数学 – JavaScript - 新しい数とその表示 - 複素数と複素平面 – 複素平面(ド・モアブルの公式と複素数のn乗根)

数学読本〈3〉

学習環境/開発環境

• Surface 3 (4G LTE)、Surface 3 タイプ カバー、Surface ペン(端末)
• Windows 10 Pro (OS)
• 数式入力ソフト(TeX, MathML): MathType
• MathML対応ブラウザ: Firefox、Safari
• MathML非対応ブラウザ(Internet Explorer, Google Chrome...)用JavaScript Library: MathJax
• OS X El Capitan - Apple (OS)
• Emacs (Text Editor)
• JavaScript (プログラミング言語)
• kjs-math-number (JavaScript Library)
• D3.js (JavaScript Library)
• Safari(Web browser)

数学読本〈3〉平面上のベクトル/複素数と複素平面/空間図形/2次曲線/数列 (松坂 和夫(著)、岩波書店)の第10章(新しい数とその表示 - 複素数と複素平面)、10.1(複素平面)、ド・モアブルの公式と複素数のn乗根、問10、11、12、13.を取り組んでみる。

問7.

$\begin{array}{l}{\cos }^3\theta +3i{\cos }^2\theta \sin \theta -3\cos \theta {\sin }^2\theta -i{\sin }^3\theta \\ \cos 3\theta ={\cos }^3\theta -3\cos \theta {\sin }^2\theta \\ \sin 3\theta =3{\cos }^2\theta \sin \theta -{\sin }^3\theta \\ \\ {\cos }^4\theta +4i{\cos }^3\theta \sin \theta -6{\cos }^2\theta {\sin }^2\theta -4i\cos \theta {\sin }^3\theta +{\sin }^4\theta \\ \cos 4\theta ={\cos }^4\theta -6{\cos }^2\theta {\sin }^2\theta +{\sin }^4\theta \\ \sin 4\theta =4{\cos }^3\theta \sin \theta -4\cos \theta {\sin }^3\theta \\ \\ {\cos }^5\theta +5i{\cos }^4\theta \sin \theta -10{\cos }^3\theta {\sin }^2\theta -10i{\cos }^2\theta {\sin }^3\theta +5\cos \theta {\sin }^4\theta -i{\sin }^5\theta \\ \cos 5\theta ={\cos }^5\theta -10{\cos }^3\theta {\sin }^2\theta +5\cos \theta {\sin }^4\theta \\ \sin 5\theta =5{\cos }^4\theta \sin \theta -10{\cos }^2\theta {\sin }^3\theta -{\sin }^5\theta \end{array}$

問11.

$\begin{array}{l}1=\cos 2k\pi +i\sin 2k\pi \\ 1^{\frac{1}{4}}=\cos \frac{2k\pi }{4}+i\sin \frac{2k\pi }{4}\\ =\cos \frac{k\pi }{2}+i\sin \frac{k\pi }{2}\\ =1,i,-1,-i\\ 1^{\frac{1}{6}}=\cos \frac{2k\pi }{6}+i\sin \frac{2k\pi }{6}\\ =\cos \frac{k\pi }{3}+i\sin \frac{k\pi }{6}\\ =1,\frac{1}{2}+\frac{\sqrt{3}}{2}i,-\frac{1}{2}+\frac{\sqrt{3}}{2}i,-1,-\frac{1}{2}-\frac{\sqrt{3}}{2}i,\frac{1}{2}-\frac{\sqrt{3}}{2}i\end{array}$

問12.

