[english] add quotes for WoT

diff --git a/spec/complex.md b/spec/complex.md
index b70737f4fccbcf3ae252a7b687272536516f1339..e72225e3b7bbf2d5a170930723ed6a9fa50e03c3 100755 (executable)
--- a/spec/complex.md
+++ b/spec/complex.md
@@ -96,12 +96,11 @@ $$|{z}|=\sqrt{a^2+b^2} \quad  \therefore z \overline{z} = |z|^2$$
 
 ### Dividing complex numbers
 
-$${{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}} \quad \text{multiplicative inverse}$$
-
-(using multiplicative inverse)
+$${{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}} \quad \text{(multiplicative inverse)}$$
 
 In practice, rationalise denominator:
-${z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}$
+
+$${z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}$$
 
 ## Argand planes
 
@@ -129,6 +128,12 @@ $$P(z) = D(z)Q(z) + R(z)$$
 
 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 $P(z)$
+
 ## 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.