複素数と複素共役の実数での剰余演算
複素数と複素共役の実数での剰余演算
\(a\in\mathbb{R}\)とする。
\(a\in\mathbb{R}\)とする。
(1)
\[ \mod\left(\alpha,1\right)=\mod\left(\Re\alpha,1\right)+i\mod\left(\Im\alpha,1\right) \](2)
\[ \mod\left(\alpha,a\right)=a\left\{ \mod\left(\Re\left(\frac{\alpha}{a}\right),1\right)+i\mod\left(\Im\left(\frac{\alpha}{a}\right),1\right)\right\} \](3)
\[ \mod\left(\overline{\alpha},1\right)=\overline{\mod\left(\alpha,1\right)}+i\left|\sgn\mod\left(\Im\alpha,1\right)\right| \](4)
\[ \mod\left(\overline{\alpha},a\right)=\overline{\mod\left(\alpha,a\right)}+ia\left|\sgn\mod\left(\Im\alpha,a\right)\right| \]-
\(\mod\left(\alpha,\beta\right)\)は剰余演算(1)
\begin{align*} \mod\left(\alpha,1\right) & =\alpha-\left\lfloor \Re\left(\alpha\right)\right\rfloor -i\left\lfloor \Im\left(\alpha\right)\right\rfloor \\ & =\Re\alpha-\left\lfloor \Re\left(\alpha\right)\right\rfloor +i\left(\Im\alpha-\left\lfloor \Im\left(\alpha\right)\right\rfloor \right)\\ & =\mod\left(\Re\alpha,1\right)+i\mod\left(\Im\alpha,1\right) \end{align*}(2)
\begin{align*} \mod\left(\alpha,a\right) & =a\mod\left(\frac{\alpha}{a},1\right)\\ & =a\left\{ \mod\left(\Re\left(\frac{\alpha}{a}\right),1\right)+i\mod\left(\Im\left(\frac{\alpha}{a}\right),1\right)\right\} \end{align*}(3)
\begin{align*} \mod\left(\overline{\alpha},1\right) & =\mod\left(\Re\overline{\alpha},1\right)+i\mod\left(\Im\overline{\alpha},1\right)\\ & =\mod\left(\Re\alpha,1\right)+i\mod\left(-\Im\alpha,1\right)\\ & =\mod\left(\Re\alpha,1\right)+i\left(-\mod\left(\Im\alpha,1\right)+\left|\sgn\mod\left(\Im\alpha,1\right)\right|\right)\\ & =\overline{\mod\left(\alpha,1\right)}+i\left|\sgn\mod\left(\Im\alpha,1\right)\right| \end{align*}(4)
\begin{align*} \mod\left(\overline{\alpha},a\right) & =a\mod\left(\frac{\overline{\alpha}}{a},1\right)\\ & =a\left(\overline{\mod\left(\frac{\alpha}{a},1\right)}+i\left|\sgn\mod\left(\Im\left(\frac{\alpha}{a}\right),1\right)\right|\right)\\ & =a\left(\overline{\frac{1}{a}\mod\left(\alpha,a\right)}+i\left|\sgn\mod\left(\Im\alpha,a\right)\right|\right)\\ & =\overline{\mod\left(\alpha,a\right)}+ia\left|\sgn\mod\left(\Im\alpha,a\right)\right| \end{align*}ページ情報
| タイトル | 複素数と複素共役の実数での剰余演算 |
| URL | https://www.nomuramath.com/kwqxg935/ |
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剰余演算の実部と虚部
\[
\mod\left(\alpha,\beta\right)=\Re\left(\beta\right)\mod\left(\Re\left(\frac{\alpha}{\beta}\right),1\right)-\Im\left(\beta\right)\mod\left(\Im\left(\frac{\alpha}{\beta}\right),1\right)+i\left\{ \Re\left(\beta\right)\mod\left(\Im\left(\frac{\alpha}{\beta}\right),1\right)+\Im\left(\beta\right)\mod\left(\Re\left(\frac{\alpha}{\beta}\right),1\right)\right\}
\]
負数の剰余演算
\[
\mod\left(-x,a,b\right)=-\mod\left(x+2b,a,b\right)+a\left|\sgn\left\{ \mod\left(x+2b,a,b\right)-b\right\} \right|+2b
\]
実数の複素数と複素共役の剰余演算
\[
\mod\left(1,\overline{\alpha}\right)=\overline{\mod\left(1,\alpha\right)}+i\overline{\alpha}\left|\sgn\mod\left(\Im\alpha,\left|\alpha\right|^{2}\right)\right|
\]
剰余演算の定数倍
\[
\frac{1}{\delta}\mod\left(\alpha,\beta,\gamma\right)=\mod\left(\frac{\alpha}{\delta},\frac{\beta}{\delta},\frac{\gamma}{\delta}\right)
\]