$z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$
+*Must include $\pm$ in solutions*
+
## 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