複素共役
Contents
呼称
- 複素共役 (complex conjugate)
定義
複素共役とは、複素数の虚部の符号を入れ替えたもの。
\[ \bar{z} = a - ib \]
性質
- $\overline{z} + \overline{w} = \overline{z+w}$
- $\overline{z} - \overline{w} = \overline{z-w}$
- $\overline{z}\cdot\overline{w} = \overline{zw}$
- $\displaystyle \frac{\overline{z}}{\overline{w}} = \overline{\left(\frac{z}{w}\right)} , ~(\forall{w}\ne 0)$
- $z$ が実数：
$z\in\R\lrArr\overline{z}=z$ - $z$ が純虚数：
$z\in\Complex \setminus \R \lrArr \overline{z}=-z$ - $\overline{\overline{z}} = z$
$\overline{z} + \overline{w} = \overline{z+w}$ の導出：
\[ \begin{aligned} \overline{z} + \overline{w} &= (a_z-ib_z) + (a_w-ib_w) \cr &= (a_z+a_w) - i(b_z+b_w) \cr &= \overline{(a_z+a_w) + i(b_z+b_w)} \cr &= \overline{(a_z+ib_z) + (a_w+ib_w)} \cr &= \overline{z+w} \cr \cr \therefore \overline{z} + \overline{w} &= \overline{z+w} \end{aligned} \]
$\overline{z} - \overline{w} = \overline{z-w}$ の導出：
\[ \begin{aligned} \overline{z} - \overline{w} &= (a_z-ib_z) - (a_w-ib_w) \cr &= (a_z-a_w) - i(b_z-b_w) \cr &= \overline{(a_z-a_w) + i(b_z-b_w)} \cr &= \overline{(a_z+ib_z) - i(a_w+ib_w)} \cr &= \overline{z-w} \cr \cr \therefore \overline{z} - \overline{w} &= \overline{z-w} \end{aligned} \]
$\overline{z}\cdot\overline{w} = \overline{zw}$ の導出：
\[ \begin{aligned} \overline{z}\cdot\overline{w} &= (a_z - ib_z)(a_w - ib_w) \cr &= a_z a_w - ia_z b_w -ia_w b_z - b_z b_w \cr &= (a_z a_w - b_z b_w) - i(a_z b_w + a_w b_z) \cr &= \overline{(a_z a_w - b_z b_w) + i(a_z b_w + a_w b_z)} \cr &= \overline{(a_z + ib_z)(a_w + ib_w)} \cr &= \overline{zw} \cr \cr \therefore \overline{z}\cdot\overline{w} &= \overline{zw} \end{aligned} \]
$\displaystyle \frac{\overline{z}}{\overline{w}} = \overline{\left(\frac{z}{w}\right)} , ~(\forall{w}\ne 0)$ の導出：
\[ \begin{aligned} \frac{\overline{z}}{\overline{w}} &= \frac{a_z-ib_z}{a_w-ib_w} \cr &= \frac{(a_z-ib_z)(a_w+ib_w)}{(a_w-ib_w)(a_w+ib_w)} \cr &= \frac{a_z a_w + b_z b_w + ia_z b_w - ia_w b_z}{a_w^2+b_w^2} \cr &= \left(\frac{a_z a_w + b_z b_w}{a_w^2 + b_w^2}\right) - i\left(\frac{a_w b_z - a_z b_w}{a_w^2 + b_w^2}\right) \cr &= \overline{\left(\frac{a_z a_w + b_z b_w}{a_w^2 + b_w^2}\right) + i\left(\frac{a_w b_z - a_z b_w}{a_w^2 + b_w^2}\right)} \cr &= \overline{\left(\frac{a_z a_w + b_z b_w - ia_z b_w + ia_w b_z}{a_w^2+b_w^2}\right)} \cr &= \overline{\left[\frac{(a_z+ib_z)(a_w-ib_w)}{(a_w+ib_w)(a_w-ib_w)}\right]} \cr &= \overline{\left(\frac{z}{w}\right)} \cr \cr \therefore \displaystyle \frac{\overline{z}}{\overline{w}} &= \overline{\left(\frac{z}{w}\right)} , ~(\forall{w}\ne 0) \end{aligned} \]
$z\in\R\lrArr\overline{z}=z$ の導出：
\[ \begin{aligned} z &= a + ib \qquad a,b\in\R \cr z\in\R &\lrArr b=0 \cr z &= a+i0 \cr &= a \cr \overline{z} &= a-i0 \cr &= a \cr \cr \therefore z\in\R&\lrArr\overline{z}=z \end{aligned} \]
$z\in\Complex \setminus \R \lrArr \overline{z}=-z$ の導出：
\[ \begin{aligned} z &= a + ib \qquad a,b\in\R \cr z \in \Complex\setminus\R &\lrArr a = 0 \cr z &= 0 + ib \cr &= ib \cr \overline{z} &= 0 - ib \cr &= -ib \cr \cr \therefore z\in\Complex \setminus \R &\lrArr \overline{z}=-z \end{aligned} \]
$\overline{\overline{z}} = z$ の導出：
\[ \begin{aligned} z &= a + ib \qquad a,b\in\R \cr \overline{z} &= a - ib \cr &= a + (-ib)\cr \overline{\overline{z}} &= a -(-ib) \cr &= a + ib \cr \cr \therefore \overline{\overline{z}} &= z \end{aligned} \]