1.1 Complex Number Algebra

Arithmetic

Addition

(α + i β) + (γ + i δ) = (a + γ) + i (β + δ)

Conjugate

\overset{―}{α + i β} = α - i β

Properties

Re (a) = \frac{a + \bar{a}}{2}, Im (a) = \frac{a - \bar{a}}{2 i}

\begin{aligned} \overset{―}{a + b} & = \bar{a} + \bar{b} \\ \overset{―}{a b} & = \bar{a} \cdot \bar{b} \end{aligned}

Let $R (a, b, c, \dots)$ stand for any rational operation applied to the complex numbers $a, b, c, \dots$ .
Then

\overset{―}{R (a, b, c, \dots)} = R (\bar{a}, \bar{b}, \bar{c}, \dots) .

As an application, consider the equation

c_{0} z^{n} + c_{1} z^{n - 1} + \dots + c_{n - 1} z + c_{n} = 0.

If $ζ$ is a root of this equation, then $\bar{ζ}$ is a root of the equation

{\bar{c}}_{0} z^{n} + {\bar{c}}_{1} z^{n - 1} + \dots + {\bar{c}}_{n - 1} z + {\bar{c}}_{n} = 0.

In particular, if the coefficients are real, $ζ$ and $\bar{ζ}$ are roots of the same equation, and we have the familiar theorem that the nonreal roots of an equation with real coefficients occur in pairs of conjugate roots.

Absolute (Norm)

$a \in C, a = α + i β$

| a | = \sqrt{α^{2} + β^{2}}

Properties

\begin{aligned} (1) & a \bar{a} & = | a |^{2} \\ | a | & = | \bar{a} | \\ ⟹ | a b |^{2} = a b \cdot \overset{―}{a d} = | a |^{2} | b |^{2} \\ (2) & ⟹ | a b | = | a | \cdot | b | \\ \frac{| a |^{2}}{| a^{2} |} & = 1 \\ ⟹ \frac{a \bar{a}}{| a |^{2}} = 1 \\ (3) & ⟹ (a)^{- 1} = \frac{\bar{a}}{| a |^{2}} \\ \frac{α + i β}{γ + i δ} = \frac{(α γ + β δ) + i (β γ - α δ)}{γ^{2} + δ^{2}} \\ (4) & | \frac{a}{b} | & = \frac{| a |}{| b |} \\ (5) & | a + b |^{2} & = | a |^{2} + | b |^{2} + 2 Re (a \bar{b}) \\ | a - b |^{2} & = | a |^{2} + | b |^{2} - 2 Re (a \bar{b}) \\ | a + b |^{2} + | a - b |^{2} & = 2 (| a^{2} | + | b |^{2}) \end{aligned}

Multiplication

(α + i β) (γ + i δ) = (α γ - β δ) + i (α δ + β γ)

Since $i^{2} = - 1$

Square Roots

Want

(x + i y)^{2} = α + i β

\begin{aligned} ⟹ (x^{2} - y^{2}) + i (2 x y) = α + i β \\ ⟹ {\begin{cases} x^{2} - y^{2} & = β ⋆ \\ 2 x y & = α \end{cases} \\ use a trick from algebra: \\ ⟹ (x^{2} + y^{2})^{2} = (x^{2} - y^{2})^{2} + (2 x y)^{2} = α^{2} + β^{2} \\ ⟹ x^{2} + y^{2} = \sqrt{α^{2} + β^{2}} \\ now cancelling from ⋆ we get \\ ⟹ {\begin{cases} x^{2} = \frac{1}{2} (α + \sqrt{α^{2} + β^{2}}) \\ y^{2} = \frac{1}{2} (- α + \sqrt{α^{2} + β^{2}}) \end{cases} \\ ⟹ \sqrt{a + i β} = \pm (\sqrt{\frac{α + \sqrt{α^{2} + β^{2}}}{2}}, i \frac{β}{| β |} \sqrt{\frac{- α + \sqrt{α^{2} + β^{2}}}{2}}) \end{aligned}

For $β = 0$ we have $\pm \sqrt{α}$ if $α \geq 0,$ otherwise $\pm i \sqrt{- a}$

Inequalities

\begin{array}{r} (6) & - | a | \leq Re (a) \leq | a | \\ - | a | \leq Im (a) \leq | a | \end{array}

Triangle Inequalities

Using (5) we get

\begin{matrix} (7) & | a + b |^{2} \leq (| a | + | b |)^{2} ⟹ | a + b | \leq | a | + | b | \end{matrix}

Results:

\begin{aligned} | a | = | (a - b) + b | \leq | a - b | + | b | \\ (8) & ⟹ & | a | - | b | \leq | a - b | \end{aligned}

\begin{aligned} | b | \leq | a - b | + | a | & ⟹ - | a - b | \leq | a | - | b | \\ ⟹ - | a - b | \leq | a | - | b | \leq | a - b | \\ (9) & ⟹ | | a | - | b | | \leq | a - b | \end{aligned}