ドゥモワブルの定理とオイラーの公式(De Moivre's theorem and Euler's formula)

ドゥモワブルの定理

定理 1..1 $(\cos{\theta} + i\sin{\theta})^{n} = \cos{n\theta} + i\sin{n\theta} \ (n$ は定数

証明帰納法を用いた証明 $n=1$

のとき，この式が成り立つのは明らか． $n=k$

で成り立つと仮定し， $n=k+1$

で成り立つことを示す．

$\displaystyle (\cos(\theta) + i\sin(\theta))^{k+1}$	$\displaystyle =$	$\displaystyle (\cos(\theta) + i\sin(\theta))^{k}(\cos(\theta) + i\sin(\theta))$
	$\displaystyle =$	$\displaystyle (\cos(k\theta)+i\sin(k\theta))(\cos(\theta) + i\sin(\theta))$
	$\displaystyle =$	$\displaystyle \cos(k\theta)\cos{\theta}-\sin(k\theta)\sin{\theta}+i\left(\sin(k\theta)\cos{\theta}+\cos(k\theta)\sin{\theta}\right)$
	$\displaystyle =$	$\displaystyle \cos(k+1)\theta + i\sin(k+1)\theta$

定理 1..2 オイラーの公式

$\displaystyle e^{i\theta} = \cos{\theta} + i \sin{\theta}$

証明 $e^{x}, \sin{x}, \cos{x}$ のマクローリン展開は

$\displaystyle e^{x}$	$\displaystyle =$	$\displaystyle 1 + x + \frac{x^{2}}{2!} + \cdots = \sum_{n=0}^{\infty}\frac{x^{n}}{n!}$
$\displaystyle \sin{x}$	$\displaystyle =$	$\displaystyle x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots = \sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}$
$\displaystyle \cos{x}$	$\displaystyle =$	$\displaystyle 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots = \sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$

であることに注意すると，実数 $\theta$ に対して

$\displaystyle e^{i\theta}$	$\displaystyle =$	$\displaystyle (1 + i\theta + \frac{i^{2}\theta^2}{2!} + \cdots )$
	$\displaystyle =$	$\displaystyle (1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots) + i(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots)$
	$\displaystyle =$	$\displaystyle \cos{\theta} + i\sin{\theta}$

が成り立つ．

例題 1..3 次の複素数を簡単にせよ．

$\displaystyle{(\frac{1+i}{\sqrt{2}})^{4}}$

解極形式に直す．

$\displaystyle \frac{1+i}{\sqrt{2}} = \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$

より， $r =\sqrt{\frac{1}{2}+\frac{1}{2}} = 1, \theta = \tan^{-1}{1} = \frac{\pi}{4}$ .　したがって，

$\displaystyle \frac{1+i}{\sqrt{2}} = e^{\frac{\pi}{4}i}$

これより，

$\displaystyle (\frac{1+i}{\sqrt{2}})^{4} = (e^{\frac{\pi}{4}i})^4 = e^{\pi i} = \cos{\pi} + i \sin{\pi} = -1$

性質

(1) $\overline{z_{1} \pm z_{2}} = \overline{z_{1}} \pm \overline{z_{2}}, \ \overline... ...}}, \ \overline{\frac{z_{1}}{z_{2}}} = \frac{\overline{z_{1}}}{\overline{z_[2}}$

(2) $\Re z = \frac{z + \overline{z}}{2}, \ \Im z = \frac{(z - \overline{z})}{2i}$

(3) $\vert z_{1}z_{2}\vert = \vert z_{1}\vert\vert z_{2}\vert, \ \vert\frac{z_{1}}{z_{2}}\vert = \frac{\vert z_{1}\vert}{\vert z_{2}\vert} \ (z_{2} \neq 0)$