実数の複素数と複素共役の剰余演算
実数の複素数と複素共役の剰余演算
\(\delta_{i,j}\)はクロネッカーのデルタ
(1)
\[ \mod\left(1,\alpha\right)=\alpha\left(\overline{\mod\left(\frac{\alpha}{\left|\alpha\right|^{2}},1\right)}+i\left|\sgn\mod\left(\Im\alpha,\left|\alpha\right|^{2}\right)\right|\right) \](2)
\[ \mod\left(a,\alpha\right)=\alpha\left(\overline{\mod\left(\frac{a\alpha}{\left|\alpha\right|^{2}},1\right)}+i\left|\sgn\mod\left(a\Im\left(\alpha\right),\left|\alpha\right|^{2}\right)\right|\right) \](3)
\[ \mod\left(1,\overline{\alpha}\right)=\overline{\mod\left(1,\alpha\right)}+i\overline{\alpha}\left|\sgn\mod\left(\Im\alpha,\left|\alpha\right|^{2}\right)\right| \](4)
\[ \mod\left(a,\overline{\alpha}\right)=\overline{\mod\left(a,\alpha\right)}+i\overline{\alpha}\left|\sgn\mod\left(a\Im\alpha,\left|\alpha\right|^{2}\right)\right| \]-
\(\mod\left(\alpha,\beta\right)\)は剰余演算\(\delta_{i,j}\)はクロネッカーのデルタ
(1)
\begin{align*} \mod\left(1,\alpha\right) & =\frac{1}{\overline{\alpha}}\mod\left(\overline{\alpha},\left|\alpha\right|^{2}\right)\\ & =\frac{1}{\overline{\alpha}}\left\{ \overline{\mod\left(\alpha,\left|\alpha\right|^{2}\right)}+i\left|\alpha\right|^{2}\left|\sgn\mod\left(\Im\alpha,\left|\alpha\right|^{2}\right)\right|\right\} \\ & =\alpha\left(\overline{\mod\left(\frac{\alpha}{\left|\alpha\right|^{2}},1\right)}+i\left|\sgn\mod\left(\Im\alpha,\left|\alpha\right|^{2}\right)\right|\right) \end{align*}(2)
\begin{align*} \mod\left(a,\alpha\right) & =a\mod\left(1,\frac{\alpha}{a}\right)\\ & =a\frac{\alpha}{a}\left(\overline{\mod\left(\frac{\alpha/a}{\left|\alpha/a\right|^{2}},1\right)}+i\left|\sgn\mod\left(\Im\left(\alpha/a\right),\left|\alpha/a\right|^{2}\right)\right|\right)\\ & =\alpha\left(\overline{\mod\left(\frac{a\alpha}{\left|\alpha\right|^{2}},1\right)}+i\left|\sgn\mod\left(a\Im\left(\alpha\right),\left|\alpha\right|^{2}\right)\right|\right) \end{align*}(3)
\begin{align*} \mod\left(1,\overline{\alpha}\right) & =\overline{\alpha}\left(\overline{\mod\left(\frac{\overline{\alpha}}{\left|\alpha\right|^{2}},1\right)}+i\left|\sgn\mod\left(\Im\overline{\alpha},\left|\alpha\right|^{2}\right)\right|\right)\\ & =\overline{\mod\left(1,\alpha\right)}+i\overline{\alpha}\left|\sgn\mod\left(-\Im\alpha,\left|\alpha\right|^{2}\right)\right|\\ & =\overline{\mod\left(1,\alpha\right)}+i\overline{\alpha}\left|\sgn\mod\left(\Im\alpha,\left|\alpha\right|^{2}\right)\right| \end{align*}(4)
\begin{align*} \mod\left(a,\overline{\alpha}\right) & =a\mod\left(1,\overline{\left(\frac{\alpha}{a}\right)}\right)\\ & =a\left\{ \overline{\mod\left(1,\frac{\alpha}{a}\right)}+i\overline{\left(\frac{\alpha}{a}\right)}\left|\sgn\mod\left(\Im\left(\frac{\alpha}{a}\right),\left|\frac{\alpha}{a}\right|^{2}\right)\right|\right\} \\ & =\overline{\mod\left(a,\alpha\right)}+i\overline{\alpha}\left|\sgn\mod\left(a\Im\alpha,\left|\alpha\right|^{2}\right)\right| \end{align*}ページ情報
| タイトル | 実数の複素数と複素共役の剰余演算 |
| URL | https://www.nomuramath.com/k742kaf0/ |
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剰余演算の実部と虚部
\[
\mod\left(\alpha,\beta\right)=\Re\left(\beta\right)\mod\left(\Re\left(\frac{\alpha}{\beta}\right),1\right)-\Im\left(\beta\right)\mod\left(\Im\left(\frac{\alpha}{\beta}\right),1\right)+i\left\{ \Re\left(\beta\right)\mod\left(\Im\left(\frac{\alpha}{\beta}\right),1\right)+\Im\left(\beta\right)\mod\left(\Re\left(\frac{\alpha}{\beta}\right),1\right)\right\}
\]
剰余演算同士の和・差
\[
\mod\left(x,a,b\right)+\mod\left(y,a,b\right)=\mod\left(x+y,a,b\right)+a\mzp_{0,1}\left(b\sgn\left(a\right),b\sgn\left(a\right)+\left|a\right|;\sgn\left(a\right)\left(\mod\left(x,a,b\right)+\mod\left(y,a,b\right)\right)\right)
\]
剰余演算の引数
\[
\mod\left(\alpha,\beta,\gamma\right):=\mod\left(\alpha-\gamma,\beta\right)+\gamma
\]
偏角と剰余の関係
\[
\Arg\alpha=\mod\left(\Arg\left(\alpha\right),-2\pi,\pi\right)
\]