From ead8a6e1392f6debe29172cae7959c7d856389bd Mon Sep 17 00:00:00 2001
From: joshuab <>
Date: Tue, 29 Jun 2010 03:18:54 +0000
Subject: Beginning of lecture 1

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-Hello world. $\frac{1}{2}$
\ No newline at end of file
+We'll begin our discussion of complex numbers with a simple picture
+of the real number line. Zero in the centre, $1,2,3,\dots$ to the
+right, $-1,-2,-3,\dots$ to the left. Think of each real number not
+as a point on the line, but as the corresponding vector, as the arrow
+emanating from $0$ and ending at, say, $2$. Then real number addition
+is vector addition; to add two numbers, say, $1$ and $2$, place
+the tail of one at the tip of the other. The sum is the vector ending
+at the final tip, $1+2=3$. This works for negative numbers too; $2+(-3)=-1$,
+$2+(-2)=0$. Multiplication is rescaling; to multiply a vector/number
+by $3$, you scale up its length by a factor of $3$, and leave the
+direction the same; $2\rightarrow6$, and $-2\rightarrow-6$. To multiply
+by a negative number $v$; you still rescale by the length of $v$,
+but you also reverse direction, rotating around $180^{\circ}$, just
+like the negative vectors are all $180^{\circ}$ away from their positive
+counterparts. So $-3\cdot-2=6$.
+
+We can extend the law of addition to all vectors in the plane containing
+the real ones. Given two vectors of whatever lengths in whatever directions,
+you add them by putting tail to tip and completing the triangle. This
+plus this equals that, this plus this equals that.
+
+To multiply a vector by a real number (a real vector), we use the
+same rule as before; if the real number is positive, pointing horizontally
+to the right, then you just rescale its target by its length, this
+times $2$ is that. If the real number is negative, pointing horizontally
+to the left, it rescales its target by its length, and then rotates
+it by $180^{\circ}$. To multiply by an arbitrary vector in the plane,
+say this one, you rescale by its length, which we call its \textbf{magnitude},
+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;
+\[
+(\nearrow)\cdot(\uparrow)=(\nwarrow)\]
+
+
+Multiplying by this halves lengths and rotates $300^{\circ}$ counterclockwise
+(or $60^{\circ}$ clockwise), taking this to that, and this to that.
+Mulitplying by $1$, the unit vector to the right, fixes the length,
+and rotates by $0^{\circ}$, ie, leaves everything the same. So $1$
+is the multiplicative identity. The unit vector pointing north is
+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 $i^{2}=-1$.
+
+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''?
+Because they satisfy all the basic rules of arithmetic:
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