+---
+geometry: margin=1.9cm
+<!-- columns: 2 -->
+graphics: yes
+tables: yes
+author: Andrew Lorimer
+classoption: twocolumn
+header-includes:
+- \usepackage{harpoon}
+- \usepackage{amsmath}
+- \pagenumbering{gobble}
+
+---
+
+
# Complex & Imaginary Numbers
## Imaginary numbers
-$i^2 = -1 \quad \therefore i = \sqrt {-1}$
+$$i^2 = -1 \quad \therefore i = \sqrt {-1}$$
### Simplifying negative surds
-$\sqrt{-2} = \sqrt{-1 \times 2}$
-$= \sqrt{2}i$
+\begin{equation}\begin{split}\sqrt{-2} & = \sqrt{-1 \times 2} \\ & = \sqrt{2}i\end{split}\end{equation}
## Complex numbers
-$\mathbb{C} = \{a+bi : a, b \in \mathbb{R} \}$
+$$\mathbb{C} = \{a+bi : a, b \in \mathbb{R} \}$$
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
-$z_1+z_2 = (a+c)+(b+d)i$
+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 $z_1−z_2=(a−c)+(b−d)i$
+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 $kz=ka+kbi$
+If $z=a+bi$ and $k \in \mathbb{R}$, then
-### Powers of $i$
-$i^0=1$
-$i^1=i$
-$i^2=-1$
-$i^3=-i$
-$i^4=1$
-$\dots$
+$$kz=ka+kbi$$
-Therefore..
+### Powers of $i$
- $i^{4n} = 1$
- $i^{4n+1} = i$
- $i^{4n+2} = -1$
- $i^{4n+3} = -i$
-Divide by 4 and take remainder.
+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 $z_1 = a+bi$ and $z_2=c+di$, then
-If $z=a+bi$, conjugate of $z$ is $\overline{z} = a-bi$ (flipped operator)
+$$z_1 \times z_2 = (ac-bd)+(ad+bc)i$$
-Also, $z \overline{z} = (a+bi)(a-bi) = a^2+b^2 = |z|^2$
+### Conjugates
-- Multiplication and addition are associative
+$$\overline{z} = a \mp bi$$
-#### Properties
+##### Properties
- $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$
- $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$
- $\overline{kz} = k \overline{z}, \text{ for } k \in \mathbb{R}$
-- $z \overline{z} = |z|^2$
+- $z \overline{z} = = (a+bi)(a-bi) = a^2+b^2 = |z|^2$
- $z + \overline{z} = 2 \operatorname{Re}(z)$
-
### Modulus
Distance from origin.
-$|{z}|=\sqrt{a^2+b^2}$
-$\therefore z \overline{z} = |z|^2$
+$$|{z}|=\sqrt{a^2+b^2} \quad \therefore z \overline{z} = |z|^2$$
-#### Properties
+###### Properties
- $|z_1 z_2| = |z_1| |z_2|$
- $|{z_1 \over z_2}| = {|z_1| \over |z_2|}$
### Multiplicative inverse
-$z^{-1} = {1 \over z} = {{a-bi} \over {a^2+B^2}} = {\overline{z} \over {|z|^2}}$
+\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
-${{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}}$
-
-(using multiplicative inverse)
+$${{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}} \quad \text{(multiplicative inverse)}$$
In practice, rationalise denominator:
-${z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}$
+
+$${z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}$$
## Argand planes
- Geometric representation of $\mathbb{C}$
-- Horizontal $= \operatorname{Re}(z)$; vertical $= \operatorname{Im}(z)$
+- horizontal $= \operatorname{Re}(z)$; vertical $= \operatorname{Im}(z)$
- Multiplication by $i$ results in an anticlockwise rotation of $\pi \over 2$
-## Solving complex quadratics
+\vfil \break
+
+## Complex polynomials
+
+**Include $\pm$ for all solutions, including imaginary**
+
+### Sum of two squares (quadratics)
+
+$$z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$$
+
+Complete the square to get to this point.
+
+#### Dividing complex polynomials
+
+$P(z) \div D(z)$ gives quotient $Q(z)$ and remainder $R(z)$:
+
+$$P(z) = D(z)Q(z) + R(z)$$
+
+#### Remainder theorem
-To solve $z^2+a^2=0$ (sum of two squares):
+Let $\alpha \in \mathbb{C}$. Remainder of $P(z) \div (z - \alpha)$ is $P(\alpha)$
-$z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$
+#### Factor theorem
+
+If $a+bi$ is a solution to $P(z)=0$, then:
+
+- $P(a+bi)=0$
+- $z-(a+bi)$ is a factor of $P(z)$
+
+#### Sum of two cubes
+
+$$a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$$
+
+## Conjugate root theorem
+
+If $a+bi$ is a solution to $P(z)=0$, then the conjugate $\overline{z}=a-bi$ is also a solution.
## Polar form
-General form:
-$z=r \operatorname{cis} \theta$
-$= r\operatorname{cos}\theta+r\operatorname{sin}\theta i$
+\begin{equation}\begin{split}z & =r \operatorname{cis} \theta \\ & = r(\operatorname{cos}\theta+i \operatorname{sin}\theta) \\ & = a + bi \end{split}\end{equation}
-where
+- $r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}$
+- $\theta=\operatorname{arg}(z)$ (on CAS: `arg(a+bi)`)
+- **principal argument** is $\operatorname{Arg}(z) \in (-\pi, \pi]$ (note capital $\operatorname{Arg}$)
-- $z=a+bi$
-- $r$ is the distance from origin, given by Pythagoras ($r=\sqrt{x^2+y^2}$)
-- $\theta$ is the argument of $z$, CCW from origin
+Each complex number has multiple polar representations:
+$z=r \operatorname{cis} \theta = r \operatorname{cis} (\theta+2 n\pi$) with $n \in \mathbb{Z}$ revolutions
-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)
+$$z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)$$
+
+$${z_1 \over z_2} = {r_1 \over r_2} \operatorname{cis}(\theta_1-\theta_2)$$
+
+## 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)
+- $\operatorname{Arg}(z) = \theta$
+- $|z+a|=|z+bi|$ where $m={a \over b}$
+- $|z+a|=|z+b| \longrightarrow 2(a-b)x=b^2-a^2$
+
+### Circle
-${z_1 \over z_2} = {r_1 \over r_2} \operatorname{cis}(\theta_1-\theta_2)$ (divide moduli, subtract angles)
+$|z-z_1|^2 = c^2 |z_2+2|^2$ or $|z-(a + bi)| = c$
-## de Moivres' Theorum
+### Locus
-$(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)$
+$\operatorname{Arg}(z) < \theta$