[spec] 4f - conjugate root theorem

diff --git a/spec/complex.md b/spec/complex.md
index b1688eb9a38e28fcc77eaa4dea24700f9299d545..959c241bfc7318f873fafb35c3ff13938e8a59f8 100755 (executable)
--- a/spec/complex.md
+++ b/spec/complex.md
@@ -108,6 +108,24 @@ $z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$
 
 *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
 
 $$\begin{equation}\begin{split}z & =r \operatorname{cis} \theta \\ & = r(\operatorname{cos}\theta+i \operatorname{sin}\theta) \\ & = a + bi \end{split}\end{equation}$$
@@ -134,3 +152,10 @@ ${z_1 \over z_2} = {r_1 \over r_2} \operatorname{cis}(\theta_1-\theta_2)$ (divid
 ## 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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