fixing latex

1 files changed, 31 insertions, 62 deletions

diff --git a/ClassJune26.page b/ClassJune26.page
index 36c2e17..eea5dff 100644
--- a/ClassJune26.page
+++ b/ClassJune26.page
@@ -30,7 +30,7 @@ and rotate counterclockwise by the angle it forms with the positive
 reals, with the horizontal, which we call its \textbf{argument}. So
 multiplying by this doubles lengths, and rotates $45^{\circ}$ counterclockwise;
 
-$(\nearrow)\cdot(\uparrow)=(\nwarrow)\$
+$(\nearrow)\cdot(\uparrow)=(\nwarrow)$
 
 Multiplying by this halves lengths and rotates $300^{\circ}$ counterclockwise
 (or $60^{\circ}$ clockwise), taking this to that, and this to that.
@@ -41,98 +41,67 @@ notable, multiplying by it just rotates things $90^{\circ}$. In particular,
 $(\uparrow)\cdot(\uparrow)=(\leftarrow)$, which is $-1$. We usually
 write $i$ for $\uparrow$, and we've just shown that $i^{2}=-1$.
 
+## Arrow Arithmetic
+
 Complex numbers are vectors in the plane, with addition given by vector
 addition, and multiplication given by dilation and rotation. Real
 numbers form the horizontal axis, and imaginary numbers form the vertical
-axis. Why does it make sense to call these arrows {}``numbers''?
+axis. Why does it make sense to call these arrows "numbers"?
 Because they satisfy all the basic rules of arithmetic:
 
-\begin{itemize}
-\item Addition is commutative.
-
-\begin{itemize}
-\item the two orders form a paralellogram, with the sum on the diagonal.
-$z+w=w+z$
-\end{itemize}
-\item Addition is associative.
+- Addition is commutative.
+    - the two orders form a paralellogram, with the sum on the diagonal, $z+w=w+z$
+- Addition is associative.
+    - $v+(z+w)=(v+z)+w$ since both compute the straight-line path from $0$ to the end of the $\overset{w}{\rightarrow}\overset{z}{\rightarrow}\overset{v}{\rightarrow}$ trail.
+- Multiplication is commutative.
+    - Multiplying $(r_{1},\theta_{1})$ by $(r_{2},\theta_{2})$ rescales by $r_{2}$ and rotates by $\theta_{2}$, giving $(r_{1}r_{2},\theta_{1}+\theta_{2})$, which is manifestly symmetric. Put another way, multiplication of complex numbers multiplies magnitudes and adds angles/arguments.
+- Multiplication is associative.
+    - use formula in polar coordinates
+- Multiplication distributes over addition.
+    - Take any two complex numbers $v,w$ and add them, making a parallelogram. To find $z(v+w)$, rotate and scale our sum. But that's the same as rotating and scaling each of $v$ and $w$ individually, to $zv$ and $zw$, then adding them together: $zv+zw=z(v+w)$. The distributive law for complex numbers is the fact that a rotated, rescaled paralellogram is still a parallelogram, or, more basically, that dilation and rotation of the plane preserve angles.
 
-\begin{itemize}
-\item $v+(z+w)=(v+z)+w$ since both compute the straight-line path from
-$0$ to the end of the $\overset{w}{\rightarrow}\overset{z}{\rightarrow}\overset{v}{\rightarrow}$
-trail.
-\end{itemize}
-\item Multiplication is commutative.
 
-\begin{itemize}
-\item Multiplying $(r_{1},\theta_{1})$ by $(r_{2},\theta_{2})$ rescales
-by $r_{2}$ and rotates by $\theta_{2}$, giving $(r_{1}r_{2},\theta_{1}+\theta_{2})$,
-which is manifestly symmetric. Put another way, multiplication of
-complex numbers multiplies magnitudes and adds angles/arguments.
-\end{itemize}
-\item Multiplication is associative.
-
-\begin{itemize}
-\item exercize
-\end{itemize}
-\item Multiplication distributes over addition.
-
-\begin{itemize}
-\item Take any two complex numbers $v,w$ and add them, making a parallelogram.
-To find $z(v+w)$, rotate and scale our sum. But that's the same as
-rotating and scaling each of $v$ and $w$ individually, to $zv$
-and $zw$, then adding them together: $zv+zw=z(v+w)$. The distributive
-law for complex numbers is the fact that a rotated, rescaled paralellogram
-is still a parallelogram, or, more basically, that dilation and rotation
-of the plane preserve angles.
-\end{itemize}
-\end{itemize}
-
-\subsection*{Coordinates}
+## Coordinates
 
