복소수
Def 1
$\mathbb{C} = \left\{a + ib \; | \; a, b \in \mathbb{R}, \; i^{2} = -1 \right\}$
$a + ib = c + id$ $\Leftrightarrow$ $a=c$, $b=d$
Let $\alpha = a + ib$.
Real part of $\alpha$ : $\mathrm{Re}(\alpha) = a$
Imaginary part of $\alpha$ : $\mathrm{Im}(\alpha) = b$
Complex conjugate of $\alpha$ : $\overline{\alpha} = a - ib$
Def 2
Let $a + ib, c+id \in \mathbb{C}$. Addition and multiplication of $\mathbb{C}$ are defined as follows.
$(a+ib) + (c + id) = (a+c) + i(b+d)$
$(a+ib) \cdot (c + id) = (ac - bd) + i(ad + bc)$
Thm 1
$+$에 대한 $a + ib$의 역원 : $-a - ib$
$\cdot$ 에 대한 $a + ib$의 역원 : $\dfrac{a}{a^{2} + b^{2}} - i \dfrac{b}{a^{2} + b^{2}}$
# $\mathbb{C}$ is field.
$(a + ib) + (-a - ib) = 0$
$(a + ib) (\dfrac{a}{a^{2} + b^{2}} - i \dfrac{b}{a^{2} + b^{2}}) = (\dfrac{a^{2}}{a^{2} + b^{2}} + \dfrac{b^{2}}{a^{2} + b^{2}}) + i(-\dfrac{ab}{a^{2} + b^{2}} + \dfrac{ab}{a^{2} + b^{2}}) = 1$
Thm 2
Let $\alpha, \beta \in \mathbb{C}$.
1. $\overline{\alpha + \beta} = \overline{\alpha} + \overline{\beta}$
2. $\overline{\alpha \beta} = \overline{\alpha} \overline{\beta}$
3. $\mathrm{Re}(\alpha) = \dfrac{\alpha + \overline{\alpha}}{2}$, $\mathrm{Im}(\alpha) = \dfrac{\alpha - \overline{\alpha}}{2i}$
pf)
Let $\alpha = a + ib$, $\beta = c + id$
$\overline{\alpha + \beta} = \overline{(a+c) + i(b+d)} = (a+c) - i(b+d) = (a - ib) + (c - id) = \overline{\alpha} + \overline{\beta}$
Let $\alpha = a + ib$, $\beta = c + id$
$\overline{\alpha \beta} = \overline{(ac - bd) + i(ad+bc)} = (ac - bd) - i(ad + bc) = (a - ib) (c - id) = \overline{\alpha} \overline{\beta}$
Let $\alpha = a + ib$
$\dfrac{\alpha + \overline{\alpha}}{2} = \dfrac{(a + ib) + (a - ib)}{2} = a = \mathrm{Re}(\alpha)$
$\dfrac{\alpha - \overline{\alpha}}{2i} = \dfrac{(a + ib) - (a - ib)}{2i} = b = \mathrm{Im}(\alpha)$
Def 3
We can map from $\mathbb{C} \to \mathbb{R}^{2}$ naturally such that $\alpha = a + ib \mapsto (a, b)$. At this time, $\mathbb{R}^{2}$ is called complex planed. So we define the absolute value of $\alpha$ as follows
$\left|\alpha \right| = \left|a+ ib \right| = \sqrt{a^{2} + b^{2}}$
Thm 3
Let $\alpha, \beta \in \mathbb{C}$.
