)\)]], " van het vlak kunnen we de straal en de hoek vinden op basis van de \ formules: " }], "Text", CellMargins->{{36, Inherited}, {Inherited, Inherited}}, CellTags->"Carthesische coordinaten"], Cell[CellGroupData[{ Cell[BoxData[{\(Clear[straal]\), RowBox[{\(Clear[hoek]\), "\n"}], StyleBox[\(straal[x_, y_] := N[Sqrt[x^2 + y^2]]\), "Input"], RowBox[{ \(hoek[x_, y_] := If[x < 0, 1, 0]\ N[Pi] + If[x < 0, \(-1\), 1]\ ArcSin[y/straal[x, y]]\), "\n"}], RowBox[{"{", RowBox[{ RowBox[{"straal", "[", StyleBox[\(\(-1\), 4\), FontColor->RGBColor[1, 0, 0]], "]"}], ",", RowBox[{"hoek", "[", StyleBox[\(\(-1\), 4\), FontColor->RGBColor[1, 0, 0]], "]"}]}], "}"}]}], "Input", CellTags->"uitrekenen van de rotatiehoek en de homothetieschaal"], Cell[OutputFormData["\<\ {4.12310562561766, 1.81577498992176}\ \>", "\<\ {4.12311, 1.81577}\ \>"], "Output"] }, Open ]], Cell[TextData[{ "De formule voor de hoek ziet er een beetje ingewikkeld uit, met de \ clausule \"If\". Het moest zo gedaan worden, omdat de functie ArcSin alleen \ maar waarden van het interval ", StyleBox["[-\[Pi]/2,\[Pi]/2]", FontSlant->"Italic"], " aanneemt. Er is dus een lichte aanpassing nodig, om alle punten van de \ cirkel te kunnen beschrijven." }], "Text", CellMargins->{{36, Inherited}, {Inherited, Inherited}}], Cell[TextData[{ "Probeer eens zelf voor een aantal punten hun straal en hoek uit te \ rekenen. Doe dat door de \"rode\" co\[ODoubleDot]rdinaten in \"", StyleBox["{straal[-1,4], hoek[-1,4]}", FontSlant->"Italic"], "\" door de door jezelf gekozen co\[ODoubleDot]rdinaten te vervangen en de \ cel te evalueren." }], "Text", CellMargins->{{36, Inherited}, {Inherited, Inherited}}], Cell[TextData[{ "De straal wordt ook ", StyleBox["MODULUS", FontColor->RGBColor[1, 0, 0]], " of afstand van een complex getal genoemd en voorgesteld door ", StyleBox["\[VerticalSeparator]", FontColor->RGBColor[1, 0, 0]], StyleBox["z", FontSlant->"Italic", FontColor->RGBColor[1, 0, 0]], StyleBox["\[VerticalSeparator]", FontColor->RGBColor[1, 0, 0]], " of ", StyleBox["Abs[z]", FontSlant->"Italic", FontColor->RGBColor[1, 0, 0]], " (absolute waarde van ", StyleBox["z", FontSlant->"Italic"], " - zijn afstand van de oorsprong). De hoek wordt ", StyleBox["ARGUMENT", FontColor->RGBColor[1, 0, 0]], " van het complexe getal genoemd. Wij schrijven: ", StyleBox["Arg[z]", FontSlant->"Italic", FontColor->RGBColor[1, 0, 0]], "." }], "Text", CellMargins->{{36, Inherited}, {Inherited, Inherited}}] }, Open ]], Cell[CellGroupData[{ Cell["Vermenigvuldiging", "Subsection", CellDingbat->"\[FilledDiamond]", CellMargins->{{36, Inherited}, {Inherited, Inherited}}, CellTags->"H2 D2"], Cell[CellGroupData[{ Cell[TextData[{ "We willen dat het punt ", StyleBox["(0,1)", FontSlant->"Italic"], " - onze \[ImaginaryI] - vermenigvuldigd met zichzelf, als resultaat ", StyleBox["-1,", FontSlant->"Italic"], " dus het punt ", StyleBox["(-1,0),", FontSlant->"Italic"], " geeft. Laat ons even op de volgende tekening kijken:\n(door de cel open \ te klikken kan je de programmatie bekijken)." }], "Text", CellMargins->{{36, Inherited}, {Inherited, Inherited}}], Cell[BoxData[ \(Show[ Graphics[{{Hue[ .5], PointSize[0.02], Point[{1, 0}]}, \n \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t{Hue[0], PointSize[0.02], Point[{0, 1}]}, \n \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t{Hue[0], Text[FontForm["\<\[ImaginaryI]\>", {"\", 20}], { 0.1, 0.95}]}, \n \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t{Hue[ .1], PointSize[0.02], Point[{\(-1\), 0}]}, \n \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t{Hue[0], Line[{{0, 0}, {0, 1}}]}, \n \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t{Hue[ .1], Line[{{0, 0}, {\(-1\), 0}}]}, \n \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t{Hue[0], Circle[{0, 0}, .1, {0, Pi/2}]}, \n \t\t\t\t\t\t\t\t\t\t\t\t\t\t\t{Hue[ .1], Circle[{0, 0}, .2, {0, Pi}]}}, \n\t\t\t\t\tAxes -> True, PlotRange -> {{\(-1\), 1}, {0, 1}}, AspectRatio -> Automatic]]\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell[GraphicsData["PostScript", "\<\ %! %%Creator: 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Zou dat niet suggereren om het vermenigvuldigen te defini", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, TextAlignment->Left, FontFamily->"Times New Roman"], "ren als rotatie rond het punt ", StyleBox["(0,0)", FontSlant->"Italic"], " over de hoek \[CurlyPhi] ? We mogen natuurlijk niet vergeten, dat we \ het vermenigvuldigen van re", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, FontFamily->"Times New Roman"], "le getallen willen ", StyleBox["uitbreiden", FontSize->14, FontColor->RGBColor[1, 0, 0]], ", dus de resultaten voor de re", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, FontFamily->"Times New Roman"], "le getallen moeten onveranderd blijven! Wat de hoek betreft, is dat in \ orde, want die is in geval van re", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, FontFamily->"Times New Roman"], "le getallen altijd nul voor positieve getallen en altijd \[Pi] voor \ negatieve getallen. Maar we zien al dat de rotatie alleen niet volstaat - bij \ re", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, FontFamily->"Times New Roman"], "le getallen worden de lengten van de stralen vermenigvuldigd! Dat \ suggereert al de volgende definitie van de vermenigvuldiging van punten van \ het vlak:" }], "Text", CellMargins->{{36, Inherited}, {Inherited, Inherited}}, TextAlignment->Left], Cell[TextData[{ StyleBox[ "om twee punten van het vlak te vermenigvuldigen, moeten we de lengten van \ hun stralen met mekaar vermenigvuldigen als re\[EDoubleDot]le getallen en de \ hoeken tussen de ", CellFrame->{{0, 0}, {0, 2}}, FontSize->14, FontColor->RGBColor[1, 0, 1]], StyleBox["stralen en het positieve deel van de ", FontSize->14, FontColor->RGBColor[1, 0, 1]], StyleBox["x", FontSize->14, FontSlant->"Italic", FontColor->RGBColor[1, 0, 1]], StyleBox["-as moeten we optellen.\n\n", FontSize->14, FontColor->RGBColor[1, 0, 1]], Cell[BoxData[ \(TraditionalForm\`\(Abs[\(z\_1\) z\_2] = \)\)], FontSize->14, FontSlant->"Italic", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\(Abs[z\_1]\ \)\)], FontSize->14, FontSlant->"Italic", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`Abs[z\_2]\)], FontSize->14, FontSlant->"Italic", FontColor->RGBColor[1, 0, 0]], StyleBox[",\n", FontSize->14], Cell[BoxData[ \(TraditionalForm\`Arg[\(z\_1\) z\_2]\)], FontSize->14, FontSlant->"Italic", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\( = Arg[z\_1]\)\)], FontSize->14, FontSlant->"Italic", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(TraditionalForm\`\(+Arg[z\_2]\)\)], FontSize->14, FontSlant->"Italic", FontColor->RGBColor[1, 0, 0]], StyleBox["\n", FontSize->14], StyleBox[ "\nVermenigvuldiging zou dus samenstelling van een rotatie en een \ homothetie (", FontSize->14, FontColor->RGBColor[1, 0, 1]], ButtonBox[ "de definitie van homothetie kan je vinden door op dit stukje blauwe tekst \ te klikken", ButtonData:>"definitie van homothetie", ButtonStyle->"Hyperlink"], StyleBox[") zijn.", FontSize->14, FontColor->RGBColor[1, 0, 1]] }], "Text", CellMargins->{{36, Inherited}, {Inherited, Inherited}}, CellTags->"terug naar de tekst - van de def. van homothetie"], Cell[TextData[{ "Dit illustreert iets wat je al lang kende: bij het vermenigvuldigen van de \ re", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, FontFamily->"Times New Roman"], "le getallen gelden volgende regels:\npositief x positief = positief \t( ", StyleBox["0+0=0 ", FontSlant->"Italic"], ")\npositief x negatief = negatief \t( ", StyleBox["0+\[Pi]=\[Pi] ", FontSlant->"Italic"], ")\nnegatief x positief = negatief \t( ", StyleBox["\[Pi]+0=\[Pi] ", FontSlant->"Italic"], ")\nnegatief x negatief = positief \t( ", StyleBox["\[Pi]+\[Pi]=2\[Pi]", FontSlant->"Italic"], " - een ge\[ODoubleDot]ri\[EDoubleDot]nteerde hoek van ", StyleBox["2\[Pi]", FontSlant->"Italic"], " op een cirkel is equivalent aan ", StyleBox["0 ", FontSlant->"Italic"], ")." }], "Text", CellMargins->{{36, Inherited}, {Inherited, Inherited}}], Cell[TextData[{ StyleBox["Als herinnering: definitie van homothetie", FontVariations->{"Underline"->True}], ": \n", StyleBox["Homothetie", FontColor->RGBColor[1, 0, 0]], " ten opzichte van punt ", StyleBox["M ", FontSlant->"Italic"], " en met de verhouding ", StyleBox["k", FontSlant->"Italic"], " (een re", StyleBox["\[EDoubleDot]", CellMargins->{{27, Inherited}, {Inherited, Inherited}}, FontFamily->"Times New Roman"], "el getal verschillend van nul) is de transformatie van het vlak die elke \ punt ", StyleBox["X", FontSlant->"Italic"], " van het vlak op een punt ", StyleBox["X' ", FontSlant->"Italic"], " afbeeldt zodat ", StyleBox[" ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`\(MX'\)\&\[RightArrow]\)], FontSlant->"Italic"], StyleBox[" = k ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`MX\&\[RightArrow]\)], FontSlant->"Italic"], ".\n\n[ ", ButtonBox["terug naar de tekst van het notebook", ButtonData:>"terug naar de tekst - van de def. van homothetie", ButtonStyle->"Hyperlink"], " ]" }], "Text", CellMargins->{{36, Inherited}, {Inherited, Inherited}}, Background->GrayLevel[0.900008], CellTags->"definitie van homothetie"], Cell[CellGroupData[{ Cell[TextData[{ "Hieronder vind je de meetkundige interpretatie van het vermenigvuldigen \ van complexe getallen. Het argument van het getal ", Cell[BoxData[ FormBox[ StyleBox[\(z\_1\), FontSlant->"Italic"], TraditionalForm]]], " (rood aangeduid) wordt opgeteld bij het argument van het getal ", Cell[BoxData[ FormBox[ StyleBox[\(z\_2\), FontSlant->"Italic"], TraditionalForm]]], " (zwart aangeduid) - de straal van het getal ", Cell[BoxData[ FormBox[ StyleBox[\(z\_2\), FontSlant->"Italic"], TraditionalForm]]], " wordt dus geroteerd over de hoek ", StyleBox["Arg[", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`z\_1\)], FontSlant->"Italic"], StyleBox["]", FontSlant->"Italic"], " rond de oorsprong van het Cartesische stelsel. Daarbij wordt de lengte \ van de straal ", StyleBox["Abs[", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`z\_2\)], FontSlant->"Italic"], StyleBox["]", FontSlant->"Italic"], " vermenigvuldigd met het positieve getal ", StyleBox["Abs[", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`z\_1\)], FontSlant->"Italic"], StyleBox["]", FontSlant->"Italic"], ". Na de rotatie wordt dus een homothetie toegepast.\nDat betekent, dat de \ twee driehoeken op de tekening hieronder - de gele [met de hoekpunten ", StyleBox["(0,0)", FontSlant->"Italic"], ", ", StyleBox["(1,0)", FontSlant->"Italic"], " en ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(z\_1\), FontSlant->"Italic"], "]"}], TraditionalForm]]], " en de blauwe [met de hoekpunten ", StyleBox["(0,0)", FontSlant->"Italic"], ", ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox[\(z\_2\), FontSlant->"Italic"], " ", "en", " ", StyleBox[\(z\_1\), FontSlant->"Italic"], StyleBox[\(z\_2\), FontSlant->"Italic"]}], "]"}], TraditionalForm]]], " gelijkvormig zijn, omdat:\n\n*\t", StyleBox["hoek ", FontSize->16, FontSlant->"Italic"], StyleBox["[", FontSize->16], StyleBox["(1,0), (0,0), ", FontSize->16, FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`\(\(z\_1]\)\ = \)\)], FontSize->16, FontSlant->"Italic"], StyleBox[" hoek ", FontSize->16, FontSlant->"Italic"], StyleBox["[", FontSize->16], 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