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# 共役な複素数の基本式

$3{\alpha}$$3{\beta}$複素数とし，それぞれの共役な複素数$3\displaystyle{\bar{{\alpha}}}$$3\displaystyle{\bar{{\beta}}}$とする．

和　$3\displaystyle{\bar{ {\alpha}+{\beta} }=\bar{{\alpha}}+\bar{{\beta}}}$ 証明

差　$3\displaystyle{\bar{ {\alpha}- {\beta} }=\bar{{\alpha}}- \bar{{\beta}}}$ 証明

積　$3\displaystyle{\bar{ {\alpha}{\beta} }=\bar{{\alpha}}\bar{{\beta}}}$ 証明

商$3\displaystyle{\bar{ \( \frac{\hspace{2}{\alpha}\hspace{2}}{\hspace{2}{\beta}\hspace{2}} \) }=\frac{\hspace{2}\bar{{\alpha}}\hspace{2}}{\hspace{2}\bar{{\beta}}\hspace{2}}\text{\hspace{5}} \( {\beta}\neq 0 \) }$ 証明

$3\displaystyle{\bar{ \(\bar{{\alpha}}\) }={\alpha}}$

## ■和の証明

$3\displaystyle{{\alpha}=a+b\text{\hspace{2}}\mathbb{{i}}}$$3\displaystyle{{\beta}=c+d\text{\hspace{2}}\mathbb{{i}}}$ とする．

$3\displaystyle{\begin{array}{lll} \bar{ {\alpha}+{\beta} } & =\bar{ \( a+b\text{\hspace{2}}\mathbb{{i}} \) + \( c+d\text{\hspace{2}}\mathbb{{i}} \) } & \vspace{6}\\ & =\bar{ \( a+c \) + \( b+d \) \mathbb{{i}} } & \vspace{6}\\ & = \( a+c \) - \( b+d \) \mathbb{{i}} & \vspace{6}\\ & = \( a- b\text{\hspace{2}}\mathbb{{i}} \) + \( c- d\text{\hspace{2}}\mathbb{{i}} \) & \vspace{6}\\ & =\bar{{\alpha}}+\bar{{\beta}} & \vspace{6}\\ \end{array}}$

## ■差の証明

$3\displaystyle{{\alpha}=a+b\text{\hspace{2}}\mathbb{{i}}}$$3\displaystyle{{\beta}=c+d\text{\hspace{2}}\mathbb{{i}}}$ とする．

$3\displaystyle{\begin{array}{lll} \bar{ {\alpha}- {\beta} } & =\bar{ \( a+b\text{\hspace{2}}\mathbb{{i}} \) - \( c+d\text{\hspace{2}}\mathbb{{i}} \) } & \vspace{6}\\ & =\bar{ \( a- c \) + \( b- d \) \mathbb{{i}} } & \vspace{6}\\ & = \( a- c \) - \( b- d \) \mathbb{{i}} & \vspace{6}\\ & = \( a- b\text{\hspace{2}}\mathbb{{i}} \) - \( c- d\text{\hspace{2}}\mathbb{{i}} \) & \vspace{6}\\ & =\bar{{\alpha}}- \bar{{\beta}} & \vspace{6}\\ \end{array}}$

## ■積の証明

$3\displaystyle{{\alpha}=a+b\text{\hspace{2}}\mathbb{{i}}}$$3\displaystyle{{\beta}=c+d\text{\hspace{2}}\mathbb{{i}}}$ とする．

$3\displaystyle{\begin{array}{lll} \bar{ {\alpha}{\beta} } & =\bar{ \( a+b\text{\hspace{2}}\mathbb{{i}} \) \( c+d\text{\hspace{2}}\mathbb{{i}} \) } & \vspace{6}\\ & =\bar{ \( ac- bd \) + \( ad+bc \) \mathbb{{i}} } & \vspace{6}\\ & = \( ac- bd \) - \( ad+bc \) \mathbb{{i}} & \vspace{6}\\ & = \( a- b\text{\hspace{2}}\mathbb{{i}} \) \( c- d\text{\hspace{2}}\mathbb{{i}} \) & \vspace{6}\\ & =\bar{{\alpha}}\bar{{\beta}} & \vspace{6}\\ \end{array}}$

## ■商の証明

$3\displaystyle{\begin{array}{lll} \bar{ \( \frac{\hspace{2}{\alpha}\hspace{2}}{\hspace{2}{\beta}\hspace{2}} \) }\cdot \bar{{\beta}} & =\bar{ \frac{\hspace{2}{\alpha}\hspace{2}}{\hspace{2}{\beta}\hspace{2}}\cdot {\beta} } & \vspace{6}\\ & =\bar{{\alpha}} & \vspace{6}\\ \end{array}}$ 共役な複素数の積より

よって，$3\displaystyle{\bar{ \( \frac{\hspace{2}{\alpha}\hspace{2}}{\hspace{2}{\beta}\hspace{2}} \) }=\frac{\hspace{2}\bar{{\alpha}}\hspace{2}}{\hspace{2}\bar{{\beta}}\hspace{2}}}$

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