---
-geometry: margin=2cm
+geometry: margin=1.9cm
<!-- columns: 2 -->
graphics: yes
tables: yes
### Conjugates
-If $z=a+bi$, conjugate is
-
-$$\overline{z} = a-bi$$
+$$\overline{z} = a \mp bi$$
##### Properties
- horizontal $= \operatorname{Re}(z)$; vertical $= \operatorname{Im}(z)$
- Multiplication by $i$ results in an anticlockwise rotation of $\pi \over 2$
-## Solving complex polynomials
+\vfil \break
-**Include $\pm$ for all solutions, including imaginary**
+## Complex polynomials
-## Solving complex quadratics
+**Include $\pm$ for all solutions, including imaginary**
-To solve $z^2+a^2=0$ (sum of two squares):
+### 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
-Dividing $P(z)$ by $D(z)$ gives quotient $Q(z)$ and remainder $R(z)$ such that:
+$P(z) \div D(z)$ gives quotient $Q(z)$ and remainder $R(z)$:
$$P(z) = D(z)Q(z) + R(z)$$
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 $P(z)$
+#### 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 $P(z)=0$, with $a, b \in \mathbb{R}$, then the conjugate $\overline{z}=a-bi$ is also a solution.
+If $a+bi$ is a solution to $P(z)=0$, then the conjugate $\overline{z}=a-bi$ is also a solution.
## Polar form
- $\theta=\operatorname{arg}(z)$ (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
+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
### Conjugate in polar form
### Multiplication and division in polar form
-$z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)$ (multiply moduli, add angles)
+$$z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)$$
-${z_1 \over z_2} = {r_1 \over r_2} \operatorname{cis}(\theta_1-\theta_2)$ (divide moduli, subtract angles)
+$${z_1 \over z_2} = {r_1 \over r_2} \operatorname{cis}(\theta_1-\theta_2)$$
## de Moivres' Theorem
## 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:** $\operatorname{Arg}(z) < \theta$
+### 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
+
+$|z-z_1|^2 = c^2 |z_2+2|^2$ or $|z-(a + bi)| = c$
+
+### Locus
+
+$\operatorname{Arg}(z) < \theta$