# Complex & Imaginary Numbers

## Imaginary numbers

$i^2 = -1 \quad \therefore i = \sqrt {-1}$

### Simplifying negative surds

$\sqrt{-2} = \sqrt{-1 \times 2}$  
$= \sqrt{2}i$


## Complex numbers

$\mathbb{C} = \{a+bi : a, b \in \mathbb{R} \}$

General form: $z=a+bi$  
$\operatorname{Re}(z) = a, \quad \operatorname{Im}(z) = b$

### Addition

If $z_1 = a+bi$ and $z_2=c+di$, then  
$z_1+z_2 = (a+c)+(b+d)i$

### Subtraction

If $z_1=a+bi$ and $z_2=c+di$, then $z_1−z_2=(a−c)+(b−d)i$

### Multiplication by a real constant

If $z=a+bi$ and $k \in \mathbb{R}$, then $kz=ka+kbi$

### Powers of $i$
$i^0=1$
$i^1=i$
$i^2=-1$
$i^3=-i$
$i^4=1$
$\dots$

Therefore..

- $i^{4n} = 1$
- $i^{4n+1} = i$
- $i^{4n+2} = -1$
- $i^{4n+3} = -i$

For $i^n$, divide $n$ by 4 and let remainder $= r$. Then $i^n = i^r$.

### Multiplying complex expressions

If $z_1 = a+bi$ and $z_2=c+di$, then  
$z_1 \times z_2 = (ac-bd)+(ad+bc)i$

### Conjugates

If $z=a+bi$, conjugate of $z$ is $\overline{z} = a-bi$ (flipped operator)

Also, $z \overline{z} = (a+bi)(a-bi) = a^2+b^2 = |z|^2$

- Multiplication and addition are associative

#### Properties

- $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$
- $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$
- $\overline{kz} = k \overline{z}, \text{ for } k \in \mathbb{R}$
- $z \overline{z} = |z|^2$
- $z + \overline{z} = 2 \operatorname{Re}(z)$

### Modulus

Distance from origin.
$|{z}|=\sqrt{a^2+b^2}$

$\therefore z \overline{z} = |z|^2$

#### Properties

- $|z_1 z_2| = |z_1| |z_2|$
- $|{z_1 \over z_2}| = {|z_1| \over |z_2|}$
- $|z_1 + z_2| \le |z_1 + |z_2|$

### Multiplicative inverse

$z^{-1} = {1 \over z} = {{a-bi} \over {a^2+B^2}} = {\overline{z} \over {|z|^2}}$

### Dividing complex numbers

${{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}}$

(using multiplicative inverse)

In practice, rationalise denominator:
${z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}$

## Argand planes

- Geometric representation of $\mathbb{C}$
- Horizontal $= \operatorname{Re}(z)$; vertical $= \operatorname{Im}(z)$
- Multiplication by $i$ results in an anticlockwise rotation of $\pi \over 2$

## Solving complex quadratics

To solve $z^2+a^2=0$ (sum of two squares):

$z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$

*Must include $\pm$ in solutions*

## Solving complex polynomials

Include $\pm$ for all solutions, including imaginary.

#### Dividing complex polynomials

Dividing $P(z)$ by $D(z)$ gives quotient $Q(z)$ and remainder $R(z)$ such that:

$$P(z) = D(z)Q(z) + R(z)$$

#### Remainder theorem

Let $\alpha \in \mathbb{C}$. Remainder of $P(z) \div (z - \alpha)$ is $P(\alpha)$

## Conjugate root theorem

If $a+bi$ is a solution to $P(z)=0$, with $a, b \in \mathbb{R}$, the the conjugate $a-bi$ is also a solution.

## Polar form

$$\begin{equation}\begin{split}z & =r \operatorname{cis} \theta \\ & = r(\operatorname{cos}\theta+i \operatorname{sin}\theta) \\ & = a + bi \end{split}\end{equation}$$

- $r=|z|$, given by Pythagoras ($r=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}$)
- $\theta=\operatorname{arg}(z)$ (on CAS: `arg(a+bi)`)
- **principal argument** is $\operatorname{Arg}(z) \in (-\pi, \pi]$ (note capital $\operatorname{Arg}$)

Note each complex number has multiple polar representations:
$z=r \operatorname{cis} \theta = r \operatorname{cis} (\theta+2 n\pi$) where $n$ is integer number of revolutions

### Conjugate in polar form

$$(r \operatorname{cis} \theta)^{-1} = r\operatorname{cis} (- \theta)$$

Reflection of $z$ across horizontal axis.

### Multiplication and division in polar form

$z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)$ (multiply moduli, add angles)

${z_1 \over z_2} = {r_1 \over r_2} \operatorname{cis}(\theta_1-\theta_2)$ (divide moduli, subtract angles)

## de Moivres' Theorem

$(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)$ where $n \in \mathbb{Z}$

## Roots of complex numbers

$n$th roots of $r \operatorname{cis} \theta$ are:  
$z={r^{1 \over n}} \cdot (\cos ({{\theta + 2k \pi} \over n}) + i \sin ({{\theta + 2 k \pi} \over n}))$

Same modulus for all solutions. Arguments are separated by ${2 \pi} \over n$