$z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$
+*Must include $\pm$ in solutions*
+
+## Solving complex polynomials
+
+#### 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
+
+Let $P(z)$ be a polynomial with real coefficients. 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 $\operatorname{Arg}(z) \in (-\pi, \pi]$
+- $\theta=\operatorname{arg}(z)$ (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
## 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 ${2 \pi} \over n$
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