1. $\left|\alpha + \beta \right| \leq \left|\alpha \right| + \left|\beta \right|$
2. $\left|\alpha \beta \right| = \left|\alpha \right| \left|\beta \right|$
3. $\left|\overline{\alpha} \right| = \left|\alpha \right|$
4. $\left|\alpha \right|^{2} = \alpha \overline{\alpha}$
5. $\left| \left|\alpha \right| - \left| \beta \right| \right| \leq \left|\alpha - \beta \right| $
pf)
Let $\alpha = a + ib$, $\beta = c + id$
$\left|\alpha + \beta \right|^{2} = \left|(a+c) + i(b+d) \right|^{2} = (a+c)^{2} + (b+d)^{2} $
$= (a^{2} + b^{2}) + 2(ac + bd) + (c^{2} + d^{2})$
$\leq (a^{2} + b^{2}) + 2 \left|ac + bd \right| + (c^{2} + d^{2})$
$= (a^{2} + b^{2}) + 2 \sqrt{(ac + bd)^{2}} + (c^{2} + d^{2})$
$= (a^{2} + b^{2}) + 2 \sqrt{a^{2}c^{2} + 2abcd + b^{2}d^{2}} + (c^{2} + d^{2})$
$\leq (a^{2} + b^{2}) + 2 \sqrt{a^{2}c^{2} + b^{2}c^{2} + a^{2}d^{2} + b^{2}d^{2}} + (c^{2} + d^{2})$
$= (a^{2} + b^{2}) + 2 \sqrt{a^{2} + b^{2}} \sqrt{c^{2} + d^{2}} + (c^{2} + d^{2}) $
$\leq \left|\alpha \right|^{2} + 2 \left|\alpha \right| \left|\beta \right| + \left|\beta \right|^{2} = (\left|\alpha \right| + \left|\beta \right|)^{2}$
$\therefore$ $\left|\alpha + \beta \right| \leq \left|\alpha \right| + \left|\beta \right|$
Let $\alpha = a + ib$, $\beta = c + id$
$\left|\alpha \beta \right|^{2} = \left|(ac - bd) + i(ad + bc) \right|^{2}$
$= (ac - bd)^{2} + (ad + bc)^{2} $
$= a^{2}c^{2} + b^{2}d^{2} + a^{2}d^{2} + b^{2}c^{2}$
$= (a^{2} + b^{2}) (c^{2} + d^{2}) = (\left|\alpha \right| \left|\beta \right|)^{2}$
$\therefore$ $\left|\alpha \beta \right| = \left|\alpha \right| \left|\beta \right|$
Let $\alpha = a + ib$
$\left|\overline{\alpha} \right|^{2} = \left|a - ib \right|^{2} = a^{2} + (-b)^{2} = a^{2} + b^{2} = \left|\alpha \right|^{2}$
$\therefore$ $\left|\overline{\alpha} \right| = \left|\alpha \right|$
Let $\alpha = a + ib$
$\alpha \cdot \overline{\alpha} = (a + ib)(a - ib) = (a^{2} + b^{2}) + i(-ab + ab) = a^{2} + b^{2} = \left|\alpha \right|$
Let $\alpha = a + ib$, $\beta = c + id$
By Thm 3 - 1,
$\left|\alpha \right| = \left|\beta + (\alpha - \beta) \right| \leq \left|\beta \right| + \left|\alpha - \beta \right|$
$\therefore$ $\left|\alpha \right| - \left|\beta \right| \leq \left|\alpha - \beta \right|$
Similarly, we can get $\left|\beta \right| - \left|\alpha \right| \leq \left|\alpha - \beta \right|$
$\therefore$ $\left|\left|\alpha \right| - \left|\beta \right| \right| \leq \left|\alpha - \beta \right|$
Def 4
When $\alpha \in \mathbb{C}$ is shown in the complex plane, the angle between the positive real axis and the half-straight line passing through $\alpha$ from the origin is called the argument of $\alpha$. If we write this as $\theta$, then
$\alpha = r(\cos \theta + i \sin \theta)$ $(r = \left|\alpha \right|)$
Thm 4
If $\alpha = r(\cos \theta + i \sin \theta)$ and $\beta = s(\cos \varphi+ i \sin \varphi)$, then
$\alpha \beta = rs(\cos (\theta + \varphi)+ i \sin (\theta + \varphi))$
$\alpha \beta = r(\cos \theta + i \sin \theta) \cdot s(\cos \varphi + i \sin \varphi)$
$= rs [(\cos \theta \cos \varphi - \sin \theta \sin \varphi) + i(\cos \theta \sin \varphi + \sin \theta \cos \varphi)]$
$= rs(\cos (\theta + \varphi)+ i \sin (\theta + \varphi)) $
Def 5
If $\alpha = r(\cos \theta + i \sin \theta)$, then we indicate
$\theta = \mathrm{arg}(\alpha)$
But $\mathrm{arg}$ is a multi-valued function. So we define a function $\mathrm{Arg}$ that is called the principal argument of $\alpha$.
$\mathrm{Arg}(\alpha) \in \mathrm{arg}(\alpha)$, $-\pi < \mathrm{Arg}(\alpha) \leq \pi$
# $\mathrm{Arg}(\alpha \beta) \equiv \mathrm{Arg}(\alpha) + \mathrm{Arg}(\beta)$ $\pmod{2\pi}$