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# Complex & Imaginary Numbers

## Imaginary numbers

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

### Simplifying negative surds

\begin{equation}\begin{split}\sqrt{-2} & = \sqrt{-1 \times 2} \\ & = \sqrt{2}i\end{split}\end{equation}


## 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 $i$

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

For $i^n$, find remainder $r$ when $n \div 4$. 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 is

$$\overline{z} = a-bi$$

##### Properties

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

### Modulus

Distance from origin.

$$|{z}|=\sqrt{a^2+b^2} \quad  \therefore z \overline{z} = |z|^2$$

###### 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

\begin{equation}\begin{split}z^{-1} & = {1 \over z} \\ & = {{a-bi} \over {a^2+B^2}} \\ & = {\overline{z} \over {|z|^2}}\end{split}\end{equation}

### Dividing complex numbers

$${{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}} \quad \text{(multiplicative inverse)}$$

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 polynomials

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

## Solving complex quadratics

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

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

#### 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 $P(\alpha)$

#### Factor theorem
If $a+bi$ is a solution to $P(z)=0$, then:

- $P(a+bi)=0$
- $z-(a+bi)$ is a factor of $P(z)$

## Conjugate root theorem

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

- $r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}$
- $\theta=\operatorname{arg}(z)$ (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

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

## Roots of complex numbers

$n$th roots of $z = r \operatorname{cis} \theta$ are

$$z={r^{1 \over n}} \operatorname{cis}({{\theta + 2 k \pi} \over n})$$

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

The solutions of $z^n=a \text{ where } a \in \mathbb{C}$ lie on circle

$$x^2 + y^2 = (|a|^{1 \over n})^2$$

## Sketching complex graphs

- **Straight line:** $\operatorname{Re}(z) = c$ or $\operatorname{Im}(z) = c$ (perpendicular bisector) or $\operatorname{Arg}(z) = \theta$
- **Circle:** $|z-z_1|^2 = c^2 |z_2+2|^2$ or $|z-(a + bi)| = c$
- **Locus:** $\operatorname{Arg}(z) < \theta$