1. 
   $\begin{array}{l}1+i=\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\\ {\left(1+i\right)}^{\frac{1}{2}}=2^{\frac{1}{4}}\left(\cos \frac{\frac{\pi }{4}+2k\pi }{2}+i\sin \frac{\frac{\pi }{4}+2k\pi }{2}\right)\\ =2^{\frac{1}{4}}\left(\cos \left(\frac{\pi }{8}+k\pi \right)+i\sin \left(\frac{\pi }{8}+k\pi \right)\right)\end{array}$
2. 
   $\begin{array}{l}i=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\\ i^{\frac{1}{3}}=\cos \frac{\frac{\pi }{2}+2k\pi }{3}+i\sin \frac{\frac{\pi }{2}+2k\pi }{3}\\ =\cos \left(\frac{\pi }{6}+\frac{2k\pi }{3}\right)+i\sin \left(\frac{\pi }{6}+\frac{2k\pi }{3}\right)\\ =\frac{\sqrt{3}}{2}+\frac{1}{2}i,-\frac{\sqrt{3}}{2}+\frac{1}{2}i,-i\end{array}$
3. 
   $\begin{array}{l}-2+2\sqrt{3}i=2\left(-1+\sqrt{3}i\right)\\ =4\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\\ {\left(-2+2\sqrt{3}\right)}^{\frac{1}{4}}=4^{\frac{1}{4}}\left(\cos \frac{\frac{2\pi }{3}+2k\pi }{4}+i\sin \frac{\frac{2\pi }{3}+2k\pi }{4}\right)\\ =\sqrt{2}\left(\cos \left(\frac{\pi }{6}+\frac{k\pi }{2}\right)+i\sin \left(\frac{\pi }{6}+\frac{k\pi }{2}\right)\right)\end{array}$
4. 
   $\begin{array}{l}-1=\cos \pi +i\sin \pi \\ {\left(-1\right)}^{\frac{1}{5}}=\cos \frac{\pi +2k\pi }{5}+i\sin \frac{\pi +2k\pi }{5}\\ =\cos \left(\frac{\pi }{5}+\frac{2k\pi }{5}\right)+i\sin \left(\frac{\pi }{5}+\frac{2k\pi }{5}\right)\end{array}$

問13.

$\begin{array}{l}\left({\omega }^n-1\right)\left({\omega }^{2n}+{\omega }^n+1\right)={\omega }^{3n}-1\\ n=3k\\ {\omega }^{2n}+{\omega }^n+1=1+1+1=3\\ n\ne 3k\\ {\omega }^{2n}+{\omega }^n+1=\frac{{\omega }^{3n}-1}{{\omega }^n-1}=0\end{array}$

問12の図示について。

number.js、D3.js を利用。

JavaScript

コード(Emacs)

(function () {
    'use strict';
    var svg,
        width = 600,
        height = 600,
        padding = 50,
        min = -2,
        max = 2,
        xscale,
        xaxis,
        yscale,
        yaxis,
        points = [],
        mag,
        arg,
        c,
        k,

        complexFromMagArg;

    complexFromMagArg = function (mag, arg) {
        var real = mag * Math.cos(arg),
            imag = mag * Math.sin(arg);

        return new Complex(real, imag);
    };
    mag = Math.pow(2, 1/4);
    arg = Math.PI / 8;
    for (k = 0; k < 4; k += 1) {
        c = complexFromMagArg(mag, arg + k * Math.PI);
        points.push([c.getReal(), c.getImag(), 'red']);
    }
    mag = 1;
    arg = Math.PI / 6;
    for (k = 0; k < 3; k += 1) {
        c = complexFromMagArg(mag, arg + k * 2 * Math.PI / 3);
        points.push([c.getReal(), c.getImag(), 'green']);
    }
    mag = Math.sqrt(2);
    arg = Math.PI / 6;
    for (k = 0; k < 4; k += 1) {
        c = complexFromMagArg(mag, arg + k * Math.PI / 2);
        points.push([c.getReal(), c.getImag(), 'blue']);
    }
    mag = 1;
    arg = Math.PI / 5;
    for (k = 0; k < 5; k += 1) {
        c = complexFromMagArg(mag, arg + k * 2 * Math.PI / 5);
        points.push([c.getReal(), c.getImag(), 'brown']);
    }
    xscale = d3.scaleLinear()
        .domain([min, max])
        .range([padding, width - padding]);
    yscale = d3.scaleLinear()
        .domain([min, max])
        .range([height - padding, padding]);
    xaxis = d3.axisBottom().scale(xscale);
    yaxis = d3.axisLeft().scale(yscale);

    svg = d3.select('#graph0')
        .append('svg')
        .attr('width', width)
        .attr('height', height);

    svg.selectAll('circle')
        .data(points)
        .enter()
        .append('circle')
        .attr('cx', function (d) {
            return xscale(d[0]);
        })
        .attr('cy', function (d) {
            return yscale(d[1]);
        })
        .attr('r', 5)
        .attr('fill', function (d) {
            return d[2];
        });

    svg.append('g')
        .attr('transform', 'translate(0, ' + (height / 2) + ')')
        .call(xaxis);
    svg.append('g')
        .attr('transform', 'translate(' + (width / 2) + ', 0)')
        .call(yaxis);
}());