ブラーマグプタ2平方恒等式
ブラーマグプタ2平方恒等式
\[ \left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=\left(ac\pm bd\right)^{2}+\left(ad\mp bc\right)^{2} \]
\[ \left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=\left(ac\pm bd\right)^{2}+\left(ad\mp bc\right)^{2} \]
\begin{align*}
\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right) & =a^{2}c^{2}+b^{2}d^{2}+a^{2}d^{2}+b^{2}c^{2}\\
& =\left(ac\pm bd\right)^{2}\mp2abcd+a^{2}d^{2}+b^{2}c^{2}\\
& =\left(ac\pm bd\right)^{2}+\left(ad\mp bc\right)^{2}
\end{align*}
\[ \left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=\left(ac-bd\right)^{2}+\left(ad+bc\right)^{2} \] \(b\rightarrow-b\)と置き換えると、
\[ \left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=\left(ac+bd\right)^{2}+\left(ad-bc\right)^{2} \] 1つにまとめて、
\[ \left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=\left(ac\pm bd\right)^{2}+\left(ad\mp bc\right)^{2} \] となる。
(0)-2
\begin{align*} \left\{ \Re^{2}\left(\alpha\right)+\Im^{2}\left(\alpha\right)\right\} \left\{ \Re^{2}\left(\beta\right)+\Im^{2}\left(\beta\right)\right\} & =\left|\alpha\right|^{2}\left|\beta\right|^{2}\\ & =\left|\alpha\beta\right|^{2}\\ & =\left|\left\{ \Re\left(\alpha\right)+\Im\left(\alpha\right)i\right\} \left\{ \Re\left(\beta\right)+\Im\left(\beta\right)\right\} i\right|^{2}\\ & =\left|\Re\left(\alpha\right)\Re\left(\beta\right)-\Im\left(\alpha\right)\Im\left(\beta\right)+i\left\{ \Re\left(\alpha\right)\Im\left(\beta\right)+\Im\left(\alpha\right)\Re\left(\beta\right)\right\} \right|^{2}\\ & =\left\{ \Re\left(\alpha\right)\Re\left(\beta\right)-\Im\left(\alpha\right)\Im\left(\beta\right)\right\} ^{2}+\left\{ \Re\left(\alpha\right)\Im\left(\beta\right)+\Im\left(\alpha\right)\Re\left(\beta\right)\right\} ^{2} \end{align*} ここで\(\Re\left(\alpha\right)=a\;,\;\Im\left(\alpha\right)=b\;,\;\Re\left(\beta\right)=c\;,\;\Im\left(\beta\right)=d\)とおくと、\[ \left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=\left(ac-bd\right)^{2}+\left(ad+bc\right)^{2} \] \(b\rightarrow-b\)と置き換えると、
\[ \left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=\left(ac+bd\right)^{2}+\left(ad-bc\right)^{2} \] 1つにまとめて、
\[ \left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=\left(ac\pm bd\right)^{2}+\left(ad\mp bc\right)^{2} \] となる。
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a^{3}\pm b^{3}=\left(a\pm b\right)\left(a^{2}\mp ab+b^{2}\right)
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a_{4}x^{4}+a_{2}x^{2}+a_{0}=\frac{1}{4a_{4}}\left(2a_{4}x^{2}+a_{2}+\sqrt{a_{2}^{\;2}-4a_{4}a_{0}}\right)\left(2a_{4}x^{2}+a_{2}-\sqrt{a_{2}^{\;2}-4a_{4}a_{0}}\right)
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a^{4}+4b^{4}=\left(a^{2}+2ab+2b^{2}\right)\left(a^{2}-2ab+2b^{2}\right)
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\]