数学 - 解析学 - 数 - 複素数(複素数体の構成、実部・虚部・共役、絶対値)

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解析入門〈1〉 (松坂和夫(著)、岩波書店)の第1章(数)、1.5(複素数)、問題1.3、1、2.を取り組んでみる。

問1

\begin{array}{l} 1 - 3 i + 3 i^{2} - i^{3} \\ = - 2 - 2 i \\ \frac{(23 + 2 i) (3 + 4 i)}{(3 - 4 i) (3 + 4 i)} \\ = \frac{61}{25} + \frac{98}{25} i \\ 1 (n = 4 k) \\ i (n = 4 k + 1) \\ - 1 (n = 4 k + 2) \\ - i (n = 4 k + 3) \end{array}

問2.

\begin{array}{l} | \frac{α}{β} | = \sqrt{\frac{α}{β} \bar{(\frac{α}{β})}} \\ = \sqrt{\frac{α}{β} \cdot \frac{\bar{α}}{\bar{β}}} \\ = \sqrt{\frac{α \bar{α}}{β \bar{β}}} \\ = \frac{\sqrt{α \bar{α}}}{\sqrt{β \bar{β}}} \\ = \frac{| α |}{| β |} \end{array}

問3.

\begin{array}{l} 2 \cdot \sqrt{10} \cdot \sqrt{20} \sqrt{2} = 40 \\ \frac{\sqrt{5} \sqrt{13}}{\sqrt{2} \sqrt{25}} = \frac{\sqrt{130}}{10} \end{array}

問4.

\begin{array}{l} {| α + β |}^{2} + {| α - β |}^{2} \\ = (α + β) \bar{(α + β)} + (α - β) \bar{(α - β)} \\ = (α + β) (\bar{α} + \bar{β}) + (α - β) (\bar{α} - \bar{β}) \\ = {| α |}^{2} + {| β |}^{2} + α \bar{β} + \bar{α} β + {| α |}^{2} + {| β |}^{2} - α \bar{β} - \bar{α} β \\ = 2 ({| α |}^{2} + {| β |}^{2}) \end{array}

問5.

\begin{array}{l} | (α - β) + β | \leq | α - β | + | β | \\ | | (α - β) + β | | - | β | \leq | α - β | \\ | α | - | β | \leq | α - β | \\ | α | - | β | \geq - | α - β | \\ | | α | - | β | | \leq | α - β | \end{array}

問6.

\begin{array}{l} | (α_{1} + α_{2} + \cdot \cdot \cdot + α_{n}) + α_{n + 1} | \\ \leq | α_{1} + α_{2} + \cdot \cdot \cdot + α_{n} | + | α_{n + 1} | \\ \leq | α_{1} | + | α_{2} | + \cdot \cdot \cdot + | α_{n + 1} | \end{array}

問7.

\begin{array}{l} | \frac{α - β}{1 - \bar{α} β} | \\ = \frac{| α - β |}{| 1 - \bar{α} β |} \\ = \frac{\sqrt{(α - β) \bar{(α - β)}}}{\sqrt{(1 - \bar{α} β) \bar{(1 - \bar{α} β)}}} \\ = \frac{\sqrt{(α - β) (\bar{α} - \bar{β})}}{\sqrt{(1 - \bar{α} β) (1 - α \bar{β})}} \\ = \frac{\sqrt{{| α |}^{2} - α \bar{β} - \bar{α} β + {| β |}^{2}}}{\sqrt{1 - α \bar{β} - \bar{α} β + {| α |}^{2} {| β |}^{2}}} \\ | α | = 1 \\ \frac{\sqrt{1 - α \bar{β} - \bar{α} β + {| β |}^{2}}}{\sqrt{1 - α \bar{β} - \bar{α} β + {| β |}^{2}}} = 1 \\ | β | = 1 \\ \frac{\sqrt{{| α |}^{2} - α \bar{β} - \bar{α} β + 1}}{\sqrt{1 - α \bar{β} - \bar{α} β + {| α |}^{2}}} = 1 \end{array}

問8.

\begin{array}{l} 1 + {| α |}^{2} {| β |}^{2} - ({| α |}^{2} + {| β |}^{2}) \\ = (1 - {| α |}^{2}) (1 - {| β |}^{2}) > 0 \\ {| α |}^{2} + {| β |}^{2} > {| α |}^{2} + {| β |}^{2} \\ | \frac{α - β}{1 - \bar{α} β} | < 1 \end{array}

問9.

\begin{array}{l} z = c + d i \\ z^{2} = c^{2} - d^{2} + 2 c d i \\ c^{2} - d^{2} = a \\ 2 c d = b \\ {(c^{2} + d^{2})}^{2} = a^{2} + b^{2} \\ c^{2} + d^{2} = \sqrt{a^{2} + b^{2}} \\ 2 c^{2} = \sqrt{a^{2} + b^{2}} + a \\ c^{2} = \frac{a + \sqrt{a^{2} + b^{2}}}{2} \\ c = \pm \sqrt{\frac{a + \sqrt{a^{2} + b^{2}}}{2}} \\ d^{2} = \frac{- a + \sqrt{a^{2} + b^{2}}}{2} \\ d = \pm \sqrt{\frac{- a + \sqrt{a^{2} + b^{2}}}{2}} \\ z = \pm (\sqrt{\frac{a + \sqrt{a^{2} + b^{2}}}{2}} + \sqrt{\frac{a + \sqrt{a^{2} + b^{2}}}{2}}) (b > 0) \\ z = \pm (\sqrt{\frac{a + \sqrt{a^{2} + b^{2}}}{2}} - \sqrt{\frac{a + \sqrt{a^{2} + b^{2}}}{2}}) (b < 0) \end{array}

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