From: Andrew Lorimer <andrew@lorimer.id.au>
Date: Tue, 12 Feb 2019 21:45:12 +0000 (+1100)
Subject: fix formatting for complex/imaginary notes
X-Git-Tag: yr12~254^2~3
X-Git-Url: https://git.lorimer.id.au/notes.git/diff_plain/c526004533c1b7a3bb623e4caf1ec609b623e1e4?hp=43f26d9be37cc8d6d917224205181840d423a97c

fix formatting for complex/imaginary notes
---

diff --git a/spec/complex.md b/spec/complex.md
index 5809170..374b793 100755
--- a/spec/complex.md
+++ b/spec/complex.md
@@ -8,9 +8,8 @@ $\therefore i = \sqrt {-1}$
 
 ### Simplifying negative surds
 
-$\sqrt{-2} = \sqrt{-1 \times 2}$
-
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$= \sqrt{2}i$
+$\sqrt{-2} = \sqrt{-1 \times 2}$  
+$= \sqrt{2}i$
 
 
 ## Complex numbers
@@ -23,21 +22,16 @@ General form: $z=a+bi$
 
 ### Addition
 
-If $z_1 = a+bi$ and $z_2=c+di$, then
-
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; $z_1+z_2 = (a+c)+(b+d)i$
+If $z_1 = a+bi$ and $z_2=c+di$, then  
+$z_1+z_2 = (a+c)+(b+d)i$
 
 ### Subtraction
 
-If $z_1=a+bi$ and $z_2=c+di$, then
-
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$z_1−z_2=(a−c)+(b−d)i$
+If $z_1=a+bi$ and $z_2=c+di$, then $z_1−z_2=(a−c)+(b−d)i$
 
 ### Multiplication by a real constant
 
-If $z=a+bi$ and $k \in \mathbb{R}$, then
-
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$kz=ka+kbi$
+If $z=a+bi$ and $k \in \mathbb{R}$, then $kz=ka+kbi$
 
 ### Powers of $i$
 $i^0=1$
@@ -48,17 +42,18 @@ $i^4=1$
 $\dots$
 
 Therefore..
+
 - $i^{4n} = 1$
 - $i^{4n+1} = i$
 - $i^{4n+2} = -1$
 - $i^{4n+3} = -i$
 
-Divide by 4 and take remainder
+Divide by 4 and take remainder.
 
 ### Multiplying complex expressions
 
-If $z_1 = a+bi$ and $z_2=c+di$, then
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$z_1 \times z_2 = (ac-bd)+(ad+bc)i$
+If $z_1 = a+bi$ and $z_2=c+di$, then  
+$z_1 \times z_2 = (ac-bd)+(ad+bc)i$
 
 ### Conjugates
 
@@ -98,8 +93,7 @@ ${z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}$
 
 To solve $z^2+a^2=0$ (sum of two squares):
 
-$z^2+a^2=z^2-(ai)^2$
-&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$=(z+ai)(z-ai)$
+$z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$
 
 ## Polar form