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複素数の商

2つの複素数$3 \hspace{1}{z}_{1}$$3 \hspace{1}{z}_{2}$の積を考える．

$3\hspace{1}{z}_{1}=\hspace{1}{x}_{1}+\hspace{1}{y}_{1}\mathbb{{i}}$$3\hspace{1}{z}_{2}=\hspace{1}{x}_{2}+\hspace{1}{y}_{2}\mathbb{{i}}$

とおくと，$3\hspace{1}{z}_{1}\neq 0$の場合，

$3\displaystyle{\begin{array}{lll} \frac{\hspace{2} \hspace{1}{z}_{1} \hspace{2}}{\hspace{2} \hspace{1}{z}_{2} \hspace{2}} & =\frac{\hspace{2} \hspace{1}{x}_{1}+\hspace{1}{y}_{1}\mathbb{{i}} \hspace{2}}{\hspace{2} \hspace{1}{x}_{2}+\hspace{1}{y}_{2}\mathbb{{i}} \hspace{2}} & \vspace{6}\\ & =\frac{\hspace{2} \( \hspace{1}{x}_{1}+\hspace{1}{y}_{1}\mathbb{{i}} \) \( \hspace{1}{x}_{2}- \hspace{1}{y}_{2}\mathbb{{i}} \) \hspace{2}}{\hspace{2} \( \hspace{1}{x}_{2}+\hspace{1}{y}_{2}\hspace{1}{\mathbb{{i}}}_{2} \) \( \hspace{1}{x}_{2}- \hspace{1}{y}_{2}\mathbb{{i}} \) \hspace{2}} & \vspace{6}\\ & =\frac{\hspace{2} \( \hspace{1}{x}_{1}\hspace{1}{x}_{2}+\hspace{1}{y}_{1}\hspace{1}{y}_{2} \) + \( \hspace{1}{x}_{2}\hspace{1}{y}_{1}- \hspace{1}{x}_{1}\hspace{1}{y}_{2} \) \mathbb{{i}} \hspace{2}}{\hspace{2} {\hspace{1}{x}_{2}}^{2}+{\hspace{1}{y}_{2}}^{2} \hspace{2}} & \vspace{6}\\ & =\frac{\hspace{2} \( \hspace{1}{x}_{1}\hspace{1}{x}_{2}+\hspace{1}{y}_{1}\hspace{1}{y}_{2} \) \hspace{2}}{\hspace{2} {\hspace{1}{x}_{2}}^{2}+{\hspace{1}{y}_{2}}^{2} \hspace{2}}+\frac{\hspace{2} \( \hspace{1}{x}_{2}\hspace{1}{y}_{1}- \hspace{1}{x}_{1}\hspace{1}{y}_{2} \) \hspace{2}}{\hspace{2} {\hspace{1}{x}_{2}}^{2}+{\hspace{1}{y}_{2}}^{2} \hspace{2}} \mathbb{{i}}& \vspace{6}\\ \end{array}}$

となります．でも，計算はできたがこの商の値がどのような意味をもつのか直感的に理解できないそこで，複素数を極形式で表現して複素数の商の意味を考えてみる．

$3\hspace{1}{z}_{1}=\hspace{1}{r}_{1} \( \cos \hspace{1}{{\theta}}_{1}+\mathbb{{i}}\sin \hspace{1}{{\theta}}_{1} \)$$3\hspace{1}{z}_{1}=\hspace{1}{r}_{2} \( \cos \hspace{1}{{\theta}}_{2}+\mathbb{{i}}\sin \hspace{1}{{\theta}}_{2} \)$

とおくと，

$3\displaystyle{\begin{array}{lll} \frac{\hspace{2} \hspace{1}{z}_{1} \hspace{2}}{\hspace{2} \hspace{1}{z}_{2} \hspace{2}} & =\frac{\hspace{2} \hspace{1}{r}_{1} \( \cos \hspace{1}{{\theta}}_{1}+\mathbb{{i}}\sin \hspace{1}{{\theta}}_{1} \) \hspace{2}}{\hspace{2} \hspace{1}{r}_{2} \( \cos \hspace{1}{{\theta}}_{2}+\mathbb{{i}}\sin \hspace{1}{{\theta}}_{2} \) \hspace{2}} & \vspace{6}\\ & =\frac{\hspace{2} \hspace{1}{r}_{1} \( \cos \hspace{1}{{\theta}}_{1}+\mathbb{{i}}\sin \hspace{1}{{\theta}}_{1} \) \( \cos \hspace{1}{{\theta}}_{2}- \mathbb{{i}}\sin \hspace{1}{{\theta}}_{2} \) \hspace{2}}{\hspace{2} \hspace{1}{r}_{2} \( \cos \hspace{1}{{\theta}}_{2}+\mathbb{{i}}\sin \hspace{1}{{\theta}}_{2} \) \( \cos \hspace{1}{{\theta}}_{2}- \mathbb{{i}}\sin \hspace{1}{{\theta}}_{2} \) \hspace{2}} & \vspace{6}\\ & =\frac{\hspace{2} \hspace{1}{r}_{1} \( \cos \hspace{1}{{\theta}}_{1}\cos \hspace{1}{{\theta}}_{2}+\sin \hspace{1}{{\theta}}_{1}\sin \hspace{1}{{\theta}}_{2} \) +\mathbb{{i}}\hspace{1}{r}_{1} \( \sin \hspace{1}{{\theta}}_{1}\cos \hspace{1}{{\theta}}_{2}- \cos \hspace{1}{{\theta}}_{1}\sin \hspace{1}{{\theta}}_{2} \) \hspace{2}}{\hspace{2} \hspace{1}{r}_{2} \( \cos \hspace{1}{{\theta}}_{2}+\sin \hspace{1}{{\theta}}_{2} \) \hspace{2}} & \vspace{6}\\ \end{array}}$
三角関数の加法定理$3{\sin }^{2}{\theta}+{\cos }^{2}{\theta}=1$ の関係を用いると
$3\displaystyle{=\frac{\hspace{2} \hspace{1}{r}_{1} \hspace{2}}{\hspace{2} \hspace{1}{r}_{2} \hspace{2}} \{ \cos \( \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \) +\sin \( \hspace{1}{{\theta}}_{1}- \hspace{1}{{\theta}}_{2} \) \} }$

この結果をよく見ると$3 \frac{\hspace{2} \hspace{1}{z}_{1} \hspace{2}}{\hspace{2} \hspace{1}{z}_{2} \hspace{2}}$ は絶対値が$3 \frac{\hspace{2} \hspace{1}{r}_{1} \hspace{2}}{\hspace{2} \hspace{1}{r}_{2} \hspace{2}}$ 偏角が$3 \hspace{1}{{\theta}}_{1}-\hspace{1}{{\theta}}_{2}$となっている．すなわち，

$3\displaystyle{ \| \frac{\hspace{2} \hspace{1}{z}_{1} \hspace{2}}{\hspace{2} \hspace{1}{z}_{2} \hspace{2}} \| =\frac{\hspace{2} \| \hspace{1}{z}_{1} \| \hspace{2}}{\hspace{2} \| \hspace{1}{z}_{2} \| \hspace{2}}\text{\hspace{5}},\text{\hspace{5}}arg \( \frac{\hspace{2} \hspace{1}{z}_{1} \hspace{2}}{\hspace{2} \hspace{1}{z}_{2} \hspace{2}} \) =arg\hspace{1}{z}_{1}-arg\hspace{1}{z}_{2}}$

になる． これを図で示すと右の図のようになる．

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