 We can't be constantly drawing arrows in the plane; to do anything
 quantitative, we need to introduce coordinates. Rectangular coordinates
 are easy, each vector has an $x$ and $y$-coordinate, given by its
 projections onto the real and imaginary axis. If we use column vector
-notation, we have \begin{eqnarray*}
-z & = & \left(\begin{array}{c}
+notation, we have
+$z = \left(\begin{array}{c}
 a\\
 b\end{array}\right)\\
- & = & a\left(\begin{array}{c}
+ = a\left(\begin{array}{c}
 1\\
 0\end{array}\right)+b\left(\begin{array}{c}
 0\\
 1\end{array}\right)\\
- & = & a\cdot1+b\cdot i\\
- & = & a+bi\end{eqnarray*}
+ = a\cdot1+b\cdot i\\
+ = a+bi\end{eqnarray*}
 Put another way, we are using $1$ and $i$ as basis vectors. For
-example, \[
+example, $
 \left(\begin{array}{c}
 3\\
-2\end{array}\right)=3(\rightarrow)+2(\uparrow)=3+2i\]
-and\[
+2\end{array}\right)=3(\rightarrow)+2(\uparrow)=3+2i$
+and $
 \overset{2}{\nwarrow}=\left(\begin{array}{c}
 -\sqrt{2}\\
-\sqrt{2}\end{array}\right)=-\sqrt{2}(\rightarrow)+\sqrt{2}(\uparrow)=-\sqrt{2}+\sqrt{2}i\]
+\sqrt{2}\end{array}\right)=-\sqrt{2}(\rightarrow)+\sqrt{2}(\uparrow)=-\sqrt{2}+\sqrt{2}i$
 
 
-Polar coordinates represent each complex number by its magnitude and
-argument, its radius and angle (in radians). So \[
-i=\uparrow=(1,\pi/2)\]
-and\[
-\overset{2}{\nwarrow}=(2,3\pi/4)\]
-
+Polar coordinates represent each complex number by its magnitude and argument, its radius and angle (in radians). So 
+$i=\uparrow=(1,\pi/2)$ and $\overset{2}{\nwarrow}=(2,3\pi/4)$.
 
 The familiar angle addition formula from trigonometry now follows
-from the distributive law applied to unit vectors.\[
-\cos(\theta+\phi)+i\sin(\theta+\phi)=(\cos\theta+i\sin\theta)(\cos\phi+i\sin\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi+i(\cos\theta\sin\phi+\sin\theta\cos\phi)\]
+from the distributive law applied to unit vectors.
+$\cos(\theta+\phi)+i\sin(\theta+\phi)=(\cos\theta+i\sin\theta)(\cos\phi+i\sin\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi+i(\cos\theta\sin\phi+\sin\theta\cos\phi)$
 
 
 In the workshop session, you'll use a similar trick to compute triple
 and quadruple angle formulas, for $\sin$ and $\cos$, deriving De
 Moivre's theorem.
 
-
-\subsection*{Functions}
+## Functions
 
 The most basic transformations of the complex plane are given by multipling
 by some fixed complex number $\rho=r(\cos\theta+i\sin\theta)=a+bi$,
@@ -159,10 +128,10 @@ b\end{array}\right)$ itself, and the second column is the image of $i=\left(\beg
 1\end{array}\right)$, which is $\rho$ rotated by $90^{\circ}$, which has coordinates
 $\left(\begin{array}{c}
 -b\\
-a\end{array}\right)$. So complex number $a+bi$ is identified with the matrix \[
+a\end{array}\right)$. So complex number $a+bi$ is identified with the matrix $
 \left(\begin{array}{cc}
 a & -b\\
-b & a\end{array}\right).\]
+b & a\end{array}\right).$
 As a sanity check, note that $i$ corresponds to $\left(\begin{array}{cc}
 0 & -1\\
 1 & 0\end{array}\right)$, which squares to $\left(\begin{array}{cc}