複素数の商

# 複素数の商

2つの複素数${z}_{1}$${z}_{2}$の積を考える．

${z}_{1}={x}_{1}+{y}_{1}i$${z}_{2}={x}_{2}+{y}_{2}i$

とおくと，${z}_{1}\ne 0$の場合，

$\frac{{z}_{1}}{{z}_{2}}$$=\frac{{x}_{1}+{y}_{1}i}{{x}_{2}+{y}_{2}i}$

$=\frac{\left({x}_{1}+{y}_{1}i\right)\left({x}_{2}-{y}_{2}i\right)}{\left({x}_{2}+{y}_{2}{i}_{2}\right)\left({x}_{2}-{y}_{2}i\right)}$

$=\frac{\left({x}_{1}{x}_{2}+{y}_{1}{y}_{2}\right)+\left({x}_{2}{y}_{1}-{x}_{1}{y}_{2}\right)i}{{{x}_{2}}^{2}+{{y}_{2}}^{2}}$

$=\frac{\left({x}_{1}{x}_{2}+{y}_{1}{y}_{2}\right)}{{{x}_{2}}^{2}+{{y}_{2}}^{2}}+\frac{\left({x}_{2}{y}_{1}-{x}_{1}{y}_{2}\right)}{{{x}_{2}}^{2}+{{y}_{2}}^{2}}i$

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．

• ${z}_{1}={r}_{1}\left(cos{\theta }_{1}+isin{\theta }_{1}\right)$

• ${z}_{1}={r}_{2}\left(cos{\theta }_{2}+isin{\theta }_{2}\right)$

とおくと，

$\frac{{z}_{1}}{{z}_{2}}$$=\frac{{r}_{1}\left(cos{\theta }_{1}+isin{\theta }_{1}\right)}{{r}_{2}\left(cos{\theta }_{2}+isin{\theta }_{2}\right)}$

$=\frac{{r}_{1}\left(cos{\theta }_{1}+isin{\theta }_{1}\right)\left(cos{\theta }_{2}-isin{\theta }_{2}\right)}{{r}_{2}\left(cos{\theta }_{2}+isin{\theta }_{2}\right)\left(cos{\theta }_{2}-isin{\theta }_{2}\right)}$

$=\frac{{r}_{1}\left(\mathrm{cos}{\theta }_{1}\mathrm{cos}{\theta }_{2}+\mathrm{sin}{\theta }_{1}\mathrm{sin}{\theta }_{2}\right)+i{r}_{1}\left(\mathrm{sin}{\theta }_{1}\mathrm{cos}{\theta }_{2}-\mathrm{cos}{\theta }_{1}\mathrm{sin}{\theta }_{2}\right)}{{r}_{2}\left(\mathrm{cos}{\theta }_{2}+\mathrm{sin}{\theta }_{2}\right)}$

$=\frac{{r}_{1}}{{r}_{2}}\left\{cos\left({\theta }_{1}-{\theta }_{2}\right)+sin\left({\theta }_{1}-{\theta }_{2}\right)\right\}$

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

• $|\frac{{z}_{1}}{{z}_{2}}|=\frac{|{z}_{1}|}{|{z}_{2}|}$

• $arg\left(\frac{{z}_{1}}{{z}_{2}}\right)=arg{z}_{1}-arg{z}_{2}$

になる． これを図で示すと図のようになる．

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