- $i^{4n+2} = -1$
- $i^{4n+3} = -i$
-Divide by 4 and take remainder.
+For $i^n$, divide $n$ by 4 and let remainder $= r$. Then $i^n = i^r$.
### Multiplying complex expressions
- $z \overline{z} = |z|^2$
- $z + \overline{z} = 2 \operatorname{Re}(z)$
-
### Modulus
Distance from origin.
$z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$
+*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
-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
-- $z=a+bi$
-- $r$ is the distance from origin, given by Pythagoras ($r=\sqrt{x^2+y^2}$)
-- $\theta$ is the argument of $z$, CCW from origin
+- $r=|z|$, 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' Theorum
+## 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 ${2 \pi} \over n$
+
+## Sketching complex graphs
-$(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)$
+- **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:** $\operatorname{Arg}(z) \lt \